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					                                   Chapter 3
Problem Summary

Prob. #                  Concepts Covered                      Level of     Notes
                                                               Difficulty
  3.1     Maximization production model                            2
  3.2     Maximization model with "", "=", ""                    5
          constraints
  3.3     Maximization production model, percentage                4
          constraints, interpretation of shadow prices and
          range of feasibility
  3.4     Maximization financial mix model, percentage             5
          constraint, determining current rate of return and
          future rate of return, shadow prices
  3.5     Maximization model, calculation of profit                4
          coefficients, sensitivity analyses
  3.6     Minimization model, determining correct RHS              4
  3.7     Maximization model, sensitivity analyses,                6
          evaluation of quantities outside the ranges of
          optimality and feasibility, many constraints
  3.8     Maximization production model, analysis of why           6
          a high profit item is not produced, sensitivity
          analyses, evaluation of purchasing additional
          resources
  3.9     Minimization model with 8 variables and 14               6
          functional constraints
 3.10     Minimization diet problem, alternate optimal             6
          solutions, interpretation of shadow prices, the
          effects of deleting a constraint
 3.11     Minimization model                                       3
 3.12     Maximization model, re-evaluation of RHS                 4
 3.13     Financial maximization model, weighted                   6
          averages, sensitivity analyses
 3.14     Multiperiod production model                             9
 3.15     Maximization agriculture model, sensitivity              7
          analyses, evaluation of an alternative based on
          shadow prices
 3.16     Supply chain model                                       7
 3.17     Supply Chain model                                       9
 3.18     Maximization finance model                               4
 3.19     Blending problem, alternate optimal solutions,           6
          shadow prices, determining the acceptability of
          the solution
 3.20     Maximization model with 9 functional                     4
          constraints
 3.21     Minimization transportation model                        5
 3.22     Classic trim loss model, definition of variables         5
 3.23     Maximization advertising model, checking the             5
          linear programming assumptions, elimination of



                                             Chapter 3- 1
           a nonbinding constraint
 3.24      Maximization retailing model, evaluating              5
           changes to objective function coefficients, the
           profitability of added resources, and the addition
           of a constraint
 3.25      Large design model, "tricky" constraints              9
 3.26      Maximization production problem, calculation of       7
           objective function coefficients, calculating
           coefficients used in the model
 3.27      Maximization advertising model, sensitivity           8
           analyses, adding constraints, changing the RHS
           values
 3.28      Integer crew scheduling model                         5
 3.29      Minimization sector assignment model                  7
 3.30      Fixed charge model                                    5
 3.31      Maximization integer model, effects of rounding,      6    Change Tolerance in Options Dialogue box
           k out of n constraints                                     to .5%; Excel may incorrectly print that part
                                                                      c is infeasible, but it gives an optimal
                                                                      solution.
 3.32      Integer Maximization model                            5    Change Tolerance in Options Dialogue box
                                                                      to .5% and check Use Automatic Scaling
 3.33      Binary model with constraints requiring binary        7
           variables
 3.34      Binary model with constraints requiring binary        6
           variables
 3.35      Minimization integer model                            4
 3.36      Maximization/Minimization integer advertising         5
           model, fixed charge
 3.37      Mixed integer model with binary variables             4
 3.38      Mixed integer financial model                         3
 3.39      Maximization integer model                            4
 3.40      Binary model, k out of n constraints                  5
 3.41      Maximization problem, calculation of net profit,      5
           evaluation of purchasing additional resources
 3.42      Maximization production model, infeasibility,         5
           sensitivity analyses, addition of constraints
 3.43      Maximization financial model                          4
 3.44      Large workforce integer model                         9
 3.45      Integer model                                         2
 3.46      Scheduling model, redundant constraints,              7
           alternate optimal solutions, shadow prices,
           adding constraints
 3.47      Fixed charge model with additional constraints        7    Change Tolerance in Options Dialogue box
           requiring binary variables                                 to .5% and check Use Automatic Scaling
 3.48      Binary model                                          2
 3.49      Data envelopment analysis model                       6
 3.50      Data Envelopment Analysis model                       5
Case 3.1   Maximization production model with input in           8
           Excel, many percentage constraints, analysis of
           output
Case 3.2   Maximization integer model, definitional              10   Solved as a linear program -- takes just a few
           variables, percentage constraints, rounding the            seconds
           linear model, "slightly" violated constraints, time        Solved as an integer program -- takes many


                                               Chapter 3- 2
           issues                                                 minutes
Case 3.3   Binary models                                     7
Case 3.4   “On the job training” model with many “linking”   9
           constraints
Case 3.5   Maximization financial mix model, linking         10
           results between two periods, percentage
           constraints, analysis of output
Case 3.6   Fixed charge model                                4




                                            Chapter 3- 3
Problem Solutions

3.1    See file Ch3.1.xls

       X1 = Number of 20-inch girls bicycles produced this week
       X2 = Number of 20-inch boys bicycles produced this week
       X3 = Number of 26-inch girls bicycles produced this week
       X4 = Number of 26-inch boys bicycles produced this week

       MAX 27X1 + 32X2 + 38X3 + 51X4
       S.T.
              X1        + X3                   200         (Min girls models)
                     X2        + X4            200         (Min boys models)
            12X1 + 12X2 + 9X3 + 9X4            4800        (Production minutes)
             6X1 + 9X2 + 12X3 + 18X4           4800        (Assembly minutes)
             2X1 + 2X2                         500         (20-inch tires)
                          2X3 + 2X4            800         (26-inch tires)
                         All X's  0




150 20-inch girls, 100 20-inch boys, 100 26-inch girls, 100 26-inch boys; profit =
$16,150


                                        Chapter 3- 4
 3.2   See file Ch3.2.xls
a.
       X1 = Number of stoves produced weekly
       X2 = Number of washers produced weekly
       X3 = Number of electric dryers produced weekly
       X4 = Number of gas dryers produced weekly
       X5 = Number of refrigerators produced weekly

       MAX 110X1 + 90X2 + 75X3 + 80X4 + 130X5
       S.T.
            5.5X1 + 5.2X2 + 5.0X3 + 5.1X4 + 7.5X5         4800 (Molding/pressing)
            4.5X1                                         1200 (Stove assembly)
                    4.5X2 + 4.0X3 + 3.0X4                 2400 (Washer/dryer assembly)
                                            9.0X5         1200 (Refrigerator assembly)
            4.0X1 + 3.0X2 + 2.5X3 + 2.0X4 + 4.0X5         3000 (Packaging)
                                All X's  0




266.6667 stoves, 448.7179 Washers, 133.33333 refrigerators; Profit = $87,051.28
Fractional values are work in progress from one week to the next.




                                      Chapter 3- 5
b. Add the following constraints:
                             X2 - X3 - X4 = 0 (Washers = Dryers)
                                   X3 - X4  100 (E. Dryers  G. Dryers + 100)
                                  -X3 + X4  100 (G. Dryers  E. Dryers + 100)




266.6667 stoves, 227.1545 washers, 63.57724 electric dryers, 163.5772 gas dryers,
133.3333 refrigerators; profit = $84,965.04




                                      Chapter 3- 6
3.3   See file Ch3.3.xls

      X1 = the number of standard Z345’s produced weekly
      X2 = the number of industrial Z345’s produced weekly
      X3 = the number of standard W250’s produced weekly
      X4 = the number of industrial W250’s produced weekly
      X5 = the total number of products produced weekly

      MAX 400X1 + 560X2 + 560X3 + 700X4
      S.T. 25X1 + 46X2 + 16X3 + 34X4                    2500   (zinc)
           50X1 + 30X2 + 28X3 + 12X4                    2800   (iron)
             X1 +    X2                                 20     (Min Z345’s)
             X1 +    X2 + X3 +        X4 -     X5      =    0   (X5 definition)
                     X2 +              X4 - .50X5          0   (Industrial min.)
             X1 +    X2                   - .75X5          0   (Max Z345’s)
                               X3 + X4 - .75X5             0   (Max W250’s)
                       X1, X2, X3, X4, X5  0




                                    Chapter 3- 7
a.       Produce 22.93578 standard Z345’s, 22.93578 standard W250’s, 45.87156
         industrial W250’s (fractional production quantities are work in progress carried
         over from one week to the next). Weekly profit = $52,752.29

b.       75% (Slack is 0 on that constraint.) If this restriction is loosened or eliminated,
         the weekly profit will increase.

c.         The shadow price of zinc is $21.10091743, which is valid for an additional
           492.1568627 pounds.
      i) 100 pounds is worth $2110.09 > $1500; yes purchase 100 additional pounds.
      ii) $2110.09 < $2600, no, 100 additional pounds should not be purchased.
      iii) Cannot tell without resolving since 800 additional pounds is outside the range of
           feasibility for the shadow price. Re-solving (not shown) with 3300 pounds of zinc
           gives a profit of $68,055.87. Since this a $15,303.87 increase, then, yes, the 800
           additional pounds should be purchased.


3.4      See file Ch3.4.xls

         X1 = amount invested in EAL stock
         X2 = amount invested in BRU stock
         X3 = amount invested in TAT stock
         X4 = amount invested in long term bonds
         X5 = amount invested in short term bonds
         X6 = amount invested in the tax deferred annuity
         X7 = the total amount invested in stocks only

MAX .15X1 + .12 X2 + .09X3 + .11X4 + .085X5 + .06X6
S.T.
       X1 +     X2 + X3 + X4 +           X5 + X6                       = 50,000 (Total)
                                               X6                       10,000 (TDA)
      X1 +     X2 +     X3                          - X7               =     0 (Stocks)
                        X3                        -.25X7                    0 (Min TAT)
                                X4 +     X5         - X7                    0(Bond  stock)
                        X3 +            X5 + X6                         12,500 (Low %)
                                  All X's  0


a.       Invest in (See following screen):
         EAL                           $ 7,500
         TAT                           $ 2,500
         Long Term Bonds               $30,000
         Tax Deferred Annuity          $10,000
                 Total return: $5,250


                                          Chapter 3- 8
b.     Rate of return for this investment = ($5,250/$50,000) = 10.5%
       The rate of return for additional funds = shadow price for the total investment
constraint (the first constraint above) = 11% which is valid to + (1E+30).

c. The return on EAL cannot fall below 12% from 15%; the return on BRU cannot increase
above 15% from 12%; the rate on long term bonds cannot increase above 13.5% from 11%.

d. Shadow price for:
 Total investment = .11 -- each additional dollar invested will earn 11%
 Minimum invested in taxed deferred annuity = -0.15 -- $0.15 lost for each extra dollar
   required to be invested in tax deferred annuities.
 Stock Definition -- a meaningless shadow price; the right hand side will not change.
 Minimum invested in TAT = -0.16 -- a $0.16 decrease in return for each extra dollar
   required to be in the low risk stock above 25%.
 Bonds >= Stocks = 0 -- no change in return for requiring bond investment to exceed
   stock investment by at least $1.
 Maximum invested in low yield investments = 0.1 -- $0.10 additional return for each
   extra dollar allowed to be invested in investments with returns less than 10%.


                                        Chapter 3- 9
3.5    See file Ch3.5.xls
                                                                    Unit Profit
       X1 = Number of full comforters produced daily         19-3(.50)-55(.20) = 6.50
       X2 = Number of queen comforters produced daily        26-4(.50)-75(.20) = 9.00
       X3 = Number of king comforters produced daily         32-6(.50)-95(.20) = 10.00

       MAX 6.50X1 + 9.00X2 + 10.00X3
       S.T.
              3X1 + 4X2 +         6X3                 2,700 (Stuffing)
             55X1 + 75X2 + 95X3                      48,000 (Fabric)
              3X1 + 5X2 +         6X3                 3,000 (Cutting minutes)
              5X1 + 6X2 +         8X3                12,000 (Sewing minutes)
                         All X's  120




  a.     240 full, 312 queen, 120 king comforters; daily profit = $5,568
         2688 pounds of stuffing, 48,000 sq. ft. of fabric, 3000 cutting minutes, 4032
         sewing minutes used.
  b.     Allowable decrease = .13636 -- Minimum selling price = $26-.14 = $25.86
  c.     (i)     $0     2688 - 
         (ii)    $0.11 46,800 - 48,200
         (iii) $0.15 2880 - 3080
         (iv)    $0     4032 - 
  d.     (See worksheet No Minimum) --300 full, 420 queen, 0 king; profit = $5,730
                                      Chapter 3- 10
3.6    See file Ch3.6.xls

       X1 = number of 8-oz. portions of steak in the diet
       X2 = number of ounces of cheese in the diet
       X3 = number of apples in the diet
       X4 = number of 8-oz. portion of milk in the diet

       MIN     51X1 +       9X2 + 1X3 +     8X4
       S.T.
              692X1 + 110X2 + 81X3 + 150X4            1410 (=1800-390 minimum
calories)
              692X1 + 110X2 + 81X3 + 150X4            1610 (=2000-390 maximum calories)
               57X1 + 6X2 + 1X3 + 8X4                   80 (=100-20 grams of protein)
                        1X2 + 22X3 + 12X4               30 (= 45-15 grams of carbs.)




                              All X's  0

       Steak = 8(1.49566474) = 11.96532 oz., Milk = 8(2.5) = 20 oz.
       Fat = 96.2789 + 29 = 125.2789 grams




                                       Chapter 3- 11
3.7    See file Ch3.7.xls

       X1 = Number of Student models produced each week
       X2 = Number of Plus models produced each week
       X3 = Number of Net models produced each week
       X4 = Number of Pro models produced each week

       MAX 70X1 + 80X2 + 130X3 + 150X4
       S.T.
                              X3                       100   (Contract)
            .4X1 + .5X2 + .6X3 + .8X4                  750   (Production Hours)
              X1 +      + X3                           700   (Celeron)
                     X2 +             X4               550   (Pentium)
              X1 + X2 + X3 +                           800   (20gb Hard Drives)
                                      X4               950   (30gb Hard Drives)
              X1 + X2 + 2X3 +         X4              1600   (Floppy Drives)
              X1 + X2 +               X4              1000   (Zip Drives)
              X1 +           X3 +     X4              1600   (CD R/W's)
                     X2 + X3 +         X4              900   (DVD's)
              X1 + X2                                  850   (15-in. monitors)
                              X3 +    X4               800   (17-in. monitors)
                    X2 + X3                           1250   (Mini-tower cases)
              X1 +                     X4              750   (Tower cases)
                          All X's  0

Given the output shown on the next page:
a.     325 Student models, 100 Plus models, 375 Net models, 425 Pro models; weekly
       profit = $143,250

b.     Since Allowable Decrease for the Plus model is 55, the optimal solution could
       change if its profit coefficient is reduced below $80 - $55 = $25. But this does not
       necessarily guarantee the production will now be 0, only that it could change. But
       if $24.99 is substituted for $80 and Solver is called again (not shown--see
       worksheet Plus Profit $24.99), there is no change to the optimal solution. Now the
       Allowable Decrease for the Plus model is $24.99, so the optimal solution could
       change if its profit coefficient is reduced below $24.99 - $24.99 = $0. Set the
       profit for the Plus model to -$0.01 and call Solver again. (See worksheet Plus
       Profit -$0.01). Now the optimal production quantity for the Plus model is 0. So,
       its profit can fall to $0, before it becomes unprofitable to produce the Plus model.

c.     Yes, the shadow price for 17-inch monitors is $25.

d.     No, production hours is not a binding constraint; DO NOT HIRE THE WORKER.




                                       Chapter 3- 12
Chapter 3- 13
3.8      See file Ch3.8.xls

       X1 = the number of Delta assemblies produced daily
       X2 = the number of Omega assemblies produced daily
       X3 = the number of Theta assemblies produced daily

       MAX 800X1 + 900X2 + 600X3
       S.T.   X1 +    X2 +     X3              7 (X70686 chips)
             2X1 +    X2 +     X3              8 (Production hours)
            80X1 + 160X2 + 80X3                480 (Quality minutes)
                   All X's  0

From the output on the next page:

a. Produce 2 Delta’s and 4 Theta’s; daily profit = $4,000. No Omega’s are produced
because the use too much quality control time compared to its slightly higher unit profit.

b. Reduced cost = -100; thus minimum profit for production = $1,000. Since costs are
$900, this implies a minimum price of $1,900.

c. 6; there is a slack of 1.

d. (i) There is slack on X70686 chips, thus additional profit is $0.
   (ii) Production hours are sunk costs. Since the shadow price for production hours is
$200 and 3 additional hours is within the range of feasibility, these 3 hours gross 3($200)
= $600 additional profit. Thus the net additional profit = $600 - $525 = $75.
  (iii) Quality control hours are also sunk costs. Since 1 additional hour = 60 additional
minutes is also within its range of feasibility, the $5 shadow price is valid for all 60
minutes grossing 60($5) = $300 additional profit. The net additional profit is $300 - $200
= $100.
                       OPTION (iii) is of the most value.




                                        Chapter 3- 14
Chapter 3- 15
3.9    See file Ch3.9.xls

       X1 = the number in group I contacted by telephone
       X2 = the number in group II contacted by telephone
       X3 = the number in group III contacted by telephone
       X4 = the number in group IV contacted by telephone
       X5 = the number in group I contacted in person
       X6 = the number in group II contacted in person
       X7 = the number in group III contacted in person
       X8 = the number in group IV contacted in person

MIN    15X1 + 12X2 + 20X3 + 18X4 + 35X5 + 30X6 + 50X7 + 40X8
S.T.      X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2000 (Total)
          X1 + X2                + X5 + X6                           1000 (W&R)
                                      X5 + X6 + X7 + X8              500 (In person)
       -.5X1                     + .5X5                                 0 (W&R,ip)
                  X2     + X4           + X6         + X8  800(Small)
             - .25X2      - .25X4+        .75X6 +     .75X8      0 (Small,ip)
          X1                     + X5                         200 (Min I)
                  X2                    + X6                         200 (Min II)
                       X3                       + X7          200 (Min III)
                              X4                   +     X8  200 (Min IV)
          X1                      + X5                               1000 (Max I)
                  X2                     + X6                 1000 (Max II)
                       X3                       + X7          1000 (Max III)
                              X4                   +    X8  1000 (Max IV)
                                   All X's 0


As shown on the next page:

       Minimum Cost = $39,800; conduct survey as follows:
                                 Telephone      Personal
                   I                500           500
                   II               600             0
                   III              200             0
                   IV               200             0




                                     Chapter 3- 16
3.10    See file Ch3.10

       X1 = the number of ounces of Multigrain Cheerios in the mixture
       X2 = the number of ounces of Grape Nuts in the mixture
       X3 = the number of ounces of Product 19 in the mixture
       X4 = the number of ounces of Frosted Bran in the mixture
       X5 = the total number of ounces in the mixture

       MIN    12X1 + 9X2 + 9X3 + 15X4
       S.T.   30X1 + 30X2 + 20X3 + 20X4           50     (Vitamin A)
              25X1 + 2X2 + 100X3 + 25X4           50     (Vitamin C)
              25X1 + 25X2 + 25X3 + 25X4           50     (Vitamin D)
              25X1 + 25X2 + 100X3 + 25X4          50     (Vitamin B6)
              45X1 + 45X2 + 100X3 + 25X4          50     (Iron)
                X1 + X2 +       X3 + X4 - X5 = 0          (Total)
                X1                        - .1X5  0      ( 10% M/G Cheerios)
                       X2                 - .1X5  0      ( 10% Grape Nuts)
                                X3        - .1X5  0      ( 10% Product 19)
                                       X4 - .1X5  0      ( 10% Frosted Bran)
                        All X's  0




                                     Chapter 3- 17
a.     From output below: Total sugar intake = 19.8 grams with this combination:

                       .2       oz.         Multigrain Cheerios
                      1.2244898 oz. Grape Nuts
                       .3755102 oz.         Product 19
                       .2        oz. Frosted Bran
                Total 2.0       oz.




The Allowable Increase for the sugar coefficient for Grape Nuts is 0, and the Allowable
Decrease for the sugar coefficient for Product 19 is 0. This indicates alternate optimal
solutions. By setting cell H6 to 19.8, then changing the target cell to MIN D4 (on
worksheet Alternate (not shown)), another optimal solution is:


                                       Chapter 3- 18
                             .2 oz.   Multigrain Cheerios
                             .8 oz.   Grape Nuts
                             .8 oz.   Product 19
                             .2 oz.   Frosted Bran
                      Total 2.0 oz.

b. Total = 2 oz. of cereal and 2 (1/2) = 1 cup of skim milk

c. Extra % above 50% required of vitamin D adds .396 grams of sugar
   Extra ounce above 10% required of Multigrain Cheerios adds 3 grams of sugar
   Extra ounce above 10% required of Frosted Bran adds 6 grams of sugar

4.6d. Product 19 has less sugar and gives percentages that are at least as large as those
for Frosted Bran for every vitamin and iron requirement. Re-solving gives the following
alternate optimal solutions with 18 grams of sugar.

             Mixture 1 (Shown) Mixture 2 (On Worksheet Alternate No 10%)
             1.53 oz. Grape Nuts             1.0 oz. Grape Nuts
               .47 oz. Product 19            1.0 oz. Product 19
       Total 2.00 oz.                  Total 2.0 oz.




                                       Chapter 3- 19
3.11          See file Ch3.11.xls

       X1 = Number of refrigerator/ovens produced
       X2 = Number of French fry makers produced
       X3 = Number of French toast makers produced

       MIN    140X1 + 50X2 + 36X3
       S.T.
              100X1 + 35X2 + 27X3           2,000,000   (Min Profit)
                 X1                             5,000   (Min Refrig/oven)
                        X2                      4,000   (Min French fry maker)
                               X3               2,300   (Min French toast maker)
                 X1                            15,000   (Max Refrig/oven)
                        X2                     15,000   (Max French fry maker)
                               X3              15,000   (Max French toast maker)

Make 14,550 refrigerator ovens, 4000 French fry makers, 15,000 French toast makers
Total Variable Cost = $2,777,000.




                                     Chapter 3- 20
3.12          See file Ch3.12.xls
a.
       X1 = Number of plates made per day
       X2 = Number of mugs made per day
       X3 = Number of steins made per day
       X4 = Total daily production

       MAX 2.50X1 + 3.25X2 + 3.90X3
       S.T.
               2X1 + 3X2 + 6X3                   1920     ((4)(8)(60) Molding min.)
               8X1 + 12X2 + 14X3                 3840     ((8)(8)(60) Finishing min.)
                        X2                       150      (Minimum mugs)
              -2X1 -   2X2 +      X3                0     (Steins  2(Plates + Mugs)
                X1 +    X2 +      X3 - X4       =    0     (Total Definition)
                X1                  - .3X4         0      (Plates  30% Total Produced)
                        All X's  0




101.8033 plates, 150 mugs, 87.54098 steins; total daily profit = $1083.42




                                       Chapter 3- 21
b.     Combine the first two constraints into one:

       10X1 + 15X2 + 20X3  5760




128 plates, 298.6667 mugs, 0 steins; total daily profit = $1290.67

This is an increase of ($1290.67 - $1083.42) =    $207.25




                                       Chapter 3- 22
3.13            See file Ch3.13.xls

       X1 = $ invested in first trust deeds
       X2 = $ invested in second trust deeds
       X3 = $ invested in third trust deeds
       X4 = $ invested in commercial trust deeds
       X5 = $ invested in a savings account
       X6 = Total $ invested in residential trust deeds
       X7 = Total $ invested in all trust deeds

MAX .0775X1 +.1125X2 +.1425X3 +.9875X4 +.0445X5
S.T.
         X1 +     X2 +     X3 +      X4 +   X5             = 68,000,000 (Total)
                                             X5             5,000,000 (Save)
         X1 +     X2 +      X3 +               - X6         =    0      (Res Tr.)
         X1 +     X2 +      X3 +     X4                 -X7 =    0      (Total Tr)
                                                  X6 - .8X7     0    (80% Res.)
         X1                                    -.6X6            0    (60% First)
        4X1 +    6X2 +     9X3 + 3X4                        340,000,000 (*)
                               All X's  0

*Average Risk Factor is found by:

                  X1            X2            X3            X4            X5
          4              6            9            3            0
              68,000,000    68,000,000    68,000,000    68,000,000    68,000,000

This expression must be  5. Multiplying both sides by 68,000,000 gives the above
constraint.

a.     As seen on the next page the optimal allocation of funds is:

       $30,240,000.00         First Trust Deeds
       $    66,666.67         Second Trust Deeds
       $20,093,333.33         Third Trust Deeds
       $12,600,000.00         Commercial Trust Deeds
       $ 5,000,000.00         Savings Account
       $68,000,000.00

Return = $6,539,400; Rate of return = $6,539,400/$68,000,000 =9.61676%

b.     Allowable Increase = .015, so could increase to 9.25%
       This would add (.015)($30,240,000) = $453,600 to the return.

Now, return = $6,539,400 + $453,600 = $6,993,000
New rate of return = $6,993,000/$68,000,000 = 10.28382%


                                       Chapter 3- 23
Output:




          Chapter 3- 24
3.14   See file Ch3.14.xls

       XJR = Motor home cabinets produced in regular time in July
       XJO = Motor home cabinets produced in overtime in July
       XAR = Motor home cabinets produced in regular time in August
       XAO = Motor home cabinets produced in overtime in August
       XJS = Motor home cabinets produced in regular time in September
       XJS = Motor home cabinets produced in overtime in September
       YJR = Mobile home cabinets produced in regular time in July
       YJO = Mobile home cabinets produced in overtime in July
       YAR = Mobile home cabinets produced in regular time in August
       YAO = Mobile home cabinets produced in overtime in August
       YJS = Mobile home cabinets produced in regular time in September
       YJS = Mobile home cabinets produced in overtime in September
       SJ = Motor home cabinets stored in July
       SA = Motor home cabinets stored in August
       SS = Motor home cabinets stored in September
       TJ = Mobile home cabinets stored in July
       TA = Mobile home cabinets stored in August
       TS = Mobile home cabinets stored in September

The objective function coefficients for the X's and Y's =
               (Material Cost) + (#Hours)(Hourly Cost for the Month and Period)

Example for mobile home cabinets in August in Overtime = $210 + 5(1.5)(16) = $330
The following tables summarize these coefficients.

Motor Home
                        Regular Time       Overtime
               July        $188              $209
               August      $194              $218
               September   $200              $227

Mobile Home
                        Regular Time       Overtime
               July        $280              $315
               August      $290              $330
               September   $300              $345

Objective function coefficients for the S's and T's

                        Motor Home (S)        Mobile Home (T)
               July         $6                      $9
               August       $6                      $9
               September    $6                      $9


                                        Chapter 3- 25
Objective Function

MIN    188XJR + 209XJO + 194XAR + 218XAO + 200 XSR + 227XSO +
       280YJR + 315YJO + 290YAR + 330YAO + 300 YSR + 345YSO +
       6SJ + 6SA + 6SS + 9TJ + 9TA + 9Ts

Constraints

Storage Constraints
 For each product: Amount Stored at the end of the month =
       (Amount Stored at the Beginning of the month) + (Production) - (Demand)

       SJ = 25 + XJR + XJO - 250       (Motor Home - July)
       SA = SJ + XAR + XAO - 250       (Motor Home - August)
       SS = SA + XSR + XSO - 150       (Motor Home - September)
       TJ = 20 + YJR + YJO - 100       (Mobile Home - July)
       TA = TJ + YAR + YAO - 300       (Mobile Home - August)
       TS = TA + YSR + YSO - 400       (Mobile Home - September)

Required For September:
      SS  10                          (Motor Home)
      TS  25                          (Mobile Home)

 Maximum Storage in any Month
     SJ + TJ        300               (July)
     SA + TA        300               (August)
     SS + TS        300               (September)

Production
      Regular Time
      3XJR + 5YJR  2100               (July)
      3XAR + 5YAR  1500               (August)
      3XSR + 5YSR  1200               (September)

       Overtime
       3XJO + 5YJO  1050              (July)
       3XAO + 5YAO  750               (August)
       3XSO + 5YSO  600               (September)

Non-negativity
      All X's, Y's, S's, and T's  0




                                        Chapter 3- 26
The optimal solution is shown in the cells C6:D8, G6:H8, and K6:L8. Total Cost = $367,969




                                           Regular Time
                                           Begin Storage + Production - End Storage = Demand
                                           Overtime
                                           September Storage -- Motor Home Cabinets September
                                           Storage -- Mobile Home Cabinets
                                           Maximum Monthly Storage




                                     Chapter 3- 27
3.15   See file Ch3.15.xls

       X1 = the number of acres of wheat planted
       X2 = the number of acres of corn planted
       X3 = the number of acres of oats planted
       X4 = the number of acres of soybeans planted

   Profit coefficients are 210($3.20) - $50 = $622, 300($2.55) - $75 = $690,
      180($1.45) - $30 = $231, and 240($3.10) - $60 = $684 respectively.

       MAX 622X1 + 690X2 + 231X3 + 684X4
       S.T.   4X1 + 5X2 + 3X3 + 10X4                    1,800      (Labor hours)
             50X1 + 75X2 + 30X3 + 60X4                  25,000     (Expenses)
              2X1 + 6X2 +      X3 + 4X4                 1,200      (Water)
            210X1                                       30,000     (Min. Wheat)
                   300X2                                30,000     (Min. Corn)
                            180X3                       25,000     (Max Oats)
               X1 + X2 +      X3 +    X4                   300     (Total acres)
                        All X's  0

a. (See Worksheet on next page) Plant 142.8571 acres of wheat, 142.8571 acres of corn
and 14.28571 acres of soybeans; profit = $197,200.

b. The net profit must rise to $675; adding the $30/acre in expenses = $705.
If selling price remains $1.45 per bushel, then yield must increase to $705 per acre/$1.45
per bushel = 486.21 bushels per acre.
If the yield remains 180 bushels per acre, then its price must rise to $705 per acre/180
bushels per acre = $3.92 per bushel.

c. (See work sheet NO CORN (not shown).) Yes, corn would still be planted; there is
currently slack on the corn constraint. If corn were not grown, the problem must be re-
solved. Plant 200 acres of wheat and 100 acres of soybeans for a profit of $192,800 -- a
decrease of $4,400.

d. The range of feasibility for acres is only valid up to 318.57 acres (18.57 additional
acres.) Thus the problem must be re-solved. See worksheet PARCEL.

Plant 181.82 acres of wheat, 101.82 acres of corn, and 56.36 acres of soybeans for a profit
of $221,898.18 -- an increase of $24,698.18. Yes Bill should lease this property for
$2,000.




                                       Chapter 3- 28
Chapter 3- 29
3.16    See file Ch3.16.xls

 PSA   = Number of professional sets produced in Sarasota and shipped to Anaheim
 PSD   = Number of professional sets produced in Sarasota and shipped to Dallas
 PST   = Number of professional sets produced in Sarasota and shipped to Toledo
 PCA   = Number of professional sets produced in Carson and shipped to Anaheim
 PCD   = Number of professional sets produced in Carson and shipped to Dallas
 PCT   = Number of professional sets produced in Carson and shipped to Toledo

 DSA   = Number of deluxe sets produced in Sarasota and shipped to Anaheim
 DSD   = Number of deluxe sets produced in Sarasota and shipped to Dallas
 DST   = Number of deluxe sets produced in Sarasota and shipped to Toledo
 DLA   = Number of deluxe sets produced in Louisville and shipped to Anaheim
 DLD   = Number of deluxe sets produced in Louisville and shipped to Dallas
 DLT   = Number of deluxe sets produced in Louisville and shipped to Toledo
 DCA   = Number of deluxe sets produced in Carson and shipped to Anaheim
 DCD   = Number of deluxe sets produced in Carson and shipped to Dallas
 DCT   = Number of deluxe sets produced in Carson and shipped to Toledo

 WLA   = Number of weekender sets produced in Louisville and shipped to Anaheim
 WLD   = Number of weekender sets produced in Louisville and shipped to Dallas
 WLT   = Number of weekender sets produced in Louisville and shipped to Toledo
 WCA   = Number of weekender sets produced in Carson and shipped to Anaheim
 WCD   = Number of weekender sets produced in Carson and shipped to Dallas
 WCT   = Number of weekender sets produced in Carson and shipped to Toledo

Objective function coefficients = (Gross Profit) - (Shipping Cost)

MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT +
 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT +
 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT
S.T.
Minimum Professional
      PSA       + PCA      480
      PSD       + PCD      320
      PST       + PCT      160
Minimum Deluxe
      DSA + DLA + DCA      640
      DSD + DLD + DCD      800
      DST + DLT + DCT      880
Minimum Weekender
           WLA + WCA       640
           WLD + WCD       1200
           WLT + WCT       800



                                       Chapter 3- 30
Maximum Professional
       PSA         + PCA      600
       PSD         + PCD      400
       PST         + PCT      200
Maximum Deluxe
       DSA + DLA + DCA        800
       DSD + DLD + DCD        1000
       DST + DLT + DCT        1100
Maximum Weekender
             WLA + WCA        800
             WLD + WCD        1500
             WLT + WCT        1000
Steel Availability
3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST                          5000
3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT                          9000
3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT  14000
Aluminum Availability
5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST                                      7000
4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT                                13000
5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT             18000
Wood Availability
5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST                          10000
4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT                          18000
5.2PCA +5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT  20000

                                       All P's, D's W's  0

Output is too large to display. Monthly Profit = $1,353,088.89

                                         Professional
                          Sarasota                            Carson
              Anaheim      Dallas      Toledo   Anaheim       Dallas   Toledo
                 0          400         200       600           0        0


                                             Deluxe
          Sarasota                          Louisville                        Carson
Anaheim    Dallas       Toledo    Anaheim    Dallas       Toledo   Anaheim    Dallas    Toledo
   0      855.5556        0          0         0           1100      800     144.4444     0


                                         Weekender
                          Louisville                          Carson
              Anaheim      Dallas      Toledo   Anaheim       Dallas   Toledo
                 0          800         1000      800          700       0




                                         Chapter 3- 31
3.17   See file Ch3.17.xls

 PSA = Number of professional sets produced in Sarasota and shipped to Anaheim
 PSD = Number of professional sets produced in Sarasota and shipped to Dallas
 PST = Number of professional sets produced in Sarasota and shipped to Toledo
 PCA = Number of professional sets produced in Carson and shipped to Anaheim
 PCD = Number of professional sets produced in Carson and shipped to Dallas
 PCT = Number of professional sets produced in Carson and shipped to Toledo
 DSA = Number of deluxe sets produced in Sarasota and shipped to Anaheim
 DSD = Number of deluxe sets produced in Sarasota and shipped to Dallas
 DST = Number of deluxe sets produced in Sarasota and shipped to Toledo
 DLA = Number of deluxe sets produced in Louisville and shipped to Anaheim
 DLD = Number of deluxe sets produced in Louisville and shipped to Dallas
 DLT = Number of deluxe sets produced in Louisville and shipped to Toledo
 DCA = Number of deluxe sets produced in Carson and shipped to Anaheim
 DCD = Number of deluxe sets produced in Carson and shipped to Dallas
 DCT = Number of deluxe sets produced in Carson and shipped to Toledo
 WLA = Number of weekender sets produced in Louisville and shipped to Anaheim
 WLD = Number of weekender sets produced in Louisville and shipped to Dallas
 WLT = Number of weekender sets produced in Louisville and shipped to Toledo
 WCA = Number of weekender sets produced in Carson and shipped to Anaheim
 WCD = Number of weekender sets produced in Carson and shipped to Dallas
 WCT = Number of weekender sets produced in Carson and shipped to Toledo

 YS = 1 if the Sarasota plant is operational
 YL = 1 if the Louisville plant is operational
 YC = 1 if the Carson plant is operational
a.
MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT +
 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT +
 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT
 -250000YS - 350000YL - 500000YC
S.T.
Minimum Professional
       PSA         + PCA        480
       PSD         + PCD        320
       PST         + PCT        160
Minimum Deluxe
       DSA + DLA + DCA          640
       DSD + DLD + DCD          800
       DST + DLT + DCT          880
Minimum Weekender
             WLA + WCA          640
             WLD + WCD          1200
             WLT + WCT          800


                                    Chapter 3- 32
Maximum Professional
       PSA         + PCA        600
       PSD         + PCD        400
       PST         + PCT        200
Maximum Deluxe
       DSA + DLA + DCA          800
       DSD + DLD + DCD          1000
       DST + DLT + DCT          1100
Maximum Weekender
             WLA + WCA          800
             WLD + WCD          1500
             WLT + WCT          1000
Steel Availability
3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST                          -5000YS  0
3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT                           -9000YL  0
3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT-
14000YC0
Aluminum Availability
5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST                             - 7000YS        0
4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT                       -13000YL        0
5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT -18000YC           0
Wood Availability
5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST                          -10000YS 0
4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT                          -18000YL 0
5.2PCA + 5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT-20000YC0
                       All P's, D's W's  0 Y's binary

Output is too large to display. (See worksheet Part a.) Monthly Profit = $437,250

                                         Professional
                          Sarasota                          Carson
              Anaheim      Dallas      Toledo   Anaheim     Dallas    Toledo
                 0           0           0        600        400       200


                                             Deluxe
          Sarasota                          Louisville                         Carson
Anaheim    Dallas       Toledo    Anaheim    Dallas       Toledo   Anaheim     Dallas   Toledo
   0         0            0          0         0           820       800        857.5     60


                                         Weekender
                          Louisville                        Carson
              Anaheim      Dallas      Toledo   Anaheim     Dallas    Toledo
                 0          1160        1000      800        340        0
Do not operate Sarasota.


                                         Chapter 3- 33
b.    Add decision variables:

      ZA = 1 if the Anaheim center is operational
      ZD = 1 if the Dallas center is operational
      ZT = 1 if the Toledo center is operational

MAX 205PSA + 218PSD + 220PST + 242PCA + 210PCD + 200PCT +
 135DSA + 147DSD + 150DST + 141DLA + 157DLD + 165DLT + 169DCA + 140DCD + 135DCT +
 170WLA + 185WLD + 191WLT + 195WCA + 170WCD + 164WCT
 -250000YS - 350000YL - 500000YC - 50000ZA - 100000ZB - 90000ZB

S.T.
Minimum Professional
       PSA         + PCA - 480ZA     0
       PSD         + PCD - 320ZD     0
       PST         + PCT -160ZT      0
Minimum Deluxe
       DSA + DLA + DCA - 640ZA       0
       DSD + DLD + DCD - 800ZD       0
       DST + DLT + DCT - 880ZT       0
Minimum Weekender
             WLA + WCA - 640ZA       0
             WLD + WCD -1200ZD  0
             WLT + WCT - 800ZT  0
Maximum Professional
       PSA         + PCA - 600ZA     0
       PSD         + PCD - 400ZD     0
       PST         + PCT - 200ZT     0
Maximum Deluxe
       DSA + DLA + DCA - 800ZA  0
       DSD + DLD + DCD -1000ZD  0
       DST + DLT + DCT -1100ZT  0
Maximum Weekender
             WLA + WCA - 800ZA  0
             WLD + WCD -1500ZD  0
             WLT + WCT -1000ZT  0
Steel Availability
3.2PSA + 3.2PSD + 3.2PST + 3.6DSA + 3.6DSD + 3.6DST                          -5000YS  0
3.6DLA + 3.6DLD + 3.6DLT + 2.8WLA + 2.8WLD + 2.8WLT                          -9000YL  0
3.2PCA + 3.2PCD + 3.2PCT + 3.6DCA + 3.6DCD + 3.6DCT +2.8WCA + 2.8WCD + 2.8WCT-
14000YC0
Aluminum Availability
5PSA + 5PSD + 5PST + 4DSA + 4DSD + 4DST                             - 7000YS       0
4DLA + 4DLD + 4DLT + 4.5WLA + 4.5WLD + 4.5WLT                       -13000YL        0


                                     Chapter 3- 34
5PCA + 5PCD + 5PCT + 4DCA + 4DCD + 4DCT +4.5WCA + 4.5WCD + 4.5WCT -18000YC         
0
Wood Availability
5.2PSA + 5.2PSD + 5.2PST + 4.8DSA + 4.8DSD + 4.8DST                         -10000YS 0
4.8DLA + 4.8DLD + 4.8DLT + 4.4WLA + 4.4WLD + 4.4WLT                         -18000YL 0
5.2PCA + 5.2PCD + 5.2PCT + 4.8DCA + 4.8DCD + 4.8DCT +4.4WCA + 4.4WCD + 4.4WCT-
20000YC0
                             All P's, D's W's  0 Y's, Z's binary

Output is too large to display. (See worksheet Part b.) Monthly Profit = $197,250

                                          Professional
                           Sarasota                          Carson
              Anaheim       Dallas      Toledo   Anaheim     Dallas    Toledo
                 0            0           0        600        400       200


                                              Deluxe
           Sarasota                          Louisville                         Carson
Anaheim     Dallas      Toledo     Anaheim    Dallas       Toledo   Anaheim     Dallas   Toledo
   0          0           0           0         0           820       800        857.5     60


                                          Weekender
                           Louisville                        Carson
              Anaheim       Dallas      Toledo   Anaheim     Dallas    Toledo
                 0           1160        1000      800        340        0


Do not operate Sarasota.




                                          Chapter 3- 35
3.18   See file Ch3.18.xls

       X1 = amount invested in Bonanza Gold
       X2 = amount invested in Cascade Telephone
       X3 = amount invested in the money market account
       X4 = amount invested in two-year treasury bonds

       MAX .15X1 + .09X2 + .07X3 + .08X4
       S.T.    X1 + X2                               50,000    (Max stocks)
               X1 + X2 + X3                          60,000    (Min potential 9%)
                                       X4            30,000    (Max treasury bonds)
            -.5X1 + .03X2 + .06X3 + .08X4            4,000     (Min return)
               X1 + X2 + X3 +          X4           = 100,000   (Total)
                          All X's  0

Invest $2,075.47 in Bonanza Gold, $47,924.53 is Cascade Telephone, $20,000 in the
money market account, $30,000 in two year treasury bonds. The expected return is
$8,424.53 (8.42453%).




                                    Chapter 3- 36
3.19   See file Ch3.19.xls

       Xij = number of gallons of crude i blended into grade j
               i = p (Pacific), g (Gulf), m (Middle East) j = r (Regular), p (Premium)
       Xj = total amount of grade j produced

Example of profit coefficient:
   Selling price of regular = $0.52, purchase cost of Pacific crude = ($14.28)/42 = $0.34;
   thus profit on a gallon of Pacific crude = $0.52 - $0.34 = $0.18

MAX .18Xpr +.16Xgr +.05Xmr +.26Xpp +.24Xgp +.13Xmp
S.T.   Xpr                 + Xpp                                      126,000 (Pacific)
               Xgr                 + Xgp                              84,000 (Gulf)
                       Xmr               + Xmp                        336,000 (M. East)
       Xpr + Xgr + Xmr                            - Xr               = 0       (Regular)
                              Xpp + Xgp + Xmp          – Xp          = 0        (Premium)
                                                    Xr                200,000 (Min Reg)
                                                        Xp            100,000 (Min Prem)
                                                    Xr + Xp           400,000 (Capacity)
     85Xpr + 87Xgr + 95Xmr                     - 87Xr                    0     (Reg Oct.)
                            85Xpp +87Xgp + 95Xmp - 91Xp                  0     (Prem Oct.)
                            All X's  0

From the sensitivity report (not shown), there are alternate optimal solutions giving a
profit of $61,620. The one represented on the screen on the next page is:

Pacific - Regular              46,000
Gulf - Regular                 84,000
Mid East - Regular             70,000
Pacific - Premium              80,000
Gulf - Premium                   0
Mid East - Premium            120,000

b. Shadow price (from Total Total (not shown)) = $0.13 per gallon; Thus 50,000
gallons is worth 50,000($0.13) = $6,500 > $5,000 -- Yes Caloco should secure the extra
refining capacity.

c. The solution uses all 190,000 gallons of Mid East Oil @ $0.47/gallon = $89,300.
   8000 barrels @ $16.80 per barrel = $134,400 -- the Middle East distributors receive
more.
       To see if it would be profitable for Caloco, the problem must be re-solved.
$16.80 per barrel = $0.40 per gallon.
 The new profit coefficients for Xmr and Xmp would be $0.12 and $0.20 respectively.
 Change the third constraint to Xmr + Xmp = 336,000
 Re-solve (See worksheet Mid East (not shown).)


                                        Chapter 3- 37
The new solution gives Caloco a profit of $67,840 -- thus it would be profitable for
Caloco to accept this offer. There is now no Gulf crude purchased and only slightly more
than 1500 barrels of Pacific crude purchased. These distributors would have to cut their
prices to stay competitive.




                                                               Minimum Requirements

                                                                 Availability
                                                              Capacity
                                                               Octane Requirements




                                      Chapter 3- 38
3.20   See file Ch3.20.xls
                                                                 Average Score
       X1 = the weight given to teaching                       (90+75+90)/3 = 85
       X2 = the weight given to research                       (60+60+75)/3 = 65
       X3 = the weight given to professional activities        (90+95+85)/3 = 90
       X4 = the weight given to service                        (80+95+95)/3 = 90

       MAX 85X1 + 65X2 + 90X3 + 90X4
       S.T. - X1 + X2                         0        (Teaching  research)
            - X1       + X3                   0        (Teaching  professional)
            - X1              + X4            0       (Teaching  service)
                    X2                        .25      (Min. research)
              X1 + X2                         .75      (Teaching + research  .75)
              X1 + X2                         .90     (Teaching + research  .90)
                           X3 - X4            0        (Service  professional)
                           X3                 .05      (Professional  .05)
              X1 + X2 + X3 + X4              = 1        (Total weights)
                        All X's  0




Teaching = .50, Research = .25, Professional = .125, Service = .125--Average = 81.25 < 85.
Professor Anna Sung should not get tenure.




                                       Chapter 3- 39
3.21   See file Ch3.21.xls -- a more compact Excel spreadsheet is discussed in Chapter
4
       Xij = the number of Matey 20 catamarans made in plant I and sold to dealership j
           i = 1 (San Diego), 2 (Santa Ana), 3 (San Jose)
           j = 1 (Newport Beach), 2 (Long Beach), 3 (Ventura), 4 (S L O), 5 (San Fran)

       MIN    1265X11 + 1285X12 + 1345X13 + 1390X14 + 1565X15 +
              1130X21 + 1130X22 + 1285X23 + 1355X24 + 1405X25 +
              1365X31 + 1340X32 + 1275X33 + 1225X34 + 1075X35

       S.T.   X11 + X12 + X13 + X14 + X15                       38 (San Diego)
              X21 + X22 + X23 + X24 + X25                45   (Santa Ana)
              X31 + X32 + X33 + X34 + X35                58   (San Jose)
              X11    + X21       + X31                =   42   (Newport Beach)
              X12    + X22       + X32                =   33   (Long Beach)
              X13    + X23       + X33                =   14   (Ventura)
              X14    + X24       + X34                =   10   (San Luis Obispo)
              X15    + X25       + X35                =   22   (San Francisco)
                     All Xij’s  0




Build 30 in San Diego -- Ship them all to Newport Beach
Build 45 in Santa Ana -- Ship 12 to Newport Beach and 33 to Long Beach
Build 46 in San Jose -- Ship 14 to Ventura, 10 to San Luis Obispo, and 22 to San Francisco



                                      Chapter 3- 40
Total building and shipping costs = $142,550

3.22   See file Ch3.22.xls

       X1 = the number cut into 4 30-in. pieces
       X2 = the number cut into 2 30-in. and 1 42-in. pieces
       X3 = the number cut into 1 30-in. and 2 42-in. pieces
       X4 = the number cut into 2 30-in. and 1 56-in. pieces
       X5 = the number cut into 1 42-in. and 1 56-in. pieces
       X6 = the number cut into 2 56-in. pieces

       MIN           18X2 + 6X3 + 4X4 + 22X5 + 8X6
       S.T.    4X1 + 2X2 + X3 + 2X4                   1,500 (30-in.)
                       X2 + 2X3        + X5      500 (42-in.)
                                     X4 + X5 + 2X6    600 (56-in.)
                             All X's  0

From the Sensitivity Report, there are alternate optimal solutions. The one shown is
312.5 (rounded to 313) cut into 4 30-in. pieces; 250 cut into 1 30-in. and 2 42-in. pieces
and 300 cut into 2 56-inch pieces.     Total waste = 3900 inches.
               Two 30-in. pieces will remain in inventory with rounded solution.




                                        Chapter 3- 41
Chapter 3- 42
3.23 See file Ch3.23.xls
a. This could violate the “no interaction” assumption of linear programming.

b.     X1 = $ spent on television advertising
       X2 = $ spent on radio advertising
       X3 = $ spent on newspaper advertising
       X4 = $ spent on its circulars

       MAX 28X1 +       18X2 + 20X3 + 15X4
       S.T.   X1 +        X2 + X3 + X4    700,000 (Budget)
              X1 +        X2              350,000 (TV/Radio)
            10X1 +       7X2 + 8X3 + 4X4  2,500,000 (Yuppies)
             5X1 +       2X2 + 3X3 + X4   1,200,000 (College)
             5X1 +       8X2 + 6X3 + 9X4  1,800,000 (Audiophiles)
              X1                          300,000 (Max TV)
                       X2                 300,000 (Max Radio)
                               X3         300,000 (Max Newspapers)
                                      X4  300,000 (Max Circulars)
                           All X's  0
$300,000 TV, $100,000 radio, $300,000 newspapers Total Exposure = 16,200,000




c. It would not affect the optimal solution; it is a non-binding constraint (there is surplus).



                                         Chapter 3- 43
3.24   See file Ch3.24

a.     X1 = number of polyester suits ordered for the season
       X2 = the number of wool suits ordered for the season
       X3 = the number of cotton suits ordered for the season
       X4 = the number of imported suits ordered for the season

       MAX 35X1 + 47X2 + 30X3 + 90X4
       S.T. .4X1 + .5X2 + .3X3 + X4  1,800 (Salesperson hours)
             2X1 + 4X2 + 3X3 + 9X4  15,000 (Advertising budget)
              X1 + 1.5X2 + 1.25X3 + 3X4  18,000 (Square footage)
                       All X's  0




Order 1500 polyester suits, 4000 cotton suits --Profit = $172,500




                                       Chapter 3- 44
b. New Objective     Within the               Change to
    Coefficient  Range of Optimality?          Solution                       Profit
(i)   $34              Yes             None (alternate optima)               $171,000
      $33              No             3000 wool, 1000 cotton*                $171,000

(ii)      $36                   Yes                       None               $174,000
          $37                   Yes                       None               $175,500

*(See worksheet Part b (i-33) (not shown)):

c.       (i) No, there is a slack on the space constraint.
        (ii) Yes, $400 will net 400(2.5) = $1000 additional profit.
       (iii) Yes, 260 salesperson hours will add 260($75) = $19,500 for a cost of $3,600.

d. Add the constraint X1 + X2 + X3 + X4  5000. (See worksheet Part d (not shown).)
   Order 1000 polyester, 1000 wool and 3000 cotton suits giving a $172,000 profit.



3.25      See file Ch3.25.xls

       X1 = Number of acres for zoo habitat attractions
       X2 = Number of acres for show areas
       X3 = Number of acres for restaurants
       X4 = Number of acres for retail establishments
       X5 = Number of acres for maintenance areas
       X6 = Number of acres for green areas
       X7 = Number of acres for roads/walkways
       X8 = Number of restaurants
       X9 = Number of retail stores

Max gross profit per hour: 1000X1 + 900X2 + 800X8 + 750X9
S.T.
Ttoal acreage = 350:          X1 + X2 + X3 + X4 + X5 + X6 + X7 = 350
Habitat  40%(350):           X1  140
Greens  that required by habitat, show, restaurant, store and walkway areas:
                              X6  .03X1 + .40X2 + .25X7 + .15X8 + .10X9
Show areas  5%(350):         X2  17.5
Habitat + Show  70%(350): X1 + X2  245
(Show + its green)  .20((Show + its green) + (Habitat + its green) :
                              1.4X2  .2(1.03X1 + 1.4X2)
Green  25%(area not for habitat and show) :          X6  .25(350-X1-X2)
Maintenance  require maintenance: X5  .01X1 + .10X2 + .08X3 + .06X4 + .02X6 + .04X7
Restaurants  20:             X8  20
Restaurants  30:             X8  30


                                          Chapter 3- 45
Restaurant acreage = .25(# restaurants)      X3 = .25X8
Stores  15                                  X9  15
Stores  25                                  X9  25
Restaurants  Stores                         X8  X9
Store acreage = .20(#stores)                 X4 = .20X9
At least 10 acres walkways:                  X7  10
At least 100 acres of non-required green: X6 -(.03X1 + .40X2 + .25X7 + .15X8 + .10X9)  100
All X's  0

Summarizing:

MAX 1000X1 + 900X2                                    + 800X8 + 750X9
S.T.       X1 +    X2 + X3 + X4 + X5 + X 6 + X7                                 = 350
           X1                                                                    140
      -.03X1 - .40X2                       + X6 -.25X7 - .15X8 - .10X9           0
                   X2                                                            17.5
           X1 +    X2                                                            245
     -2.06X1 + 1.12X2                                                            0
        .25X1 + .25X2                       + X6                                 87.5
      -.01X1 - .10X2 -.08X3 - .06X4 + X5 - .02X6 - .04X7                         0
                                                            X8                   20
                                                            X8                   30
                         X3                            -.25X8                   = 0
                                                                     X9          15
                                                                     X9          25
                                                            X8   - X9            0
                                 X4                              -.20X9         = 0
                                                       X7                        10
      -.03X1 - .40X2                       + X6 -.25X7 - .15X8 - .10X9           100
                                   All X's  0

As seen on the next page:

Habitat Area:                    180.78 acres
Show Area:                        17.50 acres
Restaurants (30)                   7.50 acres
Stores (25)                        5.00 acres
Maintenance                        7.30 acres
Green Area:                      121.92 acres
Roads/Walkways                    10.00 acres
                   TOTAL         350.00 acres

Gross Hourly Profit = $239,280 Gross Daily Profit = $2,392,800 Net Daily Profit = $392,800
Annual Net Profit = 365($392,800) = $143,372,000.


                                      Chapter 3- 46
Chapter 3- 47
3.26   See file Ch3.26.xls

       X1 = the number of 2-oz. Go bars produced daily
       X2 = the number of 2-oz. Power bars produced daily
       X3 = the number of 2-oz. Energy bars produced daily
       X4 = the number of 8-oz. Energy bars produced daily
       X5 = the total number of 2-oz. bars produced daily

Derivation of profit coefficients
The costs per 2oz. are: Protein concentrate $0.40; Sugar substitute $0.175; Carob $0.325.
Thus the unit profits are:
       2 oz. Go bars:           .68 - .03 - .2(.40) - .6(.175) - .2(.325) = $0.40
       2 oz. Power bars:        .84 - .03 - .5(.40) - .3(.175) - .2(.325) = $0.4925
       2 oz. Energy bars:       .76 - .03 - .3(.40) - .4(.175) - .3(.325) = $0.4425
       8 oz. Energy bars:     3.00 - .05 - .3(1.60) - .4(.70) - .3(1.30) = $1.80

       MAX .40X1 + .4925X2 + .4425X3 +1.80X4
       S.T.  .4X1 +     X2 +     .6X3 + 2.4X4               9,600 (Oz. protein)
            1.2X1 + .6X2 +       .8X3 + 3.2X4               16,000 (Oz. sugar sub.)
             .4X1 + .4X2 +       .6X3 + 2.4X4               12,800 (Oz. carob)
               X1 +     X2 +       X3         - X5         = 0 (Total 2-oz.)
                                                 X5         25,000 (Max 2-oz.)
                                            X4              2,000 (Max 8-oz.)
               X1                                           2,500 (Min 2-oz. Go)
                        X2                                  2,500 (Min 2-oz. Power)
                                    X3                      2,500 (Min 2-oz. Energy)
               X1                              -.5X5            0 (Max 2-oz. Go)
                        X2                     -.5X5            0 (Max 2-oz. Power)
                                    X3         -.5X5            0 (Max 2-oz. Energy)
                                   2X3 + 4X4 - X5               0 (Max Energy*)
                             All X's  0

   *The Maximum Energy Bar constraint is formulated as follows:
      Total Weight Energy Bars: 2X3 + 8X4       Total Weight All Bars: 8X4 + 2X5
             Thus, 2X3 + 8X4  .5(8X4 + 2X5) or 2X3 + 4X4 - X5  0.

       Daily production, giving daily profit of $7,273.375 (see output on next page):
                     2-oz. Go:                7,550
                     2-oz. Power:             2,500
                     2-oz. Energy:            5,050
                     8-oz. Energy               437.5




                                       Chapter 3- 48
Chapter 3- 49
3.27   See file Ch3.27.xls
a.
       X1 = $ spent on advertising ketchup only
       X2 = $ spent on advertising spaghetti sauce only
       X3 = $ spent on advertising taco sauce only
       X4 = $ spent on joint advertising

       MAX 1.20X1 + 1.12X2 + 1.10X3 + 1.05X4
       S.T.    X1 +     X2 +     X3 +      X4  2,000,000 (Total advertising)
                                           X4  400,000 (Max joint adv.)
                                           X4  100,000 (Min joint adv.)
                                 X3 +      X4  1,000,000 (Min total taco sauce)
               X1                               250,000 (Min ketchup only)
                        X2                      250,000 (Min spag sauce only)
                                 X3             750,000 (Min taco sauce only)
               X1                   +      X4  700,000 (Min total ketchup)
                        X2           +      X4  700,000 (Min total spag sauce)
              4X1 + 3.2X2 + 11X3 + 4.2X4  7,500,000 (Min bottles sold)
                                  All X's  0




                                      Chapter 3- 50
Advertising budget: $550,000 ketchup only, $450,000 spaghetti sauce only,
                    $750,000 taco sauce only, $250,000 joint advertising

Total return = $2,251,500 or (2,251,500/2,000,000) = 112.575%

b. Shadow price = $1.20

c. Shadow price = -.03. Since 700,000 is within its range of feasibility, the profit would
increase by (-50,000)(-$0.03) = $1500.




                                       Chapter 3- 51
3.28    See file Ch3.28.xls
        Xj = the number of workers that have shift j

MIN 15X1 + 25X2 + 52X3 + 22X4 + 54X5 + 24X6 + 55X7 + 23X8 + 16X9
S.T.  X1 + X2 + X3                                                                   8
             X2 + X3                                                                 10
                      X3 + X4 + X5                                                   22
                      X3 + X4 + X5                                                   15
                                    X5 + X6 + X7                                     10
                                    X5 + X6 + X7                                     20
                                                   X7 + X8                           16
                                                   X7 + X8 + X9                      8
                      X3                                                             2
                                                   X7                                2
                    .6X3 - .4X4 + .6X5                                               0
                                  .6X5 - .4X6 + .6X7                                 0
                               All X’s  0, and integer




Shift 2--7, Shift 3--3, Shift 4--13, Shift 5--6, Shift 6--12, Shift 7--2, Shift 8--14
        Total Cost = $1,661.



                                          Chapter 3- 52
3.29    See file Ch3.29.xls

        Xj = the number of patrols in sector j

    For each sector, there must be at least one patrol in that sector or one in an adjacent
sector. Thus there are 15 constraints -- one for each sector.

MIN     X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 +X11 + X12 + X13 + X14 + X15

S.T.    X1 + X2 + X9 + X10 + X11                          1    (Sector 1)
        X1 + X2 + X3 + X9                                 1    (Sector 2)
        X2 + X3 + X4 + X8 + X9                            1    (Sector 3)
        X3 + X4 + X5 + X6 + X8                            1    (Sector 4)
        X4 + X5 + X6 + X7                                 1    (Sector 5)
        X4 + X5 + X6 + X7 + X8                            1    (Sector 6)
        X5 + X6 + X7 + X8 + X13 + X14 + X15               1    (Sector 7)
        X3 + X4 + X6 + X7 + X8 + X9 + X13                 1    (Sector 8)
        X1 + X2 + X3 + X8 + X9 + X10 + X13                1    (Sector 9)
        X1 + X9 + X10 + X11 + X12 + X13                   1    (Sector 10)
        X1 + X10 + X11 + X12                              1    (Sector 11)
        X10 + X11 + X12 + X13 + X14                       1    (Sector 12)
        X7 + X8 + X9 + X10 + X12 + X13 + X14              1    (Sector 13)
        X7 + X12 + X13 + X14 + X15                        1    (Sector 14)
        X7 + X14 + X15                                    1    (Sector 15)
                              All X's binary

3 units -- Place a squad car (patrol unit) in sectors 3, 7, and 10.




                                          Chapter 3- 53
Chapter 3- 54
3.30   See file Ch3.30.xls

       X1 = Number of cars produced in Michigan
       X2 = Number of cars produced in Tennessee
       X3 = Number of cars produced in Texas
       X4 = Number of cars produced in California
       X5 = Number of vans produced in Michigan
       X6 = Number of vans produced in Tennessee
       X7 = Number of vans produced in Texas
       X8 = Number of vans produced in California
       X9 = Number of buses produced in Michigan
       X10 = Number of buses produced in Tennessee
       X11 = Number of buses produced in Texas
       X12 = Number of buses produced in California
       Y1 = Number of Michigan plants producing vehicles
       Y2 = Number of Tennessee plants producing vehicles
       Y3 = Number of Texas plants producing vehicles
       Y4 = Number of California plants producing vehicles



Note: In this formulation, since only 60 total vehicles need be produced, we use 100 as
a large enough number so that if a plant is operational, there would not be a restriction
on the number of vehicles produced at the plant.

MIN 15X1 + 15X2 + 10X3 + 14X4 + 20X5 + 28X6 + 24X7 + 15X8 + 40X9 + 29X10 +
      50X11 + 25X12 + 150Y1 + 170Y2 + 125Y3 + 500Y4

S.T.   X1 + X2 + X3 + X4                      = 30      (Cars)
       X5 + X6 + X7 + X8                      = 20      (Vans)
       X9 + X10 + X11 + X12                   = 10      (Buses)
       X1 + X5 + X9          - 100Y1           0       (Michigan)
       X2 + X6 + X10         - 100Y2           0       (Tennessee)
       X3 + X7 + X11         - 100Y3           0       (Texas)
       X4 + X8 + X12         - 100Y4           0       (California)
              All X's  0 and integer
              All Y's binary




                                        Chapter 3- 55
Build 30 cars in Texas, 20 vans in Texas and 10 buses in Tennessee; total cost
$1,365,000.




                                       Chapter 3- 56
3.31   See file Ch3.31.xls   NOTE: Change Tolerance in Options Dialogue box to .5%.

       X1 = the number of Tropic homes built
       X2 = the number of Sea Breeze homes built
       X3 = the number of Orleans homes built
       X4 = the number of Grand Key homes built

       MAX 40,000X1 + 50,000X2 + 60,000X3 + 80,000X4
       S.T.   .20X1 +    .27X2 +      .22X3 + .35X4              20     (Acres)
                 X1 +       X2                                   40     (One story)
                           X2 +         X3 +      X4             50     (3+ BR)
                 X1                                              10     (Min Trop.)
                            X2                                   10     (Min Sea Br.)
                                         X3                      10     (Min Orleans)
                                                   X4            10     (Min Gr. Key)
                          All X's  0 and integer

a. Build 29 Tropic, 11 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,610,000




                                                                       Change to .5




                                     Chapter 3- 57
b. (See worksheet Atlantic Standard - Linear (not shown))
30 Tropic, 10 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,600,000. This
solution satisfies all the constraints but is $10,000 less than the optimal solution.

c. Let Yi = 1 if the constraint holds and Yi = 0 if it does not

   Add the following constraints:
   X1 - 12Y1  0
   X2 - 12Y2  0
   X3 - 12Y3  0
   X4 - 12Y4  0
   Y1 + Y2 + Y3 + Y4  3

NOTE: Excel may incorrectly print that the problem is infeasible. But the solution
below is feasible and optimal.




Build 30 Tropic, 10 Sea Breeze, 32 Orleans, 12 Grand Key models; profit = $4,580,000.
Note: There are alternate optimal solutions.




                                        Chapter 3- 58
3.32   See file Ch3.32.xls   In Options dialogue box, Change tolerance to .5% and
                             check Use Automatic Scaling

       X1 = the number of Nissan vans Logitech should purchase
       X2 = the number of Toyota vans Logitech should purchase
       X3 = the number of Plymouth vans Logitech should purchase
       X4 = the number of Ford stretch vans Logitech should purchase
       X5 = the number of Mitsubishi minibuses Logitech should purchase
       X6 = the number of General Motors minibuses Logitech should purchase
       X7 = the total number of vehicles Logitech should purchaase

MAX      7X1 +     8X2 +      9X3 +     11X4 +    20X5 +  24X6
S.T. 26000X1 + 30000X2 + 24000X3 + 32000X4 + 50000X5 + 60000X6       250,000
      5000X1 + 3500X2 + 6000X3 + 8000X4 + 7000X5 + 11000X6           50,000
          X1 +      X2 +        X3 +        X4 +    X5 +    X6 - X7 =      0
                                                                 X7       8
                                                    X5 +    X6            1
          X1 +      X2 +        X3 +        X4                            3
                               X3 +        X4         +    X6 -.5X7     0
                         All X’s  0, and integer




                                                                       Change to .5


                                                                        CHECK




                                     Chapter 3- 59
a. Maximum Capacity = 97 using 2 Plymouth vans, 1 Ford van, 1 Mitsubishi minibus, 2
   General Motors minibuses.

b.      See worksheets b-253900, b-254000, b-249900, b-259900, b-260000 (not shown)

           Capacity Nissan   Toyota Plymouth       Ford   Mitsubishi General Motors
     (i)      97                       2            1         1            1
     (ii)     98                       3            1         3
     (iii)    96                       4                      3
     (iv) 100                          4                      2            1
     (v) 100                           4                      2            1

     Sensitivity of the right hand side gives non-smooth jumps to new optimal solutions.

c. The problem is infeasible.     The minimum van and mini-bus constraints require a
minimum budget of $122,000.




                                        Chapter 3- 60
3.33    See file Ch3.33.xls

 Yj = the number of product line j eliminated

MIN 10Y1 + 8Y2 + 20Y3 + 12Y4 + 25Y5 + 4Y6 + 15Y7 + 5Y8 + 18Y9 + 6Y10 ($1000’s)

S.T.    Y1 + Y2 +      Y3 +    Y4 +     Y5 + Y6 +            Y7 + Y8 +     Y9 + Y10     4
                                                                                ( 4 eliminated)

        50(1-Y1) + 60(1-Y2) + ....                           + 125(1-Y10)  600 (Floor space)
 or,   50Y1 + 60Y2 + ...                                        + 125Y10  225 (Floor space)

               Y2 - Y3                                                     =0     (Compaq line)
                          Y4          - Y6                                 =0     (P. Bell line)
                                 Y5                   - Y8                 =0     (Apple line)

        (1-Y1) + (1-Y2) + (1-Y3) + (1-Y4) + (1-Y5)                         2     (Computers)
 or,       Y1 + Y2 + Y3 + Y4 + Y5                                          3     (Computers)

                                     (1-Y6) + (1-Y7)                       1     (Monitors)
 or,                                     Y6 + Y7                           1     (Monitors)

                                        (1-Y8) + (1-Y9) + (1-Y10)          1     (Printers)
 or,                                                Y8 + Y9 + Y10          2     (Printers)

    15000(1-Y1) + 12000(1-Y2) + ....           + 10000(1-Y10)               75000 (Restock)
 or, 15000Y1 + 12000Y2 + ...                           + 10000Y10           93000 (Restock)

                  Y1                                               - Y10    0   (Tosh/Epson)
                               All Yi’s are binary

Eliminate the Toshiba Notebook computers, Packard Bell PC’s and monitors, Apple
Macs and printers, and Epson printers -- Cost = $62,000 (Note: There are alternate
optimal solutions giving $62,000.)




                                             Chapter 3- 61
Chapter 3- 62
3.34    See file Ch3.34.xls

   a.          Xj = the number of software application j developed (j = 1, 2, 3, 4, 5, 6)

        MAX 2X1 + 3.6X2 + 4X3 + 3X4 + 4.4X5 + 6.2X6                    (in $millions)
        S.T.  6X1 + 18 X2 + 20X3 + 16X4 + 28X5 + 34X6                   60 (Programmers)
             .4X1 + 1.1X2 + .94X3 + .76X4 + 1.26X5 + 1.8X6              3.5 (Budget in $millions)
                                  All Xj’s binary




   Korvex should develop applications 1, 2, 3, and 4; net present worth = $12,600,000.


b. Add the following constraints to the formulation in part a.

                                  X4 - X5               =0     (Proj 4 = Proj 5)
               - X1 + X2                                0     (Proj 2 only if Proj 1)
                          X3          + X6  1          (Not both Proj 3 and Proj 6)
                X1 + X2 + X3 + X4 + X5 + X6             3     (Max 3 applications)

See worksheet Part b (not shown).
Korvex should develop applications 1, 2, and 6; net present worth = $11,800,000.



                                        Chapter 3- 63
3.35   See file Ch3.35.xls

       X1 = the number of CPA’s hired
       X2 = the number of experienced accountants hired
       X3 = the number of junior accountants hired

The total number of accounts that can be serviced is 6X1 + 6X2 + 4X3. This must be
greater than or equal to 100 plus the number of corporate accounts serviced: 6X1 + 6X2 +
4X3  100 + (3X1 + X2); and the number of corporate accounts serviced must be at least
25: 3X1 + X2  25. Also the number of CPA’s and experienced accountants (X1 + X2)
must be at least two-thirds of all employees hired (X1 + X2 + X3) or X1 + X2  2/3(X1 +
X2 + X3). Rearranging terms gives the following:

       MIN    1200X1 + 900X2 + 600X3
       S.T.       3X1 + 5X2 + 4X3            100 (Service  100 personal accounts)
                  3X1 +     X2               25 (Service  25 corporate accounts)
                1/3X1 + 1/3X2 - 2/3X3        0 ( 2/3 are CPA’s or experienced)
                 All X's  0 and integer

   Hire 2 CPA’s and 19 experienced accountants; total payroll = $19,500.




                                      Chapter 3- 64
3.36   See file Ch3.36.xls

       X1 = the number of TV exposures
       X2 = the number of radio exposures
       X3 = the number of newspaper exposures

a.     MAX 500,000X1 + 50,000X2 + 200,000X3
       S.T.       X1                           250          (Max TV)
                             X2                250          (Max radio)
                                           X3  250          (Max newspapers)
             4,000X1 + 500X2 + 1,000X3  500,000             (Budget)
                        All X's  0 and integer

   Use 62 TV exposures, 4 radio exposures, and 250 newspaper exposures; total
audience reached = 81,200,000.




b. Change the objective function to: MIN 4,000X1 + 500X2 + 1,000X3 and change the
third constraint to: 500,000X1 + 50,000X2 + 200,000X3  30,000,000. See worksheet
Century -- Min Cost (not shown). Use 150 newspaper exposures only; total cost $150,000.




                                     Chapter 3- 65
c. See worksheet Part c (not shown).

MAX 500,000X1 + 50,000X2 + 200,000X3
S.T. X1                                     - 250Y1                  0
                          X2                       - 250Y2           0
                                             X3            - 250Y3   0
  4,000X1 + 500X2 + 1,000X3 + 500,000Y1 + 50,000Y2 + 100,000Y3        1,000,000
                     All X's  0 and integer All Y's binary

Use 38 TV exposures, 0 radio exposures, and 248 newspaper exposures; total audience
reached = 68,600,000.




                                       Chapter 3- 66
3.37   See file Ch3.37.xls

       X1 = the number of casings produced in Springfield
       X2 = the number of casings produced in Oak Ridge
       X3 = the number of casings produced in Westchester

       Y1 = the number of Springfield locations used
       Y2 = the number of Oak Ridge locations used
       Y3 = the number of Westchester locations used

       MIN    .224X1 + .280X2 + .245X3 + 1200Y1 + 1100Y2 + 1000Y3
       S.T.       X1                  - 65000Y1                          0 (Spring.)
                          X2                    - 50000Y2                0 (Oak R.)
                                   X3                     - 55000Y3      0 (Westch.)
                  X1 +     X2 +    X3                                   = 100,000
                     All X’s  0, All Y’s binary

Produce 65,000 in Springfield and 35,000 in Westchester. Total cost = $25,335




                                     Chapter 3- 67
3.38   See file Ch3.38.xls

       X1 = Number of shares of TCS purchased
       X2 = $ invested in MFI
       X3 = Total $ invested

       MIN                  X3
       S.T.   55X1 + X2 - X3       = 0      (Total invested)
              13X1 +.09X2           250    (Minimum expected return)
              55X1      - .4X3      0      (TCS  40% of total investment)
              55X1                  750    (Max $ in TCS)
                   X1, X2  0; X1 integer

       Buy 112 shares of TCS, invest 1044.44 in MFI for a total investment of $1704.44




                                     Chapter 3- 68
3.39   See file Ch3.39.xls

       X1 = the number of Fords leased
       X2 = the number of Chevrolets leased
       X3 = the number of Dodges leased
       X4 = the number of Macks leased
       X5 = the number of Nissans leased
       X6 = the number of Toyotas leased

MIN    2000X1 + 1000X2 + 5000X3 + 9000X4 + 2000X5
S.T.       X1 +     X2 +     X3 +      X4 +      X5 +    X6     = 5,000 (Total)
           X1 +     X2 +     X3 +      X4                        3,000 (U.S.)
        500X1 + 600X2 + 300X3 + 900X4 + 200X5 + 400X6            2,750,000 (Budget)
           X1 +     X2 + .75X3 +      5X4 + .5X5 + .75X6         10,000(Payload)
                                  All X’s 0 and integer

        Lease 1581 Fords, 8 Chevrolets, 1411 Macks, 576 Nissans, 1424 Toyotas --
capital outlay = $17,021,000.




                                     Chapter 3- 69
3.40   See file Ch3.40.xls

   X1 = 1 if 7 new police officers are hired
   X2 = 1 if the police headquarters is modernized
   X3 = 1 if two new police cars are bought
   X4 = 1 if bonuses are given to foot patrolmen
   X5 = 1 if a new fire truck and fire support equipment is purchased
   X6 = 1 if an assistant fire chief is hired
   X7 = 1 if cuts to the sports program are restored
   X8 = 1 if cuts to the music program are restored
   X9 = 1 if new computers are purchased for the high school

   Y1 = 1 if the goal of spending only $650,000 is not met
   Y2 = 1 if fewer than 3 police projects are funded
   Y3 = 1 if 7 new police officers are not hired
   Y4 = 1 if less than15 new jobs are created
   Y5 = 1 if fewer than 3 education projects are funded

MAX 4176X1 + 1774X2 + 2513X3 + 1928X4 + 3607X5 +
    962X6 + 2829X7 + 1708X8 + 3003X9

S.T.  400X1 +350X2 + 50X3 + 100X4 + 500X5 + 90X6 + 220X7 + 150X8 + 140X9  900
        7X1        + X3            + 2X5 + X6 + 8X7 + 3X8 + 2X9  10
         X1 + X2 + X3 +         X4                                              3
                        X3        +      X5                                    = 1
                                                           X7 -    X8          = 0
                                                           X7         -    X9  0
                                                                   X8 -     X9  0
400X1 +350X2 + 50X3 + 100X4 + 500X5 + 90X6 + 220X7 + 150X8 + 140X9 - MY1  650
   X1 + X2 + X3 +        X4 +    X5 + X6                                -MY2  3
   X1                                                                   +MY3  1
   X1                                                                   -MY3  1
  7X1        + X3           + 2X5 + X6 + 8X7 + 3X8 + 2X9 +MY4  15
                                                   X7 +     X8      X9 +MY5  3
                                                        Y1 + Y2 + Y3 + Y4 + Y5  2
                                 All X's and Y's binary

Fund the following projects: (Total Points = 12,943)
       2 police cars
       Bonuses for foot patrolmen
       Hire an assistant fire chief
       Restore sports funding
       Restore music funding
       Purchase new computers for the high school


                                      Chapter 3- 70
Chapter 3- 71
3.41   See file Ch3.41.xls

       X1 = the number of Turkey De-Lite sandwiches made daily
       X2 = the number of Beef Boy sandwiches made daily
       X3 = the number of Hungry Ham sandwiches made daily
       X4 = the number of Club sandwiches made daily
       X5 = the number of All Meat sandwiches made daily

       MAX 2.75X1 + 3.5X2 + 3.25X3 + 4X4 + 4.25X5
       S.T.   4X1                  + 2X4 + 3X5  384 (Turkey)
                      4X2          + 2X4 + 3X5  576 (Beef)
                               4X3 + 2X4 + 3X5  480 (Ham)
               X1 + X2 + 2X3 + 2X4                 384 (Cheese)
               X1 + X2 +        X3 + X4 +      X5  300 (Rolls)
                                All X's  0




                                    Chapter 3- 72
a.     Make 52 Turkey De-Lites, 100 Beef Boys, 76 Hungry Hams, 40 Clubs, 32 All Meat

         Total Daily Revenue = $1,036
       - Daily Supplies      = $ 700
          Net Daily Profit   = $ 336         Net Annual Profit = 200($336) = $67,200


b.     Shadow price for cheese =       $0.3333; range of feasibility = 336 - 480

    As long as the price of cheese is within its range of feasibility, its shadow price will
not change.

c. Turkey: Additional 8 lbs. = 128 oz. is within its range of feasibility, so this will add:
             128(.2292) = $29.33 to revenue - $20 cost = $9.33 net additional profit
   Beef: Additional 12 lbs. = 192 oz. is within its range of feasibility, so this will add:
             192(.4167) = $80 to revenue - $42 cost = $38 net additional profit
   Ham: Additional 10 lbs. = 160 oz. is within its range of feasibility, so this will add:
             160(.2708) = $43.33 to revenue - $30 cost = $13.33 net additional profit
   Cheese: Additional 8 lbs. = 128 oz. is NOT within its range of feasibility. The
                     problem must be re-solved.

For 8 additional pounds (128 oz.) of cheese, the new optimal revenue = $1068. (See
worksheet 128 Extra Ounces of Cheese (not shown).) This is an increase of $1068- $1036
= $32 in additional revenue - $18 cost = $14 net additional profit.

       Buy the beef.


3.42   See file Ch3.42.xls

       X1 = 100’s of men’s jackets produced in the week
       X2 = 100’s of women’s jackets produced in the week
       X3 = 100’s of men’s pants produced in the week
       X4 = 100’s of women’s pants produced in the week

       MAX 2000X1 + 2800X2 + 1200X3 + 1500X4
       S.T. 150X1 + 125X2 + 200X3 + 150X4  2500                 (Denim)
               3X1 +   4X2 +     2X3 + 2X4  36                  (Cutting)
               4X1 + 3X2 +       2X3 + 2.5X4  36                (Stitching)
             .75X1 + .75X2 + .50X3 + .50X4     8                 (Boxing)
                        All X's  0

       Produce 450 women’s jackets, 900 women’s pants; weekly profit = $26,100




                                         Chapter 3- 73
b. Add to the formulation X1  5, X2  5, X3  5, X4  5; the problem is now infeasible
(See worksheet 500 (not shown).) 500 of each requires a minimum of 55 cutting hours
and 57.5 stitching hours which exceeds the limit of 36 hours each.

c. Add to the formulation X1  3, X2  3, X3  3, X4  3. (See worksheet 300 (not
shown).) Produce 300 men’s jackets, 350 women’s jackets, 300 men’s pants, and 300
women’s pants; weekly profit = $23,900.

d. Currently all items produced are women’s items. Thus adding a constraint requiring
that at least 50% of the items produced be women’s items would be a nonbinding
constraint and the solution would not change.

To add the constraint that at least 50% of the items produced be men’s items:
   Define: X5 = total number of outfits produced weekly.
   Add:
              X1 + X2 + X3 + X4 - X5 = 0
              X1 + X3 - .5X5  0.

Produce 514 men’s jackets and 514 women’s jackets (rounded) for a profit of $24,672
(see worksheet 50% Mens (not shown)).




                                       Chapter 3- 74
3.43   See file Ch3.43.xls

       X1 = Amount invested in first trust deeds
       X2 = Amount invested in second trust deeds
       X3 = Amount invested in automobile loans
       X4 = Amount invested in business loans
       X5 = Amount invested in securities
       X6 = Total amount invested in loans

MAX .09X1 + .105X2 + .1225X3 + .1175X4 + .0675X5
S.T.                                         X5      3,333,333.33 (Max sec.)
         X1 +    X2                     -    X5      0           (Trust  sec)
         X1 +    X2 +     X3 +       X4        - X6 = 0          (Total Loans)
                                     X4      -.49X6  0         (Max bus. loan)
     -.50X1 - .50X2 +     X3                         0         (Max auto loan)
         X1 +    X2 +     X3 +       X4 +     X5    = 10,000,000      (Total)
                         All X's  0




Invest $2,537,313.43 in second trust deeds, $1,268,656.72 in auto loans, $3,656,716.42 in
business loans, $2,537,313.43 in securities. Total return = $1,022,761.19.




                                      Chapter 3- 75
3.44   See file Ch3.44.xls

       X1 = the number of operations managers kept
       X2 = the number of department managers kept
       X3 = the number of section heads kept
       X4 = the number of engineers kept
       X5 = the number of technicians kept
       X6 = the number of business support personnel kept
       X7 = the number of secretaries kept
       X8 = the total number of workers kept
       X9 = total overhead
       X10 = total direct costs

MIN    1600X1 + 1200X2 + 1000X3 + 800X4 + 600X5 + 500X6 + 350X7
S.T.   X1 + X2 + X3 + X4 + X5 + X6 + X7 - X8 = 0 (X8 definition of total workers)
       X1 + X2  120 (Managers)
       X3 = 0 (Section heads)
       X1 - .2X2  0 (Operations/Department managers)
       -20X1 - 20X2 + X4 + X5 (Technician/Management)
       X7 - .05X8  0 (Min Clerical)
       X7 - .1X8  0 (Max Clerical)
       X6 - .01X8  0 (Min Administration)
       X6 - .02X8  0 (Max Administration)
       1280X1 + 840X2 + 350X6 + 245X7 -X9 = 0 (X9 definition of total overhead)
       320X1 + 360X2 + 800X3 + 800X4 + 600X5 + 150X6 + 105X7 - X10 = 0
                                            (X10 definition of total direct costs)
       X9 - .05X10  0 (Min overhead)
       X9 - .1X10  0 (Max overhead)
       X10 = 4800000 (Fixed direct costs)
       X1  6 (Min operations managers)
       360X2 -80X3 - 60X5  0 (Department managers/ Technicians)
       X4 - 4X5  0 (Engineers/Technicians)
       X1 + X2 + X3 + X4 - .5104X8  0 (Min grade 100 personnel)
       X1 + X2 + X3 + X4 - .7656X8  0 (Max grade 100 personnel)
       X5 - .2145X8  0 (Min technicians)
       X5 - .3217X8  0 (Max technicians)
       X6 - .0107X8  0 (Min support)
       X6 - .0161X8  0 (Max support)
       X7 - .0643X8  0 (Min secretaries)
       X7 - .0965X8  0 (Max secretaries)
       X1  40        (Current operations managers)
       X2  200       (Current department managers)
       X3  900       (Current section heads)
       X4  6000      (Current engineers)


                                      Chapter 3- 76
       X5  3000     (Current technicians)
       X6  150      (Current business support personnel)
       X7  900      (Current secretaries)
              All X’s  0, X1, X2, X3, X4, X5, X6 X7 integer




Keep 9 operations managers, 109 department managers, 0 section heads, 4764 engineers,
1480 technicians, 79 business support staff, and 446 secretaries;
total weekly salary = $5,040,000.




                                      Chapter 3- 77
3.45     See file Ch3.45.xls

         X1 = the number of assistant professor positions recruited
         X2 = the number of associate professor positions recruited
         X3 = the number of full professor positions recruited
         X4 = the total number of positions recruited

         MAX 2X1 +         7X2 +    14X3
         S.T. X1 +          X2 +      X3 - X4 = 0             (Definition of X4)
                                              X4  20                (Max recruiting)
                 X1                       - .5X4  0                 (Min. assistant prof.)
                 X1 +          X2         - .7X4  0                 (Min. below full
prof.)
           55003X1 + 69885X2 + 93471X3         1,275,000 (Total salaries)
                      All X's  0 and integer

Hire 9 assistant professors, 4 associate professors, and 5 full professors
-- 116 total years of experience




                                        Chapter 3- 78
3.46   See file Ch3.46.xls
       X1 = the number of guards whose shift begins at 12:00 midnight
       X2 = the number of guards whose shift begins at 3:00AM
       X3 = the number of guards whose shift begins at 6:00AM
       X4 = the number of guards whose shift begins at 9:00AM
       X5 = the number of guards whose shift begins at 12:00 noon
       X6 = the number of guards whose shift begins at 3:00PM
       X7 = the number of guards whose shift begins at 6:00PM
       X8 = the number of guards whose shift begins at 9:00PM

MIN    X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8
S.T.   X1                          + X7 + X8  5 (mid - 1AM)             Redundant
       X1                          + X7 + X8  5 (1AM - 2AM)             Redundant
       X1                                + X8  5 (2AM - 3AM)
       X1 + X2                           + X8  5 (3AM - 4AM)            Redundant
       X1 + X2                           + X8  5 (4AM - 5AM)            Redundant
       X1 + X2                                  8 (5AM - 6AM)
       X1 + X2 + X3                             8 (6AM - 7AM)           Redundant
       X1 + X2 + X3                             12 (7AM - 8AM)          Redundant
            X2 + X3                             12 (8AM - 9AM)
            X2 + X3 + X4                        10 (9AM-10AM)           Redundant
            X2 + X3 + X4                        10 (10AM-11AM)          Redundant
                 X3 + X4                        15 (11AM - noon)
                 X3 + X4 + X5                   15 (noon - 1PM)         Redundant
                 X3 + X4 + X5                   15 (1PM - 2PM)          Redundant
                      X4 + X5                   9 (2PM - 3PM)
                      X4 + X5 + X6              9 (3PM - 4PM)           Redundant
                      X4 + X5 + X6              12 (4PM - 5PM)          Redundant
                           X5 + X6              12 (5PM - 6PM)
                           X5 + X6 + X7         12 (6PM - 7PM)          Redundant
                           X5 + X6 + X7         7 (7PM - 8PM)           Redundant
                                X6 + X7         7 (8PM - 9PM)
                                X6 + X7 +X8  7 (9PM -10PM)              Redundant
                                X6 + X7 +X8  7 (10PM -11PM)             Redundant
                                      X7 +X8  7 (11PM - mid)
                           All X’s  0 and integer

 a.     There are many solutions that give 42 total guards. This can be seen by the
        numerous adjustable cells with Allowable Increases or Decreases of 0. The
        screen on the next page shows one of them.




                                     Chapter 3- 79
b. Shadow prices give an increase in the number of officers needed per time period
given no additional changes in the constraints.

(i) Changing the number of guards required from midnight to 5AM to 7 changes the right
hand side of the first non-redundant constraint from 5 to 7. But there was 3 slack and
thus this does not require more guards.

(ii) Changing the number of officers required from 9AM to 11AM to 12 changes nothing;
X3 + X4 + X5  12 would still be a redundant constraint.




                                      Chapter 3- 80
(iii) Changing the number of officers required from 11AM to 2 PM changes the right
hand side of the fourth non-redundant constraint to 17. Re-solving (see worksheet
11AM-2PM (not shown)) changes the total guards required to 44.

(c) If the objective function coefficients had been expressed in terms of dollars, each
dollar coefficient would have been the same, say $K. From the range of optimality for
the objective function coefficients for these two shifts, we see that their ranges of
optimality would for at least one of them in each solution would be from $K to $K, i.e.
any change will change the optimal solution. Thus a change of $5 will change the
optimal solution.

(d) We do not need to do this since integer values are already obtained. If we do, we get
the same answer. (See worksheet Guardsman Services-Integer (not shown).) The shadow
prices and ranges of optimality do make sense because they are integers.




                                       Chapter 3- 81
3.47    See file Ch3.47.xls Change tolerance to .5% in Options dialogue box.

        X1 = the number of Liltrykes produced per year
        X2 = the number of Pinktrykes produced per year
        X3 = the number of Herotrykes produced per year
        X4 = the number of Robinhoods produced per year
        X5 = the number of Jeeptrykes produced per year
        X6 = the number of Monsters produced per year
        Y1 = the number of setups of Liltrykes each year
        Y2 = the number of setups of Pinktrykes each year
        Y3 = the number of setups of Herotrykes each year
        Y4 = the number of setups of Robinhoods each year
        Y5 = the number of setups of Jeeptrykes each year
        Y6 = the number of setups of Monsters each year
a.
     MAX 1.50X1 + 2.00X2 + 2.25X3 + 2.75X4 + 3.00X5 + 3.50X6
               -16,500Y1 - 18,000Y2 - 17,500Y3 - 18,000Y4 - 20,000Y5 - 17,000Y6
     S.T.
          3X1 + X2 + 2X3 + 2X4 + 2X5                      120,000 (Small wheels/year)
               2X2 + X3 + X4 + X5 + 3X6                   96,000 (Big wheels/year)
        .8X1+1.2X2 +1.5X3 +2.1X4 +1.8X5 + 3X6             108,000 (Plastic/year)
           X1                                 -1000000Y1        0
               X2                             -1000000Y2        0
                    X3                        -1000000Y3        0
                          X4                  -1000000Y4        0
                               X5           - 1000000Y5         0
                                     X6       -1000000Y6        0
                      All X’s  0, All Y’s binary




                                      Chapter 3- 82
a.      Produce 60,000 Jeeptrykes only; yearly profit = $160,000

b. In addition to the variables already defined, define Zj = 0 if goal j is met, Zj = 1 if it is not.
   Add the following.

Goal (1):       Max $70,000 for new setups
Goal (2):       If Herotryke is produced, Robinhood will not be produced
Goal (3):       At least four new models produced
Goal (4):       If Jeeptrykes are produced, Herotrykes will also be produced
Goal (5):       At least 1500 lbs. per month of plastic should be left over (at most 7500 pounds
                per month or 90,000 pounds per year used)

16500Y1 + 18000Y2 + 17500Y3 + 18000Y4 + 20000Y5 + 17000Y6 -1000000Z1  75000 (1)
                            Y3 -       Y4                -1000000Z2     1   (2)
      Y1 +       Y2 +       Y3 +       Y4 +  Y5 +      Y6 -1000000Z3     4   (3)
                                             Y5 -      Y6 -1000000Z4     0   (4)
.8X1 + 1.2X2 + 1.5X3 + 2.1X4 + 1.8X5 + 3X6                -1000000Z5  90000 (5)

and, at least 4 out of 5 goals should be satisfied: (1-Z1)+(1- Z2)+(1- Z3)+(1-Z4)+(1-Z5)  4 or,
Z1 + Z2 + Z3 + Z4 + Z5                                                              1
                                       Z’s are binary
See worksheet Little Trykes -- Part b (not shown). The solution remains the same. All
goals except goal 3 are satisfied.


                                          Chapter 3- 83
3.48   See file Ch3.48.xls

       X1 = the number of Alpha car projects implemented
       X2 = the number of Beta car projects implemented
       X3 = the number of Delta car projects implemented
       X4 = the number of Gamma car projects implemented
       X5 = the number of Kappa car projects implemented
       X6 = the number of Sigma car projects implemented

MAX 12X1 +      11X2 +     9X3 +    15X4 +     7X5 +   20X6
S.T.     X1 +     X2 +      X3 +       X4 +     X5 +     X6     3    Proj.
   15000X1 + 18000X2 + 19000X3 + 20000X4 + 8000X5 + 22000X6  125000 eng-yr1
     5000X1 + 5000X2 + 3000X3 + 4000X4 + 2000X5 + 5000X6  16000 staff-yr1
   40000X1 + 25000X2 + 19000X3 + 25000X4 + 12000X5 + 27000X6  150000 eng-yr 2
     5000X1 + 4000X2 + 3000X3 + 6000X4 + 2000X5 + 7000X6  24000 staff-yr 2
   40000X1 + 30000X2 + 19000X3 + 30000X4 + 18000X5 + 32000X6  187500 eng-yr 3
     8000X1 + 7000X2 + 3000X3 + 7500X4 + 3000X5 + 8000X6  40000 staff-yr 3
         X1 +     X2                                          1 (if alpha, no beta)
         X1                                        -      X6  0 (if sigma, alpha)
                                All X's binary




Develop Alpha, Gamma, Kappa, Sigma; total present net worth = $54 million.


                                    Chapter 3- 84
3.49   See file Ch3.49.xls            Data Envelopment Analysis

       X1 = Relative input weight applied to the campus SAT score
       X2 = Relative input weight applied to the campus faculty/student ratio
       X3 = Relative input weight applied to the campus budget
       Y1 = Relative output weight applied to the campus average GPA score
       Y2 = Relative output weight applied to the campus graduation rate
       Y3 = Relative output weight applied to the campus percent employment

For Cody
      MAX 2.63Y1 + .43Y2 + .78Y3
      S.T. 2.63Y1 + .43Y2 + .78Y3  920X1 + .068X2 + 20.3X3
           2.57Y1 + .55Y2 + .79Y3  960X1 + .059X2 + 24.6X3
           2.81Y1 + .54Y2 + .83Y3  1000X1 + .061X2 + 35.2X3
           920X1 + .068X2 + 20.3X3 = 1
                  All X's and Y's  0




              Cody efficiency rating = 1; it is efficient.

For Casper, the objective function is MAX 2.57Y1 + .55Y2 + .79Y3 and the last constraint
is 960X1 + .059X2 + 24.6X3 = 1. See Worksheet Casper. Casper is also efficient.

For Cheyenne, the objective function is MAX 2.81Y1 + .54Y2 + .83Y3 and the last constraint
is 1000X1 + .061X2 + 35.2X3 = 1. See Worksheet Cheyenne. Cheyenne is also efficient.


                                        Chapter 3- 85
3.50   See file Ch3.50.xls Data Envelopment Analysis
       X1 = Relative input weight applied to income
       X2 = Relative input weight applied to number of employees
       X3 = Relative input weight applied to size
       Y1 = Relative output weight applied to men's clothing sales
       Y2 = Relative output weight applied to women's clothing sales
       Y3 = Relative output weight applied to cosmetics sales
       Y4 = Relative output weight applied to jewelry sales

For Discount Store 1

MAX     9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4
S.T.    9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4  32987X1 + 275X2 + 23876X3
       10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4  32987X1 + 215X2 + 28755X3
       14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4  54321X1 + 185X2 + 19000X3
       15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4  54321X1 + 180X2 + 18750X3
       12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4  99765X1 + 85X2 + 11000X3
       32987X1 + 275X2 + 23876X3 = 1
                            All X's, Y's  0




Discount Store 1 is DEA efficient.




                                      Chapter 3- 86
For Discount Store 2
MAX 10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4
S.T.    9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4  32987X1 + 275X2 + 23876X3
       10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4  32987X1 + 215X2 + 28755X3
       14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4  54321X1 + 185X2 + 19000X3
       15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4  54321X1 + 180X2 + 18750X3
       12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4  99765X1 + 85X2 + 11000X3
       32987X1 + 215X2 + 28755X3 = 1
                                   All X's, Y's  0
From worksheet DISC 2 (not shown), discount store 2 is also efficient.

For Department Store 1
MAX 14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4
S.T.   9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4  32987X1 + 275X2 + 23876X3
      10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4  32987X1 + 215X2 + 28755X3
      14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4  54321X1 + 185X2 + 19000X3
      15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4  54321X1 + 180X2 + 18750X3
      12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4  99765X1 + 85X2 + 11000X3
      54321X1 + 185X2 + 19000X3 = 1
                                  All X's, Y's  0
From worksheet DEPT 1 (not shown), department store 1 is also efficient.

For Department Store 2
MAX 15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4
S.T.   9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4  32987X1 + 275X2 + 23876X3
      10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4  32987X1 + 215X2 + 28755X3
      14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4  54321X1 + 185X2 + 19000X3
      15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4  54321X1 + 180X2 + 18750X3
      12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4  99765X1 + 85X2 + 11000X3
      54321X1 + 180X2 + 18750X3 = 1
                                  All X's, Y's  0
From worksheet DEPT 2 (not shown), department store 2 is also efficient.


For the Upscale Store
MAX 12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4
S.T.     9.6Y1 + 20.3Y2 + 16.3Y3 + 4.1Y4  32987X1 + 275X2 + 23876X3
        10.3Y1 + 17.9Y2 + 15.5Y3 + 4.6Y4  32987X1 + 215X2 + 28755X3
        14.5Y1 + 55.2Y2 + 27.4Y3 + 22.3Y4  54321X1 + 185X2 + 19000X3
        15.2Y1 + 44.8Y2 + 26.8Y3 + 28.4Y4  54321X1 + 180X2 + 18750X3
        12.5Y1 + 45.9Y2 + 19.9Y3 + 35.1Y4  99765X1 + 85X2 + 11000X3
        99765X1 + 85X2 + 11000X3 = 1
                                    All X's, Y's  0
From worksheet Upscale (not shown), the upscale store 2 is also efficient.


                                     Chapter 3- 87

				
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