Solving Linear Programs (LP Section 2)

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					  Solving Linear Programs (LP Section 2)

Chapter 2 (Revisited)
 Graphical Method

Computer Solutions
 Using LINDO
 Interpreting Computer Output for LP Section 1 problems




                         Dr. C. Lightner                  1
                   Fayetteville State University
              Solving Linear Programs

The goal of a linear program is to find the values of your xj variables
that provide the largest (or smallest) possible objective value when
substituted into the objective function.
The Graphical Method can be used to solve the LP for these values
when you have one or two decision variables. This technique helps
to visualize key LP concepts.
Computer Programs are generally used to solve all LP problems.
We will learn to use the graphical method to solve small problems
and gain insight into some theory behind solving LPs. Nevertheless,
we will focus on using a computer software to solve our LP models.
                           Dr. C. Lightner                           2
                     Fayetteville State University
      Solving LPs using the Graphical Method
                                   Chapter 2


  We begin the graphical solution process by graphing each linear
  constraint of the LP model on the same graph.
  Refer to the PAR, INC example from Section 1 of your LP notes.
Maximize 10 x1 + 9 x2
Subject to:
  7/10 x1 + 1 x2 ≤ 630
   1/2 x1 + 5/6 x2 ≤ 600
     1 x1 + 2/3 x2 ≤ 708
  1/10 x1+ 1/4 x2 ≤ 135
  x1 ≥ 0, x2 ≥ 0


                                 Dr. C. Lightner                    3
                           Fayetteville State University
                   Graphical Method: PAR, INC
Constraint 1 (C1):
7/10 x1 + 1 x2 ≤ 630 Always graph constraints as an equality.

In order to graph a line we need two points that fall on the line. Arbitrarily, let x 1 =0.
Then 7/10 (0) + x2 =630. Solving this equation for x2 gives x2 =630. Therefore one point
on this line is (0, 630).
Arbitrarily, let x2 =0. Then 7/10 x1 + (0) =630. Solving this equation for x1 gives x1
=6300/7=900. Therefore a 2 nd point on this line is (900, 0). Connect the points (0, 630)
and (900,0) to draw this line.

Determine the Feasible Side of the Line:
In LP, on one side of the line all points are feasible (i.e. they satisfy the constraint) and on
the other side all points are not feasible. For instance take the point (0,0). The point x 1
=0 and x2 =0 satisfy the above constraint [7/10 * (0) + (0) is less than 630]. Thus all
points that are on the side of the line containing this point will also satisfy this constraint.
We put a red arrow to indicate the feasible side of the line.
                                     Dr. C. Lightner                                     4
                               Fayetteville State University
     Graphical Method: PAR, INC

       700


       600
x2




       500
                   C1

       400


       300


       200               x1
                   Dr. C. Lightner           5
             Fayetteville State University
              Graphical Method: PAR, INC
Constraint 2 (C2):
1/2 x1 + 5/6 x2 ≤ 600
In order to graph a line we need two points that fall on the line. Arbitrarily, let x1 =0.
Then 1/2 (0) + 5/6 x2 =600. Solving this equation for x2 gives x2 =720. Therefore
one point on this line is (0, 720).
Arbitrarily, let x2 =0. Then x1 =1200. Therefore a 2nd point on this line is (1200, 0).
Connect the points (0, 720) and (1200,0) to draw this line.
Determine the Feasible Side of the Line:
Use the point (0,0) as a test point. Since x1 =0 and x2 =0 satisfies the above
constraint [1/2 * (0) + 5/6 (0) is less than 600]. Thus all points on the side
containing the point (0,0) will satisfy this constraint. We put a red arrow to indicate
the feasible side of the line.


                                Dr. C. Lightner                                      6
                          Fayetteville State University
       Graphical Method: PAR, INC


     800

     700
                         C2
     600
x2




           C1
     500

     400

     300
                            x1
                      Dr. C. Lightner           7
     200        Fayetteville State University
              Graphical Method: PAR, INC

Constraint 3 (C3):
 1 x1 + 2/3 x2 ≤ 708
In order to graph a line we need two points that fall on the line. Arbitrarily, let x1 =0.
Then x2 =1062. Therefore one point on this line is (0, 1062).
Arbitrarily, let x2 =0. Then x1 =708. Therefore a 2nd point on this line is (708, 0).
Connect the points (0, 1062) and (708,0) to draw this line.

Determine the Feasible Side of the Line:
Using our test point point (0,0), we find that x1 =0 and x2 =0 satisfies the above
constraint [1 * (0) + 2/3 (0) is less than 708]. Thus all points on the side containing
the point (0,0) will satisfy this constraint. We put a red arrow to indicate the feasible
side of the line.


                                Dr. C. Lightner                                      8
                          Fayetteville State University
      Graphical Method: PAR, INC

     1200
            C3


     1000
x2




      800

                 C1                              C2
      600


      400
                             x1
                       Dr. C. Lightner                9
                 Fayetteville State University
              Graphical Method: PAR, INC

Constraint 4 (C4):
1/10 x1+ 1/4 x2 ≤ 135
Arbitrarily, let x1 =0. Then x2 =540. Therefore one point on this line is (0, 540).
Arbitrarily, let x2 =0. Then x1 =1350. Therefore a 2nd point on this line is (1350, 0).
Connect the points (0, 540) and (1350,0) to draw this line.
Determine the Feasible Side of the Line:
Using our test point (0,0), we find that x1 =0 and x2 =0 satisfies the above constraint
[1 * (0) + 2/3 (0) is less than 708]. Thus all points on the side containing the point
(0,0) will satisfy this constraint. We put a red arrow to indicate the feasible side of
the line.




                                Dr. C. Lightner                                    10
                          Fayetteville State University
             Graphical Method: PAR, INC

     1200
            C3

     1000
x2




            C1
      800
       C4

      600
                                       C2


      400
                              x1
                         Dr. C. Lightner           11
                   Fayetteville State University
           Graphical Method: PAR, INC

Nonnegativity Constraints
These constraints simply restrict us to the 1st quadrant of our
coordinates (i.e. only 0 or positive values of x1 and x2 are considered
feasible on our graph when we highlight our final feasible region).




                           Dr. C. Lightner                         12
                     Fayetteville State University
             Graphical Method: PAR, INC

     1200
            C3

     1000
x2




            C1
      800
       C4

      600
                                       C2


      400
                              x1
                         Dr. C. Lightner           13
                   Fayetteville State University
           Graphical Method: PAR, INC

After graphing each of the constraints and their feasible sides
individually, we must determine all points on the coordinates which
satisfy all constraints. These points make up the feasible region.




                          Dr. C. Lightner                        14
                    Fayetteville State University
     PAR, INC GRAPHICAL METHOD
           FEASIBLE REGION

     1200
        C3


     1000
        C1
x2




                     C4
      800


      600


      400
                          x1
                   Dr. C. Lightner           15
             Fayetteville State University
                FEASIBLE REGION

Once the feasible region is drawn, we must determine the point in
this region that maximizes (or minimizes) the objective function.




                          Dr. C. Lightner                       16
                    Fayetteville State University
   Extreme Points and the Optimal Solution

The corners or vertices of the feasible region are referred to as the
extreme points.
Fundamental Theorem of Linear Programming
An optimal solution to an LP problem can be found at an extreme
point of the feasible region.
Optimal point occurs on the objective function line corresponding to
the optimal objective function value
When looking for the optimal solution, you do not have to evaluate all
feasible solution points.
You have to consider only the extreme points of the feasible region.

                          Dr. C. Lightner                        17
                    Fayetteville State University
     PAR, INC GRAPHICAL METHOD
 FEASIBLE REGION AND EXTREME POINTS

     1200


     1000
x2




      800


      600


      400
                        x1
                  Dr. C. Lightner           18
            Fayetteville State University
                Finding the Optimal Point

How to find the optimal point
  Randomly draw objective function line.
  Push the line in the “direction of decrease” (if minimization problem)
  or the “direction of increase” (if maximization problem)
  The last point that you encounter in the feasible region is the optimal
  point




                              Dr. C. Lightner                        19
                        Fayetteville State University
          Drawing an Objective Function Line

1. Randomly select a point in the feasible region.
2. Substitute the coordinates of this point into the objective function to
   obtain an objective function value.
3. Plot the line obtained by setting the original objective function equal
   to the objective function value obtained in 2.
PAR, INC Example
Step 1: The point (200,0) is in the feasible region.
Step 2: PAR INC OBJECTIVE FUNCTION 10 x1 + 9 x2 . (200,0) gives
    10 (200) + 9 (0)= 2000.
Step 3: Plot the line 10 x1 + 9 x2 =2000.
       Connect the points (0, 222.22) and (200, 0)
                              Dr. C. Lightner                         20
                        Fayetteville State University
               PAR, INC GRAPHICAL METHOD
           FEASIBLE REGION AND EXTREME POINTS

                  1200


                  1000
         x2




                   800

Obj. Func. line
                   600


                   400
                                     x1
                               Dr. C. Lightner           21
                         Fayetteville State University
     FINDING THE DIRECTION OF INCREASE
           (For Maximization problems)
     Maximization problem: Finding the direction of increase
1.   Let v1= the coefficient of x1 in the objective function.
2.   Let v2= the coefficient of x2 in the objective function.
3.   Plot the point (v1, v2) on the graph.
4.   Draw an arrow from the origin (the point x1=0, x2=0) to the point
     (v1,v2). This is the direction of increase. Push the objective function
     in this direction until you encounter the last point in the feasible
     region. This point is optimal.




                               Dr. C. Lightner                         22
                         Fayetteville State University
     FINDING THE DIRECTION OF DECREASE
           (For Minimization problems)
     Minimization problem: Finding the direction of decrease
1.   Let v1= - (the coefficient of x1 in the objective function).
2.   Let v2= - (the coefficient of x2 in the objective function).
3.   Plot the point (v1, v2) on the graph.
4.   Draw an arrow from the origin (the point x1=0, x2=0) to the point
     (v1,v2). This is the direction of decrease. Push the objective
     function in this direction until you encounter the last point in the
     feasible region. This point is optimal.




                               Dr. C. Lightner                              23
                         Fayetteville State University
        PAR INC GRAPHICAL METHOD

PAR INC objective function is
Maximize 10 x1 + 9 x2
Thus v1=10 and v2=9. I can draw an arrow from the origin in the
direction of the point (10,9). This is the direction of increase.




                          Dr. C. Lightner                           24
                    Fayetteville State University
               PAR, INC GRAPHICAL METHOD
           FEASIBLE REGION AND EXTREME POINTS

                  1200


                  1000
         x2




                                       Direction of Increase
                   800                                   This is the optimal point.
Obj. Func. line

                   600


                   400               x1
                               Dr. C. Lightner                                  25
                         Fayetteville State University
     PAR, INC GRAPHICAL METHOD
 FEASIBLE REGION AND EXTREME POINTS

     1200


     1000
x2




                          Slide 15 shows that the optimal point is
      800                 formed by the intersection of Constraint 1
                          and Constraint 3.

      600


      400               x1
                  Dr. C. Lightner                               26
            Fayetteville State University
          PAR INC GRAPHICAL SOLUTION

    Constraint 1: 7/10 x1 + 1 x2= 630
    Constraint 3: 1 x1 + 2/3 x2 = 708
Find the point where these two lines intersect.
Solve the first equation for x2:
   x2= 630-7/10 x1
Substitute the above equation for x2 in the 2nd equation.
    1 x1 + 2/3(630-7/10 x1)=708
Solving the above equation for x1 gives, x1 = 540. Thus
   x2= 630-7/10 (540)= 252.
The optimal point denoted x1*=540, x2*=252.
                              Dr. C. Lightner               27
                        Fayetteville State University
       PAR INC GRAPHICAL SOLUTION

If we substitute the optimal point x1*=540, x2*=252 into the objective
function, 10 (540)+ 9 (252)= 7668.
Thus PAR INC should produce 540 Standard bags and 252 Deluxe
bags to receive a maximum profit of $7668.




                           Dr. C. Lightner                         28
                     Fayetteville State University
     Discussion about the Graphical Method

Refer to Additional Graphical Method Problems link in LP Section 2 folder and/ or
your textbook for more practice solving problems via the graphical method.

You will have one homework (containing 2 problems) dedicated to this approach.
YOU WILL NOT HAVE TO SOLVE A PROBLEM USING THIS TECHNIQUE ON
AN EXAM.
This approach to solving Linear Programming models allows us to visualize the
process that computer software use to soft LP models.
Computer software use the knowledge that the optimal point is an extreme point of
the feasible region (There are characteristics of extreme points that allow the
computer to recognize them). The software will “intelligently” search these
extreme points to determine the optimal solution to an LP model.


                              Dr. C. Lightner                                 29
                        Fayetteville State University
                Computer Solutions

Computer programs designed to solve LP problems are now widely
available.
Most large LP problems can be solved with just a few minutes of
computer time.
Small LP problems usually require only a few seconds.
We will use LINDO to solve our LP problems.




                         Dr. C. Lightner                    30
                   Fayetteville State University
                Using LINDO to Solve

Look in the LP Section 2 folder and click on the link Downloading
Free LINDO Software and follow the instructions.
The Par Inc file located in this folder is a MS Word file which explains
how to enter (and solve) a program in LINDO.
Enter the other examples presented in this chapter into LINDO.




                           Dr. C. Lightner                         31
                     Fayetteville State University
          Interpretation of Computer Output

We will discuss the following output:
 –   objective function value
 –   values of the decision variables
 –   reduced costs
 –   slack/surplus




                               Dr. C. Lightner           32
                         Fayetteville State University
                          Reduced Cost

The reduced cost for a decision variable whose value is 0 in the optimal solution is
the amount the variable's objective function coefficient would have to improve
(increase for maximization problems, decrease for minimization problems) before
this variable could assume a positive value.
The reduced cost for a decision variable with a positive value is 0.
Example
  Consider the following objective function:
  Min 2 x1 + 5 x2 + 4 x3
 Suppose the optimal value of x1 is zero, with a reduced cost of 1.2. Since this is a
 minimization problem, this tells us that the current coefficient of x1 , which is 2,
 must be decreased by 1.2 in order for the optimal value of x1 to be nonzero. Thus
 if the objective function coefficient of x1 was 0.8 (or less), resolving the LP would
 yield a nonzero value of x1.
                               Dr. C. Lightner                                    33
                         Fayetteville State University
                       SLACK/SURLUS

The slack for “less than or equal to” constraints is the difference between the right
hand side of an equation and the value of the left hand side after substituting the
optimal values of the decision variables.
The slack represents the amount of unused units of the right hand side resources.
The surplus for “greater than or equal to” constraints is the difference between the
right hand side of an equation and the value of the left hand side after substituting
the optimal values of the decision variables.
The surplus represents the number of units in which the optimal solution causes the
constraint to exceed the right hand side lower limit.




                               Dr. C. Lightner                                   34
                         Fayetteville State University
                          Par INC LINDO Solution
      LP OPTIMUM FOUND AT STEP              2
          OBJECTIVE FUNCTION VALUE
          1)   7667.942
       VARIABLE       VALUE                  REDUCED COST
         X1           539.984253             0.000000
         X2           252.011032             0.000000
           ROW        SLACK OR               DUAL PRICES
Constraint 1          SURPLUS
              2)      0.000000               4.374566           Constraint 2
              3)      120.007088             0.000000
              4)      0.000000               6.937804
Constraint 3  5)      17.998819              0.000000              Constraint 4
              6)      539.984253             0.000000           X1>=0 Constraint
              7)      252.011032             0.000000
       NO. ITERATIONS=      2
                                                         X2>=0 Constraint
                                      Dr. C. Lightner                              35
                                Fayetteville State University
             Par INC LINDO Solution
RANGES IN WHICH THE BASIS IS UNCHANGED:

                      OBJ COEFFICIENT RANGES
VARIABLE      CURRENT      ALLOWABLE       ALLOWABLE
              COEF         INCREASE        DECREASE
   X1         10.000000    3.499325        3.700000
   X2         9.000000     5.285714        2.333000

              RIGHTHAND SIDE RANGES
  ROW         CURRENT     ALLOWABLE                ALLOWABLE
              RHS        INCREASE                  DECREASE
   2          630.000000 52.358864                 134.400009
   3          600.000000 INFINITY                  120.007088
   4          708.000000 192.000000                127.986000
   5          135.000000  INFINITY                 17.998819
   6          0.000000    539.984253               INFINITY
   7          0.000000    252.011032               INFINITY
                         Dr. C. Lightner                        36
                   Fayetteville State University
                              PAR INC

Recall that we rounded values (instead of entering fractions) when entering our
model. Therefore, our LINDO output will be slightly different from the actual
solution.
From the spreadsheet we see that the maximum that PAR can achieve while
meeting the constraints is 7667.942 or 7668.
The optimal solution is to produce X1 =539.984253 or 540 standard bags,
X2=252.011032 or 252 deluxe bags.
The reduced costs for both decision variables is zero since their optimal values are
nonzero.
Row 1 represents the objective function.
Row 2 represents constraint 1, row 3 represents constraint 2, row 4 represents
constraint 3, etc…
Refer to the order in which you entered the model into LINDO.
                               Dr. C. Lightner                                 37
                         Fayetteville State University
                 PAR INC (Slack/Surplus)

Constraint 1 is a “less than or equal to constraint” (0.7 x1+ x2 ≤630)
Plugging the optimal values into the left hand side of constraint 1 gives
       0.7(540) + 1 (252)2 = 630
Since the left hand side (with optimal values substituted) equals the left hand side,
the slack is 0.
Constraint 2 is a “less than or equal to constraint” (0.5 x1+ 0.8333 x2 ≤600)
Plugging the optimal values into the left hand side of constraint 2 gives
       0.5 (540) + 0.8333(252) = 480
The difference between this value and 600 is 120. The spreadsheet lists 120 as
the slack for this constraint.
Recall from LP Section 1that Constraint 2 represents the amount of sewing hours
available for use. This slack indicates that the optimal solution leaves 120 sewing
hours unused. Suggesting that less workers may be used to achieve the optima
profit since so many hours will be wasted.

                               Dr. C. Lightner                                  38
                         Fayetteville State University
             PAR INC (Slack/Surplus)

The output also reveals that 18 hours of Inspection and packing will
also go unused.
The second output sheet gives information about sensitivity analysis.
We will discuss this topic in the next section.




                          Dr. C. Lightner                       39
                    Fayetteville State University
           Floataway Tours LINDO Solution
OBJECTIVE FUNCTION VALUE
   1) 5040.000

VARIABLE        VALUE       REDUCED COST
  X1            28.000000    0.000000
  X2            0.000000     2.000000
  X3            0.000000     12.000000
  X4            28.000000     0.000000

   ROW   SLACK OR SURPLUS     DUAL PRICES
   2)     0.000000                 0.012000
   3)     6.000000                 0.000000
   4)     0.000000                 -2.000000
   5)    52.000000                 0.000000
   6)    28.000000                 0.000000
   7)     0.000000                 0.000000
   8)     0.000000                 0.000000
   9)    28.000000                 0.000000
                              Dr. C. Lightner           40
                        Fayetteville State University
              Floataway Tours LINDO Solution
                     OBJ COEFFICIENT RANGES
VARIABLE      CURRENT         ALLOWABLE ALLOWABLE
               COEF           INCREASE      DECREASE
  X1           70.000000      44.999996    1.875001
  X2          80.000000       2.000001     INFINITY
  X3          50.000000       12.000001   INFINITY
  X4          110.000000      INFINITY    16.363636

                      RIGHTHAND SIDE RANGES
  ROW        CURRENT      ALLOWABLE     ALLOWABLE
               RHS        INCREASE      DECREASE
   2       420000.000000    INFINITY    45000.000000
   3       50.000000        6.000000    INFINITY
   4       0.000000         70.000000   30.000000
   5       200.000000       52.000000    INFINITY
   6       0.000000         28.000000    INFINITY
   7       0.000000         0.000000     INFINITY
   8       0.000000         0.000000     INFINITY
   9       0.000000         28.000000    INFINITY

                                   Dr. C. Lightner           41
                             Fayetteville State University
Floataway Tours LINDO Solution Discussion

The maximum daily profit that they can achieve (while meeting constraints) is
$5040.
The optimal solution is x1=28 or purchase 28 Speedhawks and x2=28 or purchase
28 Classys.
The reduced costs for Speedhawks and Classys is 0 since their optimal values are
nonzero.
The reduced cost for Silverbirds (x2) is 2. This means that the objective function
coefficient for x2 (currently 80) must be improved by 2 in order for the optimal value
of x2 to become nonzero (i.e. in order for it to become profitable to order
Silverbirds). Since this is a maximization problem, improved means increased.
Therefore if the daily expected profit for Silverbirds was 82 opposed to 80, it would
be profitable to order Silverbirds.
Similarly if the objective function coefficient for Catmans (x3 ) was 62 (instead of 50)
it would be profitable to order these type of boats.

                                Dr. C. Lightner                                   42
                          Fayetteville State University
Floataway Tours LINDO Solution Discussion

The surplus for constraint 2 (the requirement that at least 50 boats be
purchased) shows that they purchase 6 boats over the 50 minimum
requirement.
The surplus for constraint 4 ( the requirement that they have a
seating capacity of at least 200) shows that they can seat 52
passengers over the 200 minimum requirement.




                           Dr. C. Lightner                        43
                     Fayetteville State University
           Police Scheduling LINDO Solution
OBJECTIVE FUNCTION VALUE
   1) 19.00000
VARIABLE       VALUE          REDUCED COST
  X1           0.000000        0.000000
  X2           6.000000        0.000000
  X3           4.000000        0.000000
  X4           3.000000        0.000000
  X5           1.000000        0.000000
  X6           5.000000        0.000000
  ROW SLACK OR SURPLUS          DUAL PRICES
  2)   0.000000                     0.000000
  3)   0.000000                     -1.000000
  4)   0.000000                     0.000000
  5)   0.000000                     -1.000000
  6)   0.000000                     0.000000
  7)   0.000000                     -1.000000

                                Dr. C. Lightner           44
                          Fayetteville State University
             Police Scheduling LINDO Solution
                  OBJ COEFFICIENT RANGES
VARIABLE      CURRENT      ALLOWABLE     ALLOWABLE
              COEF        INCREASE      DECREASE
  X1           1.000000     INFINITY  0.000000
  X2           1.000000    0.000000   1.000000
  X3           1.000000    0.000000   0.000000
  X4           1.000000    0.000000   0.000000
  X5           1.000000    0.000000   0.000000
  X6           1.000000    0.000000   0.000000

                     RIGHTHAND SIDE RANGES
  ROW        CURRENT      ALLOWABLE     ALLOWABLE
              RHS         INCREASE      DECREASE
   2       5.000000         0.000000   3.000000
   3       6.000000         INFINITY   0.000000
   4       10.000000        0.000000   INFINITY
   5       7.000000         INFINITY   0.000000
   6       4.000000         0.000000   3.000000
   7       6.000000         3.000000   0.000000

                                  Dr. C. Lightner           45
                            Fayetteville State University
       Police Scheduling LINDO Solution

The department needs 19 officers to meet the daily requirements.
The optimal solution is to hire 0 officers to work the shift from 8AM –
4PM (x1=0), 6 officers to work the shift from Noon – 8PM (x2=6), 4
officers to work the shift from 4PM – Midnight (x3=4), 3 officers to
work the shift from 8PM – 4AM (x4=3), 1 officers to work the shift
from Midnight – 8AM (x5=1), and 5 officers to work the shift from
4AM – Noon (x6=5).
The surplus for all constraints is 0. Thus each shift has exactly the
number of required officers for each shift.



                           Dr. C. Lightner                         46
                     Fayetteville State University
                  THE END




       See your textbook for more
 examples and detailed explanations
of all topics discussed in these notes.



                    Dr. C. Lightner           47
              Fayetteville State University