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C H A P T E R
Linear Programming: The Simplex Method
TEACHING SUGGESTIONS Teaching Suggestion 9.1: Meaning of Slack Variables. Slack variables have an important physical interpretation and represent a valuable commodity, such as unused labor, machine time, money, space, and so forth. Teaching Suggestion 9.2: Initial Solutions to LP Problems. Explain that all initial solutions begin with X1 0, X2 0 (that is, the real variables set to zero), and the slacks are the variables with nonzero values. Variables with values of zero are called nonbasic and those with nonzero values are said to be basic. Teaching Suggestion 9.3: Substitution Rates in a Simplex Tableau. Perhaps the most confusing pieces of information to interpret in a simplex tableau are “substitution rates.” These numbers should be explained very clearly for the first tableau because they will have a clear physical meaning. Warn the students that in subsequent tableaus the interpretation is the same but will not be as clear because we are dealing with marginal rates of substitution. Teaching Suggestion 9.4: Hand Calculations in a Simplex Tableau. It is almost impossible to walk through even a small simplex problem (two variables, two constraints) without making at least one arithmetic error. This can be maddening for students who know what the correct solution should be but can’t reach it. We suggest two tips: 1. Encourage students to also solve the assigned problem by computer and to request the detailed simplex output. They can now check their work at each iteration. 2. Stress the importance of interpreting the numbers in the tableau at each iteration. The 0s and 1s in the columns of the variables in the solutions are arithmetic checks and balances at each step. Teaching Suggestion 9.5: Infeasibility Is a Major Problem in Large LP Problems. As we noted in Teaching Suggestion 7.6, students should be aware that infeasibility commonly arises in large, real-world-sized problems. This chapter deals with how to spot the problem (and is very straightforward), but the real issue is how to correct the improper formulation. This is often a management issue.
ALTERNATIVE EXAMPLES Alternative Example 9.1: Simplex Solution to Alternative Example 7.1 (see Chapter 7 of Solutions Manual for formulation and graphical solution).
1st Iteration Cj l b 0 0
Solution Mix
3 X1
9 X2
0 S1
0 S2
Quantity
S1 S2
1 1
4 2
1 0
0 1
24 16
Zj Cj Zj
0 3
0 9
0 0
0 0
0
3 X1
9 X2
0 S1
0 S2
⁄4 1 ⁄2
1 0
⁄4 1⁄2
0 1
6 4
⁄4 ⁄4
9 0
9 ⁄4 9⁄4
0 0
54
2nd Iteration Cj l b 9 0
Solution Mix X2 S2
1
Zj Cj Zj
9
3
1
Quantity
This is not an optimum solution since the X1 column contains a positive value. More profit remains ($C\v per #1). 3rd/Final Iteration Cj l Solution b Mix
3 X1
9 X2
0 S1
0 S2
Quantity
9 3
X2 X1
0 1
1 0
⁄2 13⁄2
⁄2 23⁄2
4 8
Zj Cj Zj
3 0
9 0
⁄2 ⁄2
⁄2 3⁄2
60
1
3 3
1
3
This is an optimum solution since there are no positive values in the Cj Zj row. This says to make 4 of item #2 and 8 of item #1 to get a profit of $60. Alternative Example 9.2: Set up an initial simplex tableau, given the following two constraints and objective function: Minimize Z 8X1 6X2 Subject to:
2X1 4X2 8
3X1 2X2 6 The constraints and objective function may be rewritten as: Minimize 8X1 6X2 0S1 0S2 MA1 MA2 2X1 4X2 1S1 0S2 1A1 0A2 8 3X1 2X2 0S1 1S2 0A1 1A2 6
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The first tableau would be: Cj l b
Solution Mix
8 X1
6 X2
0 S1
0 S2
M A1
M A2
Quantity
A1 A2
2 3
4 2
1 0
0 1
1 0
0 1
8 6
Zj
5M
6M
M
M
M
M
14M
Cj Zj
8 5M
6 6M
M
M
0
0
M M
The second tableau: Cj l b
Solution Mix
8 X1
6 X2
0 S1
0 S2
M A1
M A2
Quantity
⁄2 2
1 0
1⁄4 1 ⁄2
0 1
⁄4 1⁄2
0 1
2 2
Zj
3 2M
6
3⁄2 1⁄2M
M
⁄2 1⁄2M
M
12 2M
Cj Zj
5 2M
0
⁄2 ⁄2M
0
6 M
X2 A2
1
⁄2 ⁄2M
3
M
1
1
3
3
3
The third and final tableau: Cj l Solution b Mix
8 X1
6 X2
0 S1
0 S2
M A1
M A2
6 8
X2 X1
0 1
1 0
3⁄8 1 ⁄4
⁄4 1⁄2
⁄8 1⁄4
1⁄4 1 ⁄2
Zj
8
6
1⁄4
5⁄2
Cj Zj
0
0
⁄4
1
1
⁄2
5
3
⁄4
Quantity ⁄2 1
3
⁄2
1
17
5
M ⁄4 1
M ⁄2 5
Printout for Alternate Example 9-3 A minimal, optimum cost of 17 can be achieved by using 1 of a type #1 and C\x of a type #2.
Simplex Tableau : 2 \Cj
3.000
9.000
0.000
0.000
x1
x2
s1
s2
Cb\
Basis
Bi
9.000
x2
4.000
0.000
1.000
0.500
0.500
3.000
x1
8.000
1.000
0.000
1.000
2.000
Zj
60.000
3.000
9.000
1.500
1.500
0.000
0.000
1.500
1.500
Cj Zj
Maximize Profit $3X1 $9X2 Subject to: 1X1 4X2 24 clay 1X1 2X2 16 glaze where X1 small vases made X2 large vases made
Final Optimal Solution
The optimal solution was X1 8, X2 4. Profit $60. Using software (see the printout to the left), we can perform a variety of sensitivity analyses on this solution.
Z 60.000 Variable
Value
Reduced Cost
x1
8.000
0.000
x2
4.000
0.000
Constraint
Slack/Surplus
Shadow Price
C1
0.000
1.500
C2
0.000
1.500
Alternative Example 9.4: Levine Micros assembles both laptop and desktop personal computers. Each laptop yields $160 in profit; each desktop $200. The firm’s LP primal is: Maximize profit $160X1 $200X2 subject to: 1X1 2X2 20 labor hours
Objective Coefficient Ranges Lower
Current
Variables
Limit
x1
2.250
x2
6.000
Upper
Allowable
Allowable
Values
Limit
Increase
Decrease
3.000
4.500
1.500
0.750
9.000
12.000
3.000
3.000
Right-Hand-Side Ranges Lower
Current
Upper
Allowable
Allowable
Limit
Values
Limit
Increase
Decrease
C1
16.000
24.000
32.000
8.000
8.000
C2
12.000
16.000
24.000
8.000
4.000
Constraints
Alternative Example 9.3: Referring back to Hal, in Alternative Example 7.1, we had a formulation of:
9X1 9X2 108 RAM chips 12X1 6X2 $120 royalty fees where X1 no. laptops assembled daily X2 no. desktops assembled daily
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Here is the primal optimal solution and final simplex tableau. Cjl b
Solution Mix
$160 X1
$200 X2
0 S1
0 S2
200 160 0
X2 X1 S3 Zj Cj Zj
0 1 0 160 0
1 0 0 200 0
1 1 6 40 40
1⁄9 2 ⁄9 2 131⁄3 131⁄3
0 S3 Quantity 0 0 1 0 0
8 4 24 $2,240
or X1 4, X2 8, S3 $24 in slack royalty fees paid Profit $2,240/day Here is the dual formulation: subject to: 1y1 9y2 12y3 160 2y1 9y2 6y3 200 Here is the dual optimal solution and final tableau. Solution Mix
20 y1
108 y2
120 y3
0 S1
0 S2
108 20
y2 y1 Zj Cj Zj
0 1 20 0
1 0 108 0
2 6 96 24
2⁄9 12⁄9 42⁄9 42⁄9
⁄9 1 8 8 1
Artificial variables have no physical meaning but are used with the constraints that are or . They carry a high coefficient, so they are quickly removed from the initial solution. 9-4. The number of basic variables (i.e., variables in the solution) is always equal to the number of constraints. So in this case there will be eight basic variables. A nonbasic variable is one that is not currently in the solution, that is, not listed in the solution mix column of the tableau. It should be noted that while there will be eight basic variables, the values of some of them may be zero. 9-5. Pivot column: Select the variable column with the largest positive Cj Zj value (in a maximization problem) or smallest negative Cj Zj value (in a minimization problem). Pivot row: Select the row with the smallest quantity-tocolumn ratio that is a nonnegative number. Pivot number: Defined to be at the intersection of the pivot column and pivot row.
Minimize Z 20y1 108y2 120y3
Cj l b
117
Quantity 131⁄3 40 $2,240
This means y1 marginal value of one more labor hour $40 y2 marginal value of one more RAM chip $13.33 y3 marginal value of one more $1 in royalty fees $0
SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 9-1. The purpose of the simplex method is to find the optimal solution to LP problems in a systematic and efficient manner. The procedures are described in detail in Section 9.6. 9-2. Differences between graphical and simplex methods: (1) Graphical method can be used only when two variables are in model; simplex can handle any dimensions. (2) Graphical method must evaluate all corner points (if the corner point method is used); simplex checks a lesser number of corners. (3) Simplex method can be automated and computerized. (4) Simplex method involves use of surplus, slack, and artificial variables but provides useful economic data as a by-product. Similarities: (1) Both methods find the optimal solution at a corner point. (2) Both methods require a feasible region and the same problem structure, that is, objective function and constraints. The graphical method is preferable when the problem has two variables and only two or three constraints (and when no computer is available). 9-3. Slack variables convert constraints into equalities for the simplex table. They represent a quantity of unused resource and have a zero coefficient in the objective function. Surplus variables convert constraints into equalities and represent a resource usage above the minimum required. They, too, have a zero coefficient in the objective function.
9-6. Maximization and minimization problems are quite similar in the application of the simplex method. Minimization problems usually include constraints necessitating artificial and surplus variables. In terms of technique, the Cj Zj row is the main difference. In maximization problems, the greatest positive Cj Zj indicates the new pivot column; in minimization problems, it’s the smallest negative Cj Zj. The Zj entry in the “quantity” column stands for profit contribution or cost, in maximization and minimization problems, respectively. 9-7. The Zj values indicate the opportunity cost of bringing one unit of a variable into the solution mix. 9-8. The Cj Zj value is the net change in the value of the objective function that would result from bringing one unit of the corresponding variable into the solution. 9-9. The minimum ratio criterion used to select the pivot row at each iteration is important because it gives the maximum number of units of the new variable that can enter the solution. By choosing the minimum ratio, we ensure feasibility at the next iteration. Without the rule, an infeasible solution may occur. 9-10. The variable with the largest objective function coefficient should enter as the first decision variable into the second tableau for a maximization problem. Hence X3 (with a value of $12) will enter first. In the minimization problem, the least-cost coefficient is X1, with a $2.5 objective coefficient. X1 will enter first. 9-11. If an artificial variable is in the final solution, the problem is infeasible. The person formulating the problem should look for the cause, usually conflicting constraints. 9-12. An optimal solution will still be reached if any positive Cj Zj value is chosen. This procedure will result in a better (more profitable) solution at each iteration, but it may take more iterations before the optimum is reached. 9-13. A shadow price is the value of one additional unit of a scarce resource. The solutions to the Ui dual variables are the primal’s shadow prices. In the primal, the negatives of the Cj Zj values in the slack variable columns are the shadow prices. 9-14.
The dual will have 8 constraints and 12 variables.
9-15. The right-hand-side values in the primal become the dual’s objective function coefficients.
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The primal objective function coefficients become the righthand-side values of dual constraints. The transpose of the primal constraint coefficients become the dual constraint coefficients, with constraint inequality signs reversed. 9-16. The student is to write his or her own LP primal problem of the form:
d. With the additional change, the optimal corner point in part B is still the optimal corner point. Profit doesn’t change. Once the right-hand side went beyond 240, another constraint prevented any additional profit, and there is now slack for the first constraint. 9-18.
a. See the table below. b. 14X1 4X2 3,360
maximize profit C1X1 C2X2
10X1 12X2 9,600
subject to A11X1 A12X2 B1
X1, X2 0
A21X1 A22X2 B2
c. Maximize profit 900X1 1,500X2
and for a dual of the nature:
d. Basis is S1 3,360, S2 9,600.
minimize cost B1U1 B2U2
e. X2 should enter basis next.
subject to A11U1 A21U2 C1 9-17.
S2 will leave next.
f.
A12U1 A22U2 C2
g. 800 units of X2 will be in the solution at the second tableau.
a.
h. Profit will increase by (Cj Zj)(units of variable entering the solution)
X2 60
(1,500)(800) 1,200,000 Table for Problem 9-18 Cjl b
Solution Mix
$900 X1
0
S1
14
0
S2
10
$0 S1
$0 S2
Quantity
4
1
0
3,360
12
0
1
9,600
0
0
0
0
0
900
1,500
0
0
Zj
20
Cj Zj
120
X1
b. The new optimal corner point is (0,60) and the profit is 7,200. c. The shadow price (increase in profit)/(increase in right-hand side value)
$1,500 X2
9-19. a. Maximize earnings 0.8X1 0.4X2 1.2X3 0.1X4 0S1 0S2 MA1 MA2 subject to X1 2X2 X3 5X4 S1 150
(7,200 2,400)/(240 80)
X2 4X3 8X4 A1 70
4,800/160
6X1 7X2 2X3 X4 S2 A2 120
30
c. S1 150, A1 70, A2 120, all other variables 0 Table for Problem 9-19b Solution Mix
0.8 X1
0.4 X2
1.2 X3
⫺0.1 X4
0 S1
0 S2
⫺M A1
⫺M A2
Quantity
0
S1
1
2
1
5
1
0
0
0
150
M
A1
0
1
4
8
0
0
1
0
70
M
A2
6
7
2
1
0
1
0
1
120
Zj
6M
8M
2M
7M
0
M
M
M
190M
Cj Zj
0.8 6M
0.4 8M
1.2 2M
0.1 7M
0
M
0
0
Cj l b
9-20.
First tableau:
Cj l Solution b Mix
$3 X1
$5 X2 0
1
$0 S1
$0 S2
1
0
Quantity
$0
S1
6
$0
S2
3
2
0
1
18
Zj
$0
$0
$0
$0
$0
Cj Zj
$3
$5
$0
$0
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Second tableau:
b.
Cj l Solution b Mix
$3 X1
$5 X2
$0 S1
$0 S2
Quantity
Cj l Solution b Mix
10 X1
8 X2
0 S1
0 S2
Quantity
$5
X2
0
1
1
0
6
0
S1
4
2
1
0
80
$0
S2
3
0
2
1
6
0
S2
1
2
0
1
50
Zj
0
0
0
0
0
Cj Zj
10
8
0
0
Zj
$0
$5
$5
$0
Cj Zj
$3
$0
$5
$0
$30
This represents the corner point (0,0).
Third and optimal tableau: Cj l Solution b Mix
$3 X1
c. The pivot column is the X1 column. The entering variable is X1.
$5 X2
$0 S1
$0 S2 0
6
Row 2: 50/1 50
⁄3
2
These represent the points (20,0) and (50,0) on the graph.
$36
$5
X2
0
1
1
$3
X1
1
0
2⁄3
Zj
$3
$5
$3
$1
Cj Zj
$0
$0
$3
$1
d. Ratios: Row 1: 80/4 20
Quantity
1
X1 2, X2 6, S1 0, S2 0, and profit $36
e. The smallest ratio is 20, so 20 units of the entering variable (X1) will be brought into the solution. If the largest ratio had been selected, the next tableau would represent an infeasible solution since the point (50,0) is outside the feasible region.
Graphical solution to Problem 9-20:
f.
9 Second Corner Point of Simplex
g. Second iteration
(Optimal Corner Point of Simplex) (X1 = 2, X2 = 6; Profit = $36) 6
b
Cj l Solution b Mix
c
X2
0 S2
Quantity
X1
1
0.5
0.25
0
20
S2
0
1.5 0.25
1
30
Zj
10
5
2.5
0
200
Cj Zj
0
3
2.5
0
a 3
6
9
8 X2
0 S1
0 S2
Quantity
X1
1
0
0.3333
0.3333
10
8
X2
0
1
0.1667
0.6667
20
Zj
10
8
2
2
260
Cj Zj
0
0
2
2
a.
h. The second iteration represents the corner point (20,0). The third (and final) iteration represents the point (10,20).
Constraints
9-22.
10, 20.
10 X1
10
X2 40
25
0 S1
0
Cj l Solution b Mix
X1
9-21.
8 X2
Third iteration
First Corner Point of Simplex
0
10 X1
10
3
0
The leaving variable is the solution mix variable in row with the smallest ratio. Thus, S1 is the leaving variable. The value of this will be 0 in the next tableau.
Basis for first tableau: A1 80 A2 75
Isoprofit line
(X1 0, X2 0, S1 0, S2 0) Second tableau:
A1 55 X1 25
20
X1
50
(X2 0, S1 0, S2 0, A2 0)
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Graphical solution to Problem 9-22:
b. The variable X2 has a Cj Zj value of $0, indicating an alternative optimal solution exists by inserting X2 into the basis. c. The alternative optimal solution is found in the tableau in the next column to be X1 C\m 0.42, X2 ZX\m 1.7, ROI $6. Tableau for Problem 9-25c
80 (X1 = 0, X2 = 75)
60
Cjl b X2 40
(X1 = 14, X2 = 33) (Optimal Solution)
Solution Mix
2 X1
3 X2
3
X2
0
1
2
X1
1
0
Zj
2
3
Cj Zj
0
0
20
0 S1 ⁄7
⁄21
1
⁄21 1
⫺M A1
0 S2 2
12
⁄7
3
⁄7 1
3
⁄3
0
0
1⁄3
0
M
1
Quantity
2⁄7
⁄7
⁄7
$6
(X1 = 80, X2 = 0)
d. The graphical solution is shown below. 0
3 0
20
40 X1
60
80
9X1 + 3X2 ≥ 9
X1 14
Third tableau:
X2 33
(X
2
(S1 0, S2 0, A1 0, A2 0) Cost 221 at optimal solution
1
9-24. At the second iteration, the following simplex tableau is found:
X2
6X1 + 9X2 ≤ 18 1
6 X1
3 X2
6
X1
1
1
1
0
S2
0
0
1
⁄2
1
2
Zj
6
6
3
0
6
Cj Zj
0
9
3
0
0 S1
0 S2
Quantity
⁄2
0
1
(X1 = 1, X2 = 0) 0
9-25. a. The optimal solution using simplex is X1 3, X2 0. ROI $6. This is illustrated in the problem’s final simplex tableau:
0 S1
0 S2
⫺M A1
Quantity
⁄2
1
1
6
⁄2
0
0
3
13⁄2
0
0
$6
0
M
0
S1
0
⁄2
3
2
X1
1
3
⁄2
1
Zj
2
3
Cj Zj
0
0
13⁄2
2
b 3
X1
9-26. This problem is degenerate. Variable X2 should enter the solution next. But the ratios are as follows:
5 X 3 row 5 1 X 1 row
12 unacceptable 3
S2 row
10 5 2
Tableau for Problem 9-25a
7
c 1
0
Alternative optimum at a and b, Z $6.
At this point, X2 should enter the basis next. But the two ratios are 1/1 negative and 2/0 undefined. Since there is no nonnegative ratio, the problem is unbounded.
3 X2
(X1 = 3, X2 = 0)
Feasible Region
Cj l Solution b Mix
2 X1
)
a
9-23. This problem is infeasible. All Cj Zj are zero or negative, but an artificial variable remains in the basis.
Cj l Solution b Mix
= 3/7, X2 = 12/7
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Since X3 and S2 are tied, we can select one at random, in this case S2. The optimal solution is shown below. It is X1 27, X2 5, X3 0, profit $177. Cj l Solution b Mix
6 X1
3 X2
5 X3
$5
X3
0
0
1
1
$6
X1
1
0
0
3
$3
X2
0
1
0
1
Zj
6
3
5
Cj Zj
0
0
0
9-27. MA2
0 S1
0 S2
⁄2
⁄2
0 S3
0
⁄2
⁄2
27
⁄2
1⁄2
⁄2
3
⁄2
1
13 ⁄2 3
Quantity
⁄2
1
7
1
8 ⁄2
5
13 ⁄2
3
$177
3
13 ⁄2 8 ⁄2 13 ⁄2 3
3
3
Minimum cost 50X1 10X2 75X3 0S1 MA1
subject to 1X1 1X2 0X3 0S1 1A1 0A2 1,000 0X1 2X2 2X3 0S1 0A1 1A2 2,000 1X1 0X2 0X3 1S1 0A1 0A2 1,500 First iteration: Cj l b
Solution Mix
50 X1
10 X2
75 X3
0 S1
M A1
M A2
Quantity
M
A1
1
1
0
0
1
0
1,000
M
A2
0
2
2
0
0
1
2,000
0
S1
1
0
0
1
0
0
1,500
Zj
M
M
2M
0
M
M
3,000M
Cj Zj
M 50
M 10
2M 75
0
0
0
Solution Mix
50 X1
10 X2
75 X3
0 S1
M A1
M A2
Quantity
M
A1
1
1
0
0
1
0
1,000
75
X3
0
1
1
0
0
0
S1
1
0
0
1
0
0
1,500
Zj
M
M 75
75
0
M
37 ⁄2
1,000M 75,000
Cj Zj
M 50
M 65
0
0
0
M 371⁄2
Second iteration: Cj l b
⁄2
1
1
1,000
Third iteration: Cj l b
Solution Mix
50 X1
10 X2
75 X3
0 S1
M A1
M A2
Quantity
50
X1
1
75
X3
0
1
0
0
1
0
1,000
1
1
0
0
0
S1
0
1
0
1
1
0
Zj
50
25
75
0
50
37 ⁄2
Cj Zj
0
15
0
0
M 50
M 371⁄2
⁄2
1
1
1,000 500 $125,000
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Fourth and final iteration: Cj l b
Solution Mix
50 X1
10 X2
75 X3
0 S1
M A1
M A2
Quantity
0
1,500
50
X1
1
0
0
1
0
75
X3
0
0
1
1
1
10
X2
0
1
0
1
1
0
Zj
50
10
75
15
65
37 ⁄2
Cj Zj
0
0
0
15
M 65
M 37 ⁄2
⁄2
500
1
500 $117,500
1
1
X1 1,500, X2 500, X3 500, Z $117,500 9-28.
X1 number of kilograms of brand A added to each batch X2 number of kilograms of brand B added to each batch Minimize costs 9X1 15X2 0S1 0S2 MA1 MA2 subject to X1 2X2 S1 A1 30 X1 4X2 S2 A2 80
Cj l b
Solution Mix
$9 X1
$15 X2
$0 S1
$0 S2
M A1
M A2
M
A1
1
2
1
0
1
0
30
M
A2
1
4
0
1
0
1
80 110M
Zj
2M
6M
M
M
M
M
Cj Zj
2M 9
6M 15
M
M
0
0
Quantity
First iteration: Cj l b
Solution Mix
15
X2
M
A2
$9 X1
$15 X2
$0 S1
$0 S2
⁄2
1
⁄2
0
1
0
2
1
⁄2 M
15
⁄2 2M
M
⁄2 M
0
⁄2 2M
M
1
Zj
15
Cj Zj
3
1
15
15
M A1
M A2
Quantity
⁄2
0
15
2
1
20
⁄2 2M
M
225 20M
3M ⁄2
0
1
15
15
Second iteration: Cj l b
Solution Mix
$9 X1
$15 X2
$0 S1
$0 S2
M A1
15
X2
⁄4
1
0
⁄4
0
1
0
S1
1⁄2
0
1
1⁄2
1
1
Zj
⁄4
15
4
⁄4
0
15
Cj Zj
⁄4
0
0
⁄4
M
1
15
21
Third and final iteration: X1 0 kg, X2 20 kg, cost $300 9-29.
X1 number of mattresses X2 number of box springs Minimize cost 20X1 24X2 subject to X1 X2 30 X1 2X3 40 X1, X2 0
1
15
15
M A2
Quantity
⁄4
20
⁄2
10
⁄4
$300
M ⁄4 15
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Initial tableau: Cj l b
Solution Mix
$20 X1
$24 X2
$0 S1
$0 S2
M A1
M A2
Quantity
M
A1
1
1
1
0
1
0
30
M
A2
1
2
0
1
0
1
40
Zj
2M
3M
M
M
M
M
70M
Cj Zj
2M 20
3M 24
M
M
0
0
Second tableau: Cj l b
Solution Mix
$20 X1
$24 X2
$0 S1
⁄2
0
1
⁄2
1
0
⁄2M 12
24
M
1⁄2M 12
0
M
$0 S1
$0 S2
M A1
M A2
Quantity
M
A1
1
$24
X2
1
Zj
1
Cj Zj
$0 S2
M A1
M A2
Quantity
⁄2
10
⁄2
20
⁄2
1
1⁄2
0
⁄2M 12
0
1⁄2M 12
0
1
1
1
1
⁄2M 12
10M 480
1
⁄2M 12
3
Final tableau: Cj l b
Solution Mix
$20 X1
$24 X2
$20
X1
1
0
2
1
2
1
20
$24
X2
0
1
1
1
1
1
10
Zj
20
24
16
4
16
4
$640
Cj Zj
0
0
16
4
M 16
M4
X1 20, X2 10, cost $640 9-30.
Maximize profit 9X1 12X2 subject to X1 X2 10 X1 2X2 12 X1, X2 0
Initial tableau: Cj l Solution b Mix
$9 X1
$12 X2
$0 S1
$0 S2
Quantity
$0
S1
1
1
1
0
10
$0
S2
1
2
0
1
12
Zj
0
0
0
0
$0
Cj Zj
9
12
0
0
Final tableau: Cj l Solution b Mix
$9 X1
$12 X2
$0 S1
$0 S2
⁄2
0
1
1⁄2
4
Quantity
$12 X2
$0 S1
$0 S2
0
2
1
Quantity
$4
X1
1
$12
X2
0
1
1
1
2
Zj
9
12
6
3
$96
Cj Zj
0
0
6
3
Second tableau: Cj l Solution b Mix
$9 X1
X1 8, X2 2, profit $96
$0
S1
1
$12
X2
1
⁄2
1
0
1⁄2
6
6
12
0
6
subject to 2X1 X2 3X3 120
Zj
$72
Cj Zj
0
0
6
2X1 6X2 4X3 240
3
9-31.
Maximize profit 8X1 6X2 14X3
X1, X2 0
8
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Initial tableau: Solution Mix
$8 X1
$6 X2
$14 X3
0 S1
⫺M A1
Quantity
0
S1
2
1
3
1
0
120
M
A2
2
6
4
0
1
240 240M
Cj l b
Zj
2M
6M
4M
0
M
Cj Zj
8 2M
6 6M
14 4M
0
0
Second tableau: Cj l b
Solution Mix
$14 X3
0 S1
⫺M A1
⁄3
1
1⁄6 ⁄6
$8 X1
$6 X2
⁄3
0
7
Quantity
$0
S1
5
$6
X2
1
⁄3
1
2
80
⁄3
0
Zj
2
6
4
0
1
Cj Zj
6
0
10
0
M 1
40
1
$240
Final tableau: Cj l b
Solution Mix
$8 X1
$6 X2
$14 X3
0 S1
⫺M A1
$14
X3
⁄7
0
1
$6
X2
1⁄ 7
1
Zj
⁄7
Cj Zj
1.1
5
64
X 1 0, X 2
Quantity
⁄7
1
⁄14
240
0
2⁄ 7
3⁄14
120
6
14
⁄7
⁄7
0
0
⁄7
M ⁄ 7
3
30
⁄7 ⁄7
$582 ⁄ 7
2
30
6
2
120 240 , X3 , profit = $582 N\m 7 7
(which is X1 0, X2 17.14, X3 34.29, profit $582.86) 9-32.
b. Maximize profit 8,000X1 6,000X2 5,000X3 3,500X4 0S1 0S2 0S3 0S4 0S5 0S6 0S7 0S8 MA1 MA2 subject to
a. X1 number of deluxe one-bedroom units converted X2 number of regular one-bedroom units converted X3 number of deluxe studios converted
1,100X1 1,000X2 600X30, 500X4 S1
35,000
700X10, 600X20, 400X30, 300X4 S2
28,000
2,000X1 1,600X2 1,200X3 900X4 S3
45,000
1,000X1 400X20, 900X30, 200X4 S4
19,000
X4 number of efficiencies converted Objective: maximum profit 8,000X1 6,000X2 5,000X3 3,500X4 subject to 1,100X1 1,000X2 600X30, 500X4 $35,000 700X10, 600X20, 400X30, 300X4 $28,000 2,000X1 1,600X2 1,200X3 900X4 $45,000 1,000X1 400X20, 900X30, 200X4 $19,000 X1 X2
X3
X4
50
X1 X2
X3
X4
25
X1 X2 0.40(X1 X2 X3 X4) X1 X2 0.70(X1 X2 X3 X4) The above constraints can be rewritten as: 0.3X1 0.3X2 0.7X3 0.7X4 0
a
0.6X1 0.6X2 0.4X3 0.4X4 0
9-33.
X1 X2
X3
X4
S5
50
X1 X2
X3
X4
S6 A1 25
0.6X1 0.6X2 0.4X3 0.4X4 S7 A2
0
0.3X1 0.3X2 0.7X3 0.7X4 S8 A2
0
a. The initial formulation is
minimize cost $12X1 18X2 10X3 20X4 7X5 8X6 subject to 3X3
X1
100
25X2 X3 2X4 8X5 2X1
X2
4X4
18X1 15X2 2X3 X4 15X5
900
X6 250 150 25X6 300
2X4 6X5
70
b. Variable X5 will enter the basis next. (Its Cj Zj value is the smallest negative number, that is, 21M 7.) Variable A3 will leave the basis because its ratio (150/15) is the smallest of the three positive ratios.
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9-34. a. We change $10 (the Cj coefficient for X1) to $10 and note the effect on the Cj Zj row in the table below. Simplex table for Problem 9-34 Solution Mix
$10 ⫹ ⌬ X1
$30 X2
$0 S1
$0 S2
Quantity
$10
X1
1
4
2
0
160
$0
S2
0
6
7
1
200
Zj
10
40 4
20 2
0
$1,600 160
Cj Zj
0
10 4
20 2
0
Cj l b
From the X2 column, we require for optimality that 10 4 0
or
21⁄2
From the S1 column, we require that 20 2 0
or
10
Since the 2 ⁄2 is more binding, the range of optimality is 1
$71⁄2 Cj (for X1) 앝 b.
The range of insignificance is 앝 Cj (for X2) $40
c. One more unit of the first scarce resource is worth $20, which is the shadow price in the S1 column. d. Another unit of the second resource is worth $0 because there are still 200 unused units (S2 200). e. This change is within the range of insignificance, so the optimal solution would not change. If the 30 in the Cj row were changed to 35, the Cj Zj would still be positive, and the current solution would still be optimal. f. The solution mix variables and their values would not change, because $12 is within the range of optimality found in part a. The profit would increase by 160(2) 320, so the new maximum profit would be 1,600 320 1,920. g. The right-hand side could be decreased by 200 (the amount of the slack) and the profit would not change. 9-35. a. The shadow prices are: 3.75 for constraint 1; 22.5 for constraint 2; and 0 for constraint 3. The shadow price is 0 for constraint 3 because there is slack for this constraint. This means there are units of this resource that are available but are not being utilized. Therefore, additional units of this could not increase profits. b. Dividing the RHS values by the coefficients in the S1 column, we have 37.5/0.125 300 so we can reduce the right-hand-side by 300 units; and 12.5/(0.125) 100, so we can increase the right-hand-side by 100 units and the same variables will remain in the solution mix. c. The right-hand-side of this constraint could be decreased by 10 units. The solution mix variable in this row is slack variable S3. Thus, the right-hand-side can be decreased by this amount without changing the solution mix. 9-36.
a. Produce 18 of model 102 and four of model H23. b. S1 represents slack time on the soldering machine; S2 represents available time in the inspection department. c. Yes—the shadow price of the soldering machine time is $4. Clapper will net $1.50 for every additional hour he rents.
d. No—the profit added for each additional hour of inspection time made available is only $1. Since this shadow price is less than the $1.75 per hour cost, Clapper will lower his profit by hiring the part-timer. 9-37. a. The first shadow price (in the S1 column) is $5.00. The second shadow price (in the S2 column) is $15.00. b. The first shadow price represents the value of one more hour in the painting department. The second represents the value of one additional hour in the carpentry department. c. The range of optimality for tables (X1) is established from Table 9-37c on the next page. 5 15 0
or
3.333 from S1 column
15 5 0
or
30 from S2 column
Hence the Cj for X1 must decrease by at least $3.33 to change the optimal solution. It must increase by $30 to alter the basis. The range of optimality is $66.67 Cj $100.00 for X1. d. The range of optimality for X2. See Table 9-37d. 5 2 0 15 0
or or
2.5 from S1 column 5 from S2 column
The range of optimality for profit coefficient on chairs is from $35 ( 50 15) to $52.50 ( 50 2.5). e. Ranging for first resource—painting department Quantity 30 40
S1
Ratio
⁄2
20
2
20
3
Thus the first resource can be reduced by 20 hours or increased by 20 hours without affecting the solution. The range is from 80 to 120 hours. f. Ranging for second resource—carpentry time. Quantity
S2
Ratio
30
1
⁄2
60
40
1
40
Range is thus from 200 hours to 300 hours (or 240 40 to 240 60).
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Table for Problem 9-37c Solution Mix
70 ⫹ ⌬ X1
50 X2
70
X1
1
0
50
X2
0
1
Zj
70
Cj Zj
0
Solution Mix
70
X1
50
X2
0
1
2
1
40
Zj
70
50
5 2
15
$4,100 40
Cj Zj
0
0
5 2
15
Cj l b
0 S1
0 S2
Quantity
⁄2
1⁄2
30
2
1
40
50
5 ⁄2
15 ⁄2
0
5 3⁄2
15 1⁄2
70 X1
50 ⫹ ⌬ X2
0 S1
0 S2
Quantity
1
0
⁄2
⁄2
30
3
3
1
$4,100 30
Table for Problem 9-37d Cj l b
3
1
9-38. Note that artificial variables may be omitted from the sensitivity analysis since they have no physical meaning. a. Range of optimality for X1 (phosphate): Solution Mix
$5 ⫹ ⌬ X1
$6 X2
$0 S1
$0 S2
Quantity
$0
S2
0
0
1
1
550
$5
X1
1
0
1
0
300
$6
X2
0
1
1
0
700
Zj
5
6
1
0
$5,700 300
Cj Zj
0
0
1
0
Cj l b
1 0
or
1
If the Cj value for X1 increases by $1, the basis will change. Hence 앝 Cj (for X1) $6. Range of optimality for X2 (potassium): Solution Mix
5 X1
6⫹⌬ X2
0 S1
0 S2
Quantity
0
S2
0
0
1
1
550
5
X1
1
0
1
0
300
6
X2
0
1
1
0
700
Zj
5
6
1
0
$5,700 700
Cj Zj
0
0
1
0
Cj l b
1 0
or
1
If the Cj value for X2 decreases by $1, the basis will change. The range is thus $5 Cj (for X2) 앝. b. This involves right-hand-side ranging on the slack variables S1 (which represents number of pounds of phosphate under the 300-pound limit). Quantity
S2
Ratio
550
1
550
300
1
300
700
1
700
This indicates that the limit may be reduced by 300 pounds (down to zero pounds) without changing the solution. The question asks if the resources can be increased to 400 pounds without affecting the basis. The smallest negative ratio (550) tells us that the limit can be raised to 850 pounds without changing the solution mix. However, the values of X1, X2, and S2 would change. X1 would now be 400, X2 would be 600, and S2 would be 450. This is best seen graphically in Figure 9.3.
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9-39.
LINEAR PROGRAMMING: THE SIMPLEX METHOD
Thus $435.85 is the maximum the laboratory should be willing to pay an outside resource to conduct the 120 test 1’s, 115 test 2’s, and 116 test 3’s per day. 8U1 4U2 9U3 is the value of 8, 4, and 9 of tests 1, 2, and 3, respectively, performed per hour by a biochemist. This means that the prices U1, U2, and U3 need to be such that their total value does not exceed the cost per hour to the lab for using one of its own biochemists. Similarly, 4U1 6U2 4U3 is the value of 4, 6, and 4 of tests 1, 2, and 3, respectively, performed per hour by a biophysicist. Again, the prices U1, U2, and U3 need to be such that the total value does not exceed the cost per hour for the lab to use one of its own biophysicists.
Minimize cost 4U1 8U2 subject to 1U1 2U2 80 3U1 5U2 75 U1, U2 0
The dual of the dual is the original primal. 9-40.
Maximize profit 50U1 4U2 subject to 12U1 1U2 120 20U1 3U2 250 U1, U2 0
9-41.
U1 $80, U2 $40, cost $1,000
9-42.
Primal objective function:
9-46. a. There are 8 variables (2 decision variables, 3 surplus variables, and 3 artificial variables) and 3 constraints. b. The dual would have 2 constraints and 5 variables (3 decision variables and 2 slack variables). c. The dual problem would be smaller and easier to solve.
maximize profit 0.5X1 0.4X2 primal constraints: 2X1 1X2 120 2X1 3X2 240 X1, X2 0 primal solution: X1 30, X2 60, profit $39 9-43.
Maximize profit 10X1 5X2 31X3 28X4 17X5 subject to
X1 X2
12X5 28
2X2 2X3
53 5X4 2X5
X2 5X3
X1
X5
70 18
X1, X2, X3, X4, X5 0 9-44. a. Machine 3, as represented by slack variable S3, still has 62 hours of unused time. b. There is no unused time when the optimal solution is reached. All three slack variables have been removed from the basis and have zero values. c. The shadow price of the third machine is the value of the dual variable in column 6. Hence an extra hour of time on machine 3 is worth $0.265. d. For each extra hour of time made available at no cost on machine 2, profit will increase by $0.786. Thus 10 hours of time will be worth $7.86. 9-45.
The dual is maximize Z 120U1 115U2 116U3 subject to
8U1
4U2 9U3 23
4U1
6U2 4U3 18 U1, U2, U3 0
U1 $2.07 is the price of each test 1 U2 $1.63 is the price of each test 2 U3 $0
is the price of each test 3
Using the dual objective function: Z 120U1 115U2 116U3 120(2.07) 115(1.63) 116(0) $248.4 $187.45 $0 $435.85
127
9-47. a. X1 27.38 tables, X2 37.18 chairs daily, profit $3775.78. b. Not all resources are used. Shadow prices indicate that carpentry hours and painting hours are not fully used. Also, the 40-table maximum is not reached. c. The shadow prices relate to the five constraints: $0 value to making more carpentry and painting time available; $63.38 is the value of additional inspection/rework hours; $1.20 is the value of each additional foot of lumber made available. d. More lumber should be purchased if it costs less than the $1.20 shadow price. More carpenters are not needed at any price. e. Flair has a slack (X4) of 8.056 hours available daily in the painting department. It can spare this amount. f. Carpentry hours range: 221 to infinity. Painting hours range: 92 to infinity. Inspection/rework hours range: 19Z\x to 41. g. Table profit range: $41.67 to $160 Chair profit range: $21.87 to $84. 9-48. Printout 1 on the right illustrates the model formulation (see the next page). a. Printout 2 provides the optimal solution of $9,683. Only the first product (A158) is not produced. b. Printout 2 also lists the shadow prices. The first, for example, deals with steel alloy. The value of one more pound is $2.71. c. There is no value to adding more workers, since all 1,000 hours are not yet consumed. d. Two tons of steel at a total cost of $8,000 implies a cost per pound of $2.00. It should be purchased since the shadow price is $2.71. e. Printout 3 (also on the next page) illustrates that profit declines to $8,866 with the change to $8.88. f. Printout 4 (on page 129) shows the new constraints. Profit drops to $9,380, and none of the products remain. Previously, only A158 was not produced.
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Printout 1 for Problem 9-48
Printout 2 for Problem 9-48 ***** Program Output *****
Problem Title: DATASET PROBLEM 9-48
Final Optimal Solution at Simplex Tableau : 18
***** Input Data *****
Z $9,683.228
Max. Z 18.79X1 6.31X2 8.19X3 45.88X4 63.00X5 4.10X6 81.15X7 50.06X8 12.79X9 15.88X10 17.91X11 49.99X12 24.00X13 88.88X14 77.01X15
Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
Subject to C1 4X2 6X3 10X4 12X5 10X7 5X8 1X9 1X10 2X12 10X14 10X15 980 C2 .4X1 .5X2 .4X4 1.2X5 1.4X6 1.4X7 1.0X8 .4X9 .3X10 .2X11 1.8X12 2.7X13 1.1X14 400 C3 .7X1 1.8X2 1.5X3 2.0X4 1.2X5 1.5X6 7.0X7 5.0X8 1.5X12 5.0X13 5.8X14 6.2X15 600 C4 5.8X1 10.3X2 1.1X3 8.1X5 7.1X6 6.2X7 7.3X8 10X9 11X10 12.5X11 13.1X12 15X15 2500 C5 10.9X1 2X2 2.3X3 4.9X5 10X6 11.1X7 12.4X8 5.2X9 6.1X10 7.7X11 5X12 2.1X13 1X15 1800 C6 3.1X1 1X2 1.2X3 4.8X4 5.5X5 .8X6 9.1X7 4.8X8 1.9X9 1.4X10 1X11 5.1X12 3.1X13 7.7X14 6.6X15 1000 C7 1X1 0 C8 1X2 20 C9 1X3 10 C10 1X4 10 C11 1X5 0 C12 1X6 20 C13 1X7 10 C14 1X8 20 C15 1X9 50 C16 1X10 20 C17 1X11 20 C18 1X12 10 C19 1X13 20 C20 1X14 10 C21 1X15 10
(d) Cost is $2.00/lb for more steel; we should do it.
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS 9-49.
Maximize 20X1 10X2 0S1 0S2 Subject to: 5X1 4X2 S1 250 2X1 5X2 S2 150 X1, X2 0
Value
Reduced Cost
0.000 20.000 10.000 10.000 11.507 20.000 10.000 20.000 50.000 20.000 20.000 54.946 20.000 12.202 10.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
Constraint
Slack/Surplus
Shadow Price
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 C21
0.000 113.866 0.000 0.000 258.885 8.530 0.000 0.000 0.000 0.000 11.507 0.000 0.000 0.000 0.000 0.000 0.000 44.946 0.000 2.202 0.000
2.712 0.000 10.649 2.183 0.000 0.000 1.324 46.187 26.455 2.535 0.000 27.370 34.041 32.676 11.749 10.842 9.374 0.000 29.243 0.000 48.870
Cj l Solution b Mix
20 X1
10 X2
0 S1
0 S2
Quantity
0
S1
5
4
1
0
250
0
S2
2
5
0
1
150
Zj
0
0
0
0
0
Cj Zj
20
10
0
0
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Printout 3 for Problem 9-48 Problem Title: DATASET PROBLEM 9-48 ***** Input Data ***** Max. Z 18.79X1 6.31X2 8.19X3 45.88X4 63.00X5 4.10X6 81.15X7 50.06X8 12.79X9 15.88X10 17.91X11 49.99X12 24.00X13 8.88X14 77.01X15
129
LINEAR PROGRAMMING: THE SIMPLEX METHOD
9-50. The shadow prices are 3/10 for constraint 1; 0 for constraint 2; and 3 for constraint 3. A zero shadow price means that additional units of that resource will not affect profit. This occurs because there is slack available. In this problem, constraint 2 has 425 units of slack (S2 425), so additional units of this resource would simply increase the slack. 9-51.
a. Maximize 10X1 8X2 Subject to: 2X1 1X2 24 2X1 4X2 36 X1, X2 0 S1 24; S2 26; X1 0; X2 0. Profit 0.
b. c.
***** Program Output ***** Final Optimal Solution At Simplex Tableau
Cj l Solution b Mix
Z $8865.500 Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
0 S1
0 S2
Quantity
0
S1
2
1
1
0
24
0.000 20.000 10.000 16.993 7.056 20.000 10.000 20.000 50.000 20.000 20.000 57.698 20.000 10.000 10.000
0
S2
2
4
0
1
36
Zj
0
0
0
0
0
Cj Zj
10
8
0
0
The pivot column is the X1 column. d. Variable X1 will enter the solution mix. Profit will increase $10 for each unit of this that is brought into the solution. e. ratio for row 1 24/2 12; ratio for row 2 36/2 18. The pivot row is row 1 (it has the smallest ratio). f. The variable in the pivot row will leave the solution mix. This is S1. g. The ratio for the pivot row is 12, so 12 units of X1 will be in the next solution. h. The total profit will increase by ($10 per unit) (12 units) $120. 9-52. a. Maximize profit 20X1 30X2 15X3 0S1 0S2 MA2 MA3 Subject to: 3X1 5X2 2X3 S1 120
Final Optimal Solution at Simplex Tableau : 21
2X1 X2 2X3 S2 A2 250 X1 X2 X3 A3 180
Z $9,380.234
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
8 X2
Value
Printout 4 for Problem 9-48
Variable
10 X1
Value 0.000 0.000 0.000 0.000 28.723 20.000 10.000 37.517 50.000 20.000 33.941 37.485 20.000 10.000 10.277
b. S1 120; A2 250; Profit 430M. 9-53.
X1, X2, X3 0 A3 180; all
others 0.
a. S1 12; X2 16; X1 4; all others 0. b. The dual prices are 0 for constraint 1 (department A), 3 for constraint 2 (department B), and 4.5 for constraint 3 (department C). c. The company would be willing to pay up to the dual price for additional hours. This is $0 for department A, $3 for department B, and $4.50 for department C. d. The profit on product #3 would have to increase by $1 (the negative of the Cj Zj value).
