Linear Programming Models Graphical and Computer Methods by ols15776

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									              Linear Programming Models: Graphical and
                                                        C H A P T E R
                                                                     7
                          Computer Methods


TEACHING SUGGESTIONS                                                        model). Here, the issue is the source of data. When accountants tell
                                                                            you a profit contribution is $8.50 per unit, is that figure accurate
Teaching Suggestion 7.1: Draw Constraints for a
                                                                            within 10% or within 10¢? The solution to an LP problem can
Graphical LP Solution.
                                                                            change dramatically if the input parameters are not exact. Mention
Explain constraints of the three types ( , , ) carefully the first
                                                                            that sensitivity analysis also has other names, such as right-hand-
time you present an example. Show how to find the X1, X2 inter-
                                                                            side ranging, post-optimality analysis, and parametric programming.
cepts so a straight line can be drawn. Then provide some practice
in determining which way the constraints point. This can be done
by picking a few X1, X2 coordinates at random and indicating                ALTERNATIVE EXAMPLES
which direction fulfills the constraints.                                    Alternative Example 7.1: Hal has enough clay to make 24
Teaching Suggestion 7.2: Feasible Region Is a Convex Polygon.               small vases or 6 large vases. He only has enough of a special glaz-
Explain Dantzing’s discovery that all feasible regions are convex           ing compound to glaze 16 of the small vases or 8 of the large
(bulge outward) polygons (many-sided figures) and that the opti-             vases. Let X1 the number of small vases and X2 the number of
mal solution must lie at one of the corner points. Draw both con-           large vases. The smaller vases sell for $3 each, while the larger
vex and concave figures to show the difference.                              vases would bring $9 each.
Teaching Suggestion 7.3: Using the Iso-Profit Line Method.                                                (a) Formulate the problem.
This method can be much more confusing than the corner point ap-                                         (b) Solve graphically.
proach, but it is faster once students feel comfortable drawing the         SOLUTION:
profit line. Start your first line at a profit figure you know is lower                                      (a)        Formulation
than optimal. Then draw a series of parallel lines, or run a ruler paral-
                                                                            OBJECTIVE FUNCTION:
lel, until the furthest corner point is reached. See Figures 7.6 and 7.7.
                                                                                                          Maximize $3X1               $9X2
Teaching Suggestion 7.4: QA in Action Boxes in the LP Chapters.
There are a wealth of motivating tales of real-world LP applica-                                          Subject to: Clay constraint: 1X1              4X2   24
tions in Chapters 7–9. The airline industry in particular is a major                                      Glaze constraint: 1X1              2X2   16
LP user.                                                                                                  (b)        Graphical solution
Teaching Suggestion 7.5: Feasible Region for the
Minimization Problem.                                                                                    15
Students often question the open area to the right of the constraints
in a minimization problem such as that in Figure 7.10. You need
to explain that the area is not unbounded to the right in a mini-
mization problem as it is in a maximization problem.
                                                                            X2 = Number of Large Vases




Teaching Suggestion 7.6: Infeasibility.
                                                                                                         10
This problem is especially common in large LP formulations since
many people will be providing input constraints to the problem.                                               (0, 8)
This is a real-world problem that should be expected.                                                                             Glaze Constraint
Teaching Suggestion 7.7: Alternative Optimal Solutions.                                                       (0, 6)
This issue is an important one that can be explained in a positive                                       B
way. Managers appreciate having choices of decisions that can be                                         5                         (8, 4)                Clay Constraint
made with no penalty. Students can be made aware that alternative
                                                                                                                                  C
optimal solutions will arise again in the transportation model, as-
signment model, integer programming, and the chapter on net-                                                        Feasible Region
work models.                                                                                                        A                                                (24, 0)
Teaching Suggestion 7.8: Importance of Sensitivity Analysis.                                                      (0, 0)                      D (16, 0)
                                                                                                         0
Sensitivity analysis should be stressed as one of the most important                                          0            5         10        15        20                25
LP issues. (Actually, the issue should arise for discussion with every                                                         X1 = Number of Small Vases

                                                                                                                                                                               85
86                                     CHAPTER 7         LINEAR PROGRAMMING MODELS: GRAPHICAL       AND   COMPUTER METHODS



  Point                           X1             X2             Income            plied to minimization problems. Conceptually, isoprofit and iso-
                                                                                  cost are the same.
                   A               0              0                 $0
                                                                                       The major differences between minimization and maximiza-
                   B               0              6                 54
                                                                                  tion problems deal with the shape of the feasible region and the di-
                   C               8              4                 60*
                   D              16              0                 48            rection of optimality. In minimization problems, the region must
                                                                                  be bounded on the lower left, and the best isocost line is the one
*Optimum income of $60 will occur by making and sell-                             closest to the zero origin. The region may be unbounded on the top
ing 8 small vases and 4 large vases.                                              and right and yet be correctly formulated. A maximization prob-
Draw an isoprofit line on the graph on page 85 from (20, 0) to                     lem must be bounded on the top and to the right. The isoprofit line
(0, 6X\c) as the $60 isoprofit line.                                               yielding maximum profit is the one farthest from the zero origin.
Alternative Example 7.2: A fabric firm has received an order                         7-2. The requirements for an LP problem are listed in Section
for cloth specified to contain at least 45 pounds of cotton and 25                 7.2. It is also assumed that conditions of certainty exist; that is, co-
pounds of silk. The cloth can be woven out on any suitable mix of                 efficients in the objective function and constraints are known with
two yarns, A and B. Material A costs $3 per pound, and B costs $2                 certainty and do not change during the period being studied. An-
per pound. They contain the following proportions of cotton and                   other basic assumption that mathematically sophisticated students
silk (by weight):                                                                 should be made aware of is proportionality in the objective func-
                                                                                  tion and constraints. For example, if one product uses 5 hours of a
  Yarn                           Cotton (%)              Silk (%)                 machine resource, then making 10 of that product uses 50 hours of
                                                                                  machine time.
                   A                   30                  50
                                                                                        LP also assumes additivity. This means that the total of all ac-
                   B                   60                  10
                                                                                  tivities equals the sum of each individual activity. For example, if
                                                                                  the objective function is to maximize P 6X1 4X2, and if X1
What quantities (pounds) of A and B yarns should be used to mini-
                                                                                  X2 1, the profit contributions of 6 and 4 must add up to produce
mize the cost of this order?
                                                                                  a sum of 10.
                       Objective function: min. C     3A     2B
                                                                                   7-3. Each LP problem that has been formulated correctly does
                       Constraints:    0.30A     0.60B     45 lb (cotton)         have an infinite number of solutions. Only one of the points in the
                                       0.50A     0.10B     25 lb (silk)           feasible region usually yields the optimal solution, but all of the
Simultaneous solution of the two constraint equations reveals that                points yield a feasible solution. If we consider the region to be
A 39 lb, B 55 lb.                                                                 continuous and accept noninteger solutions as valid, there will be
    The minimum cost is C $3A $2B 3(39) (2)(55)                                   an infinite number of feasible combinations of X1 and X2.
$227.                                                                              7-4. If a maximization problem has many constraints, then it can
                                                                                  be very time consuming to use the corner point method to solve it.
                   300                                                            Such an approach would involve using simultaneous equations to
                                                                                  solve for each of the feasible region’s intersection points. The iso-
                                                                                  profit line is much more effective if the problem has numerous con-
                   250                                                            straints.
                                                                                    7-5. A problem can have alternative optimal solutions if the
                                                                                  isoprofit or isocost line runs parallel to one of the problem’s con-
                   200
Pounds of Yarn B




                                                                                  straint lines (refer to Section 7-8 in the chapter).
                                                                                    7-6. This question involves the student using a little originality
                   150                                                            to develop his or her own LP constraints that fit the three condi-
                                                                                  tions of (1) unboundedness, (2) infeasibility, and (3) redundancy.
                                                                                  These conditions are discussed in Section 7.8, but each student’s
                   100                                                            graphical displays should be different.
                                       min C                                        7-7. The manager’s statement indeed had merit if the manager
                       50                                                         understood the deterministic nature of linear programming input
                                                                                  data. LP assumes that data pertaining to demand, supply, materi-
                                                                                  als, costs, and resources are known with certainty and are constant
                                                                                  during the time period being analyzed. If this production manager
                                 50         100    150       200          250     operates in a very unstable environment (for example, prices and
                                              Pounds of Yarn A                    availability of raw materials change daily, or even hourly), the
                                                                                  model’s results may be too sensitive and volatile to be trusted. The
SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS                                    application of sensitivity analysis might be trusted. The applica-
 7-1. Both minimization and maximization LP problems employ                       tion of sensitivity analysis might be useful to determine whether
the basic approach of developing a feasible solution region by                    LP would still be a good approximating tool in decision making.
graphing each of the constraint lines. They can also both be solved                7-8. The objective function is not linear because it contains the
by applying the corner point method. The isoprofit line method is                  product of X1 and X2, making it a second-degree term. The first,
used for maximization problems, whereas the isocost line is ap-                   second, fourth, and sixth constraints are okay as is. The third and
                                            CHAPTER 7         LINEAR PROGRAMMING MODELS: GRAPHICAL             AND    COMPUTER METHODS                                             87


fifth constraints are nonlinear because they contain terms to the                                    Let: X1      number of air conditioners to be produced
second degree and one-half degree, respectively.                                                        X2       number of fans to be produced
  7-9. For a discussion of the role and importance of sensitivity                                   Maximize profit              25X1         15X2
analysis in linear programming, refer to Section 7.9. It is needed
                                                                                                    subject to 3X1               2X2     240 (wiring)
especially when values of the technological coefficients and con-
tribution rates are estimated—a common situation. When all                                                           2X1         1X2     140 (drilling)
model values are deterministic, that is, known with certainty, sen-                                                         X1, X2       0
sitivity analysis from the perspective of evaluating parameter ac-
                                                                                                               Profit at point a (X1                 0, X2      0)     $0
curacy may not be needed. This may be the case in a portfolio se-
lection model in which we select from among a series of bonds                                                  Profit at point b (X1 0, X2 120)
whose returns and cash-in values are set for long periods.                                                                      25(0)  (15)(120)                          $1,800
7-10. Sensitivity analysis is important in all quantitative model-                                             Profit at point c (X1                40, X2          60)
ing techniques. Especially common is the analysis of inventory                                                                          25(40)         (15)(60)           $1,900
model results in which we test the model’s sensitivity to changes
                                                                                                               Profit at point d (X1                70, X2      0)
in demand, lead time, cost, and so on.
                                                                                                                                         25(70)             (15)(0)       $1,750
7-11. The computer is valuable in (1) solving LP problems
quickly and accurately; (2) solving large problems that might take                      The optimal solution is to produce 40 air conditioners and 60 fans
days or months by hand; (3) performing extensive sensitivity                            during each production period. Profit will be $1,900.
analysis automatically; and (4) allowing a manager to try several                       7-15.
ideas, models, or data sets.
                                                                                             140
7-12. The student is to create his or her own data and LP formula-
tion. (a) The meaning of the right-hand-side numbers (resources) is
to be explained. (b) The meaning of the constraint coefficient (in                            120
terms of how many units of each resource that each product re-
quires) is also to be explained. (c) The problem is to be solved
graphically. (d) A simple sensitivity analysis is to be conducted by                         100
changing the contribution rate (Cj value) of the X1 variable. For ex-                                                       (26.67, 80)
ample, if C1 was $10 as the problem was originally formulated, the
                                                                                                           b
student should resolve with a $15 value and compare solutions.                               80
                                                                                                                 e
7-13. A change in a technological coefficient changes the feasi-                         X2
ble solution region. An increase means that each unit produced re-                                                          c           Optimal Solution
quires more of a scarce resource (and may lower the optimal                                  60
profit). A decrease means that because of a technological advance-
ment or other reason, less of a resource is needed to produce 1
                                                                                             40
unit. Changes in resource availability also change the feasible re-
gion shape and can increase or decrease profit.
                                                                                                                     Feasible
7-14.                                                                                        20                      Region

                     140
                                                                                                           a                                    d
                                                                                                0
                           b                                                                        0       20             40           60             80           100      120
                     120                                                                                                                X1

                                               Drilling Constraint                                  Maximize profit              25X1         15X2
                     100
                                                                                                    subject to             3X1         2X2      240
Number of Fans, X2




                                                                                                                           2X1         1X2      140
                     80                                Optimal Solution                                                     X1                  20
                                                      (X1 = 40, X2 = 60)                                                                X2      80
                                                  c                                                                               X1, X2        0
                     60
                                                                                                    Profit at point a (X1              20, X2         0)
                                                                                                                                             25(20)          (15)(0)       $500
                     40
                                                                     Wiring                         Profit at point b (X1              20, X2         80)
                                   Feasible Region                   Constraint
                                                                                                                                        25(20)         (15)(80)           $1,700
                     20


                               a                              d
                      0
                           0           20      40        60       80       100    120
                                            Number of Air Conditioners, X1
88                        CHAPTER 7              LINEAR PROGRAMMING MODELS: GRAPHICAL                           AND    COMPUTER METHODS


            Profit at point c (X1        40, X2         60)                                 300
                                         25(40)         (15)(60)            $1,900
            Profit at point d (X1        70, X2         0)
                                                                                            250
                                         25(70)         (15)(0)             $1,750
            Profit at point e (X1        26.67, X2           80)
                                         25(26.67)            15(80)        $1,867          200
Hence, even though the shape of the feasible region changed from
Problem 7-14, the optimal solution remains the same.
7-16.                                                                                    X2 150

     300
                                                                                                     a
                                                                                            100
                                                                                                                                              Optimal Solution,
     250                                                                                                                                       $2862.50 Profit
                                                                                             50
             a                                                                                            Feasible Region                             b
     200

                                                                                                 0                                                          c
                                                                                                     0     50     100       150      200       250        300   350      400
X2 150                      b                                                                                                        X1

                                                                                         7-18.
     100
                                                                                            60
                                                                                                                  X1 + X2 = 60
                    Feasible
     50             Region
                                                                                            50

                                                                                                                                                      Feasible
                                                   c                                                                                                  Region
        0                                                                                   40
            0       50      100      150        200          250      300      350
                                           X1
                                                                                         X2 30                                       b
                                                                                                                X2 = 20
            X1      number of model A tubs produced
            X2      number of model B tubs produced                                         20
                                                                                                                                                  a                Optimal
            Maximize profit          90X1     70X2
                                                                                                                                                                   Solution
            subject to 125X1          100X2         25,000 (steel)                                                    X1 = 30
                                                                                            10
                             20X1        30X2       6,000 (zinc)
                                      X1, X2        0
                     Profit at point a (X1         0, X2        200)     $14,000              0
                                                                                                 0          10          20            30          40            50            60
            Profit at point b (X1      85.71, X2           142.86)       $17,714.10                                                    X1
                                                                                     a




                     Profit at point c (X1         200, X2 = 0)          $18,000
                                                                                                     X1     number of undergraduate courses
                                                              optimal solution
                                                                                                     X2     number of graduate courses
7-17.       X1      number of benches produced
                                                                                                     Minimize cost          $2,500X1          $3,000X2
            X2      number of tables produced
            Maximize profit          $9X1     $20X2                                                   subject to        X1                30
                                                                                                                                X2       20
            subject to     4X1       6X2     1,200 hours
                                                                                                                       X1       X2       60
                         10X1       35X2     3,500 pounds
                                                                                                     Total cost at point a           (X1       40, X2        20)
                                  X1, X2     0
                                                                                                                                     2,500(40)            (3,000)(20)
                 Profit at point a (X1      0, X2         100)        $2,000
                                                                                                                                     $160,000
            Profit at point b (X1        262.5, X2           25)      $2,862.50
                 Profit at point c (X1      300, X2           0)      $2,700
                               CHAPTER 7                     LINEAR PROGRAMMING MODELS: GRAPHICAL                       AND   COMPUTER METHODS                                      89


            Total cost at point b                  (X1        30, X2      30)                       60
                                                   2,500(30)           (3,000)(30)
                                                   $165,000
            Point a is optimal.
7-19.                                                                                                                                      a
                                                                                                    40
   40

                                                                                               X2                                         b                    Feasible
                                                                                                                                                               Region
                                                                                                                              Optimal
                                                                                                                              Solution
   30
                                                                                                    20
                                            Optimal Solution

                              a                       Feasible Region is
                                                     Heavily Shaded Line
X2 20

                                                                                                       0                                                                     c
                                                      b                                                    0                    20                        40                     60
                                                                                                                                               X1
   10
                                                                                               7-21.

                                                                                                    250,000
                                                                                                                           X1 + X2 = 100,000
        0
            0                  10                    20                 30             40
                                                     X1                                             200,000
                                                                                                                                                          X1 = 125,000

                X1     number of Alpha 4 computers
                X2     number of Beta 5 computers                                                   150,000
            Maximize profit                    $1,200X1         $1,800X2                        X2                         X2 = 100,000

            subject to         20X1           25X2        800 hours                                                                                       a
                                                                                                    100,000
                                                          (total hours 5 workers                                                                                     Feasible
                                                              160 hours each)                                                                                        Region
                                                                                                                                                                   is this Line
                                       X1     10
                                                                                                     50,000
                                       X2     15
            Corner points:              a(X1         10, X2       24), profit        $55,200
                                        b(X1             1
                                                     21 , X2         15), profit     $52,500                                                                      b
                                                         4                                                     0
                                                                                                                   0       50,000     100,000 150,000 200,000 250,000
            Point a is optimal.
                                                                                                                                             X1
7-20.       Let: X1           number of pounds of compost in each bag
                                                                                                           X1          $ invested in Treasury notes
                      X2      number of pounds of sewage waste in each bag
                                                                                                           X2          $ invested in bonds
            Minimize cost                   5X1    4X2 (in cents)
                                                                                                           Maximize ROI             0.08X1          0.09X2
            subject to X1                    X2    60 (pounds per bag)
                                                                                                                                 X1                 $125,000
                                  X1               30 (pounds compost per bag)
                                                                                                                                         X2         $100,000
                                             X2    40 (pounds sewage per bag)
                                                                                                                                 X1      X2         $250,000
            Corner point a:
                                                                                                                                      X1, X2        0
            (X1        30, X2           40) ⇒ cost           5(30)      (4)(40)      $3.10
                                                                                                                                                                                      a




                                                                                                           Point a (X1        150,000, X2           100,000), ROI         $21,000
            Corner point b:
                                                                                                                                                                optimal solution
            (X1        30, X2           30) ⇒ cost           5(30)      (4)(30)      $2.70
                                                                                                           Point b (X1        250,000, X2           0),         ROI       $20,000
            Corner point c:
                (X1        60, X2           0) ⇒ cost        5(60)      (4)(0)     $3.00
90                        CHAPTER 7                 LINEAR PROGRAMMING MODELS: GRAPHICAL                                        AND   COMPUTER METHODS


7-22.                                                                                                      7-23.    Point a lies at intersection of constraints (see figure below):
                                                                                                               3X1        2X2     120
                                                                                                                   X1     3X2     90
     50
                                                                                                           Multiply the second equation by 3 and add it to the first (the
                     5X1 + 3X2 ≤ 150                                                                       method of simultaneous equations):
                                                                                                                    3X1    2X2    120
     40
                           Isoprofit Line Indicates                                                                 3X1    9X2         270
                            that Optimal Solution                                                                          7X2         150 ⇒ X2   21.43 and X1       25.71
                               Lies at Point a
                                                                                                                   Cost    $1X1        $2X2   $1(25.71)     ($2)(21.43)
     30
                                         3           3                                                                     $68.57
X2                              (X1 = 18 4 , X2 = 18 4 , Profit = $150)
                                                                                                           7-24.     X1     $ invested in Louisiana Gas and Power
     20                                                      X1 – 2X2 ≤ 10
                                                                                                                     X2     $ invested in Trimex Insulation Co.
                                    a
                                                                                                                    Minimize total investment      X1     X2
                  Feasible                                                                                          subject to    $0.36X1     $0.24X2     $720
                  Region                                             3X1 + 5X2 ≤ 150
     10                                                                                                                           $1.67X1     $1.50X2     $5,000
                                                                                                                                  0.04X1 0.08X2           $200
                                                                                                                    Investment at a is $3,333.
        0                                                                                                           Investment at b is $3,179. k optimal solution
            0        10             20                     30              40                 50
                                               X1                                                                   Investment at c is $5,000.
                                                                                                                    Short-term growth is $927.27.
                                                                                                                    Intermediate-term growth is $5,000.
Note that this problem has one constraint with a negative sign.
This may cause the beginning student some confusion in plotting                                                     Dividends are $200.
the line.                                                                                                           See graph on page 91.


Figure for Problem 7-23.
     80
                  8X1 + 2X2 ≥ 160                                                                       X2 ≤ 70



     60


                                                                     Feasible Region


X2 40
                                         Iso
                                            cos
                                                    t Li
                                                        ne
                                                                =$
                                                                  100
                                                                          =1
                                                                               X
                                                                               1   +2
     20                         a                                                       X
                                                                                          2


                                                                                                        X1 + 3X2 ≥ 90
                3X1 + 2X2 ≥ 120

     0
            0              20                        40                            60                 80                  100
                                                                     X1                 Isoprofit Line Indicates
                                                                                         That Optimal Solution
                                                                                            Lies At Point a
                                   CHAPTER 7              LINEAR PROGRAMMING MODELS: GRAPHICAL                    AND   COMPUTER METHODS                                  91


Figure for Problem 7-24.
   4,000



                        a(X1 = 0, X2 = 3,333)

   3,000                                                                Feasible Region




X2 2,000                                            b(X1 = 1,359, X2 = 1,818.18)




   1,000
                                                                                            c (X1 = 5,000, X2 = 0)




                0
                    0                  1,000               2,000                  3,000       4,000            5,000
                                                                        X1


7-25.                                                                                                   Point c (X1      25, X2       20), exposure       875,000
                                                                                                                                                             400,000
                           3,000X1 + 1,250X2 ≤ 100,000
   80                                                                                                                                                     1,275,000
                                     X1 ≥ 5                                  X1 ≤ 25                    Point d (X1      25, X2       10), exposure       875,000
                               b
                                     Optimal Exposure
                                                                                                                                                             200,000
   60                                     Rating                                                                                                          1,075,000
                                                                                                7-26.   Let: X1      number of barrels of pruned olives
                                                                                                             X2      number of barrels of regular olives
X2 40                                                                                                   Maximize profit         $20X1      $30X2
                                                                                                        subject to 5X1         2X2      250 (labor hours)
                                   Feasible Region                                                                    1X1      2X2      150 (acres)
                                                                         X2 ≥ 10
   20                                                               c
                                                                                                                        X1              40 (barrels)
                                                                                                                             X1, X2     0
                              a                                     d
                                                                                                        a. Corner point a         (X1     0, X2    0), profit       0
        0
            0              5         10        15         20       25        30        35                    Corner point b       (X1     0, X2    75), profit       $2,250
                                                     X1                                                      Corner point c       (X1 25, X2 62Z\x), profit
                                                                                                                                  $2,375 k optimal profit
            Let: X1                number of TV spots
                                                                                                             Corner point d       (X1     40, X2      25), profit       $1,550
                         X2        number of newspaper ads
                                                                                                             Corner point e       (X1    40, X2       0), profit    $800
            Maximize exposures                  35,000X1        20,000X2
                                                                                                        b. Produce 25 barrels of pruned olives and 62Z\x barrels
            subject to 3,000X1                  1,250X2        $100,000                                    of regular olives.
                                          X1                   5                                        c. Devote 25 acres to pruning process and 125 acres to
                                          X1                   25                                          regular process.
                                                          X2   10
                Point a (X1            5, X2        10), exposure       375,000
                                                                        (optimal)
                Point b (X1            5, X2        68), exposure       175,000
                                                                             1,360,000
                                                                        1,535,000
92                             CHAPTER 7          LINEAR PROGRAMMING MODELS: GRAPHICAL           AND   COMPUTER METHODS


     125                                                                    X1 2X2 2 line—this is also on the same slope as the isoprofit
                                                                            line X1 2X2 and hence there will be more than one optimal solu-
                                                                            tion.
     100                                                                          As a matter of fact, every point along the heavy line will pro-
                                                                            vide an “alternate optimum.”
              b                                                             Formulation 3:
         75
X2                         c                                                     5
                                     Optimal Solution

         50                                                                                                   Unbounded Region

                                                                                 4
                   Feasible      d
         25        Region
                                                                                 3
              a                 e
         0                                                                  X2
              0           25         50      75         100     125   150
                                             X1                                  2



7-27.                                                                            1
Formulation 1:

     8                                                                           0
                                                                                     0       1         2      3        4        5         6
                                                                                                              X1

     6                                                                      Formulation 4:
                                                        Infeasible
                                                         Solution
                                                                                 8
                                                         Region
X2 4

                                                                                 6
     2

                                                                            X2 4
     0
         0            2          4         6            8       10    12                          Feasible Region
                                           X1
                                                                                 2
                                                                                 1
Formulation 2:
                                                                                 0
                                                                                     0   1   2    3    4      6        8        10       12
     2
                                                                                                              X1

                                                                            Formulation 4 appears to be proper as is. Note that the constraint
                                                                            4X1 6X2 48 is redundant.


X2 1


                                            Line For X1 + 2X2
                  Feasible
                  Region

     0
         0                       1                      2              3
                                            X1
                   CHAPTER 7            LINEAR PROGRAMMING MODELS: GRAPHICAL                   AND   COMPUTER METHODS                               93


7-28. Using the isoprofit line or corner point method, we see that                8
point b (where X1     37.5 and X2      75) is optimal if the profit
   $3X1 $2X2. If the profit changes to $4.50 per unit of X1, the
optimal solution shifts to point c. If the objective function be-
comes P $3X1 $3X2, the corner point b remains optimal.                           6

     150
                 Profit Line for 3X1 + 3X2                                                     Profit = 4X1 + 6X2 = $21.71

                                                                              X2 4

                                                                                                           Optimal solution at
             a            Profit Line for 4.50X1 + 2X2
     100                                                                                                    X1 = 26/7, X2 = 15/7
                                                                                 2
                                                                                                                              1X1 + 3X2 = 8
X2
                   b


     50                                                                          0
                                             Profit Line for                          0         2            4                  6              8
                                              3X1 + 2X2
                                                                                                             X1

                                                                              7-30.       a.
                                                                                100
                                    c
         0
             0          50                   100                   150
                                 X1

7-29. The optimal solution of $26 profit lies at the point X1             2,      75
X2 3.                                                                                            Isoprofit Line for
                                                                                                1X1 + 1X2 = $66.67
     8
                                                                                          a
                                                                              X2 50
                                                                                                           (X   1   = 331/3, X2 = 331/3   )
     6                                                                                               b
                                                                                 25



X2 4
                                                                                                            c
                                                                                  0
                             Profit = 4X1 + 6X2 = $26                                 0         25              50              75            100
                                                                                                                X1
     2




     0
         0        2            4                6              8
                               X1

If the first constraint is altered to 1X1 3X2 8, the feasible re-
gion and optimal solution shift considerably, as shown in the next
column.
94                       CHAPTER 7             LINEAR PROGRAMMING MODELS: GRAPHICAL                                               AND       COMPUTER METHODS


            b.                                                                 7-32.
     150                                                                                          12


                                                                                                  10
                           Isoprofit Line for
                          3X1 + 1X2 = $150
     100                                                                                                  8
                                                                                                                                  6X1 + 4X2 = 36


X2                                                                             X2 6

                                                                                                                                             (X  1   = 5, X2 = 11/2 ; Profit = $29       )
      50                                                                                                  4


                                                                                                          2                                                              1X1 + 2X2 = 8
                                   Optimal Solution is
                                       Now Here

        0                                                                                                 0
            0                 50                   100                   150                                  0           2                 4            6               8        10         12
                                         X1                                                                                                             X1

                                                                               Using the corner point method, we determine that the optimal so-
            c.   If X1’s profit coefficient was overestimated, but should
                                                                               lution mix under the new constraint yields a $29 profit, or an in-
                 only have been $1.25, it is easy to see graphically that
                                                                               crease of $3 over the $26 profit calculated.
                 the solution at point b remains optimal.
                                                                               7-33.                               Let: X1         number of coconuts carried
7-31.
                                                                                                                        X2         number of skins carried
     100
                                                                                                                   Maximize profit                    60X1         300X2 (in rupees)
                                                                                                                   subject to       5X1           15X2        300 pounds
                                                                                                                                        1
                                                                                                                                        8   X1       1X2      15 cubic feet
     75                                                                                                                                          X 1, X 2     0
                                                                                                                      At point a: (X1                0, X2        15), P         4,500 rupees
                                                                                                                      At point b: (X1                24, X2        12), P        1,440       3,600
                                                                                                                                                                                 5,040 rupees
X2 50                        (X
                              1   = 426/7, X2 = 142/7; Profit = $571/7   )                                            At point c: (X1                60, X2        0),       P   3,600 rupees
                                                                               The three princes should carry 24 coconuts and 12 lions’ skins.
                                                                               This will produce a wealth of 5,040 rupees.
     25                                                                                                   20
                                              Optimal Solution
                                                Remains at
                                    b             Point b

      0                                                                                                           a
           0            25              50               75          100                                  15
                                                                               Number of Lion Skins, X2




                                        X1                                                                                                  Optimal Solution

                                                                                                                              b
The optimal solution is at point b, but profit has decreased from                                          10
$662 to $571, and the solution has changed considerably.
   3       7




                                                                                                                       Feasible
                                                                                                          5            Region




                                                                                                                                                              c
                                                                                                          0
                                                                                                               0                   30          60          90                                 120
                                                                                                                                     Number of Coconuts, X1
                          CHAPTER 7             LINEAR PROGRAMMING MODELS: GRAPHICAL                      AND     COMPUTER METHODS                            95


7-34.     a.   Let: X1       number of pounds of stock X purchased per                  0.2X1        0.4X2        0.1X3     0.2X4      26,000 (hours of assembly
                             cow each month                                                                                            time available)
                     X2      number of pounds of stock Y purchased per                  0.5X1        0.1X2        0.5X3     0.5X4      1,200 (hours of inspection
                             cow each month                                                                                            time)
                     X3      number of pounds of stock Z purchased per                          X1        150 (units of XJ201)
                             cow each month
                                                                                                X2        100 (units of XM897)
Four pounds of ingredient Z per cow can be transformed to:
                                                                                                X3        300 (units of TR29)
       4 pounds      (16 oz/lb)         64 oz per cow
                                                                                                X4        400 (units of BR788)
                      5 pounds          80 oz
                                                                                    7-36.     Maximize Z           [220 (0.45)(220) 44 20]X1
                       1 pound          16 oz                                                                        [175 (0.40)(175) 30 20]X2
                      8 pounds          128 oz                                                                     57X1         55X2
               3X1     2X2      4X3       64 (ingredient A requirement)                       Constraints:
               2X1     3X2      1X3       80 (ingredient B requirement)                                      X1     X2      390     production limit
               1X1     0X2      2X3       16 (ingredient C requirement)                              2.5X1        2.4X2     960     labor hours
               6X1     8X2      4X3       128 (ingredient D requirement)                      Corner points:
                                 X3       5 (stock Z limitation)                                     X1      384, X2       0,       profit   $21,888
Minimize cost        $2X1       $4X2       $2.50X3                                                   X1      0,     X2     390, profit       $21,450
          b. Cost         $80                                                                        X1      240, X2       150, profit       $21,930
                X1        40 lbs. of X                                              Students should point out that those three options are so close in
                X2        0 lbs. of Y                                               profit that production desires and sensitivity of the RHS and cost
                X3        0 lbs. of Z                                               coefficient are important issues. This is a good lead-in to the dis-
                                                                                    cussion of sensitivity analysis. As a matter of reference, the
7-35.     Let: X1     number units of XJ201 produced                                right-hand side ranging for the first constraint is a production
               X2     number units of XM897 produced                                limit from 384 to 400 units. For the second constraint, the hours
               X3     number units of TR29 produced                                 may range only from 936 to 975 without affecting the solution.
                                                                                          The objective function coefficients, similarly, are very sensi-
               X4     number units of BR788 produced
                                                                                    tive. The $57 for X1 may increase by 29 cents or decrease by $2.
Maximize profit            9X1   12X2       15X3      11X4                           The $55 for X2 may increase by $2 or decrease by 28 cents.
subject to                                                                          7-37.     a. Let: X1          number of MCA regular modems made and
       0.5X1   1.5X2       1.5X3        0.1X4    15,000 (hours of wiring                                          sold in November
                                                 time available)                                           X2      number of MCA intelligent modems made
       0.3X1   0.1X2       0.2X3        0.3X4    17,000 (hours of drilling                                         and sold in November
                                                 time available)                    Data needed for variable costs and contribution margin (refer to
                                                                                    the table on the bottom of this page):


Table for Problem 7-37(a)
                                                               MCA REGULAR MODEM                                                  MCA INTELLIGENT MODEM
                                                       Total                       Per Unit                                Total                        Per Unit
    Net sales                                        $424,000                       $47.11                                $613,000                        $58.94
    Variable costsa
     Direct labor                                      60,000                         6.67                                  76,800                          7.38
     Indirect labor                                     9,000                         1.00                                  11,520                          1.11
     Materials                                         90,000                        10.00                                 128,000                         12.31
     General expenses                                  30,000                         3.33                                  35,000                          3.37
     Sales commissions                               $231,000                       $23.44                                $360,000                        $25.76
    Total variable costs                             $220,000                       $24.44                                $311,320                        $29.93
    Contribution margin                              $204,000                       $22.67                                $301,680                        $29.01
a
Depreciation, fixed general expense, and advertising are excluded from the calculations.
96                      CHAPTER 7             LINEAR PROGRAMMING MODELS: GRAPHICAL               AND   COMPUTER METHODS


Hours needed to produce each modem:                                          Solution: Produce:
                                 5,000 hours                                     1,100 units of W75C—backorder 300 units
          MCA regular                                0.555 hour/modem
                                9,000 modems                                       250 units of W33C—backorder 0 units
                                 10,400 hours                                          0 units of W5X—backorder 1,510 units
       MCA intelligent                                1.0 hour/modem
                                10,400 modems
                                                                                   600 units of W7X—backorder 516 units
                Maximize profit         $22.67X1      $29.01X2
                                                                             Maximized profit will be $59,900. By addressing quality problems
       subject to    0.555X1      1.0X2       15,400 (direct labor hours)    listed earlier, we could increase our capacity by up to 3% reducing
                                       X2     8,000 (intelligent modems)     our backorder level.
          b.                                                                 2. Bringing in temporary workers in the Drawing Department
                                                                             would not help. Drawing is not a binding constraint. However, if
     X2                                                                      these former employees could do rework, we could reduce our re-
                                                                             work inventory and fill some of our backorders thereby increasing
15,400
                                                                             profits. We have about a third of a month’s output in rework inven-
                                                                             tory. Expediting the rework process would also free up valuable cash.
                                            P = $534,339                     3. The plant layout is not optimum. When we install the new equip-
                                                                             ment, an opportunity for improving the layout could arise. Exchang-
 8,000                                                                       ing the locations for packaging and extrusion would create a better
                                  b
                                                                             flow of our main product. Also, as we improve our quality and re-
                                                               Optimal       duce our rework inventory, we could capture some of the space now
                                                            P = $629,000
                                                                             used for rework storage and processing and put it to productive use.
                                                                                   Our machine utilization of 63% is quite low. Most manufac-
                                                                 c           turers strive for at least an 85% machine utilization. If we could
                                                             27,750   X1     determine the cause(s) of this poor utilization, we might find a key
          c. The optimal solution suggests making all MCA                    to a dramatic increase in capacity.
          regular modems. Students should discuss the implications
          of shipping no MCA intelligent modems.                             INTERNET CASE STUDY:
7-38.     Minimize cost         12X1    9X2      11X3      4X4               Agri-Chem Corporation
               subject to X1     X2     X3      X4   50                      This case demonstrates an interesting use of linear programming
                                                                             in a production setting.
                           X1    X2     X3      X4   7.5
                                                                                        Let X1     ammonia
                           X1    X2     X3      X4   22.5
                                                                                           X2      ammonium phosphate
                           X1    X2     X3      X4   15.0
                                                                                           X3      ammonium nitrate
Solution:
                                                                                           X4      urea
                           X1    7.5 pounds of C-30
                                                                                           X5      hydrofluoric acid
                           X2    15 pounds of C-92
                                                                                           X6      chlorine
                           X3    0 pounds of D-21
                                                                                           X7      caustic soda
                           X4    27.5 pounds of E-11
                                                                                           X8      vinyl chloride monomer
Cost      $3.35.
                                                                             Objective function:
SOLUTION TO MEXICANA WIRE WORKS CASE                                              Maximize Profit          80X1 120X2 140X3        140X4     90X5
                                                                                                            70X6 60X7 90X8
1.   Maximize P         34 W75C         30 W33C         60 W5X      25 W7X
                                                                             Subject to the following constraints:
     subject to:
                                                                                  X1    1,200                 X5   560
     1 W75C          1,400
                                                                                  X2    540                   X6   1,200
     1 W33C          250
                                                                                  X3    490                   X7   1,280
     1 W5XC          1,510
                                                                                  X4    160                   X8   840
     1 W7XC          1,116
                                                                             Current natural gas usage 85,680 cu. ft. 103/day
     1 W75C          2 W33C       0 W5X         1 W7X       4,000
                                                                             20 percent curtailment 68,554 cu. ft. 103/day
     1 W75C          1 W33C       4 W5X         1 W7X       4,200
     1 W75C          3 W33C       0 W5X         0 W7X       2,000
     1 W75C          0 W33C       3 W5X         2 W7X       2,300
     1 W75C                                                 150
                                                1 W7X       600
                     CHAPTER 7         LINEAR PROGRAMMING MODELS: GRAPHICAL                AND   COMPUTER METHODS                      97


Hence, the ninth constraint is:                                                After 8 simplex iterations, optimal solution is reached. The
     8X1    10X2     12X3     12X4    7X5     18X6    20X7     14X7       following is the production schedule:
                                                               68,544         X1    1200              X5   560
     The following is the production schedule (tons/day);                     X2     540              X6   720
     X1    1,200              X5     560                                      X3     490              X7   0
     X2    540                X6     1,200                                    X4     160              X8   840
     X3    490                X7     425                                      Objective function value: $428,200
     X4    160                X8     840                                       The caustic soda production is eliminated completely and the
                                                                          chlorine production is reduced from 1,200 to 720 tons/day.
     Objective function value      $487,300
     Because of the natural gas curtailment, the caustic soda pro-
duction is reduced from 1280 tons/day to 425 tons/day.
     For a 40 percent natural gas curtailment, the ninth constraint is:
     8X1    10X2     12X3     12X4    7X5     18X6    20X7     14X8
                                                               51,408

								
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