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9
A
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Linear Programming: The Simplex Method
TEACHING SUGGESTIONS Teaching Suggestion 9.1: Meaning 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.
1st Iteration C j l b 0 0
Teaching Suggestion 9.2: Initial 9.2: Initial Solutions to LP Problems. Explain that all initial solutions begin with X 1 0, X 2 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: 9.3: Substitution 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. substitution. Teaching Suggestion 9.4: Hand 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. Encourag Encouragee students students to also also solve solve the assigned assigned proble problem m by computer and to request the detailed simplex output. They can now check their work at each iteration. 2. Stress Stress the importanc importancee of interpret interpreting ing the the numbers 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 9.5: Infeasibility Is a Major Problem in Large LP Problems. 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 Simplex Solution to Alternative Example 7.1 (see Chapter 7 of Solutions Manual Manual for formulation and graphical solution). Alternative Example 9.1:
Solution Mix
3 X 1
9 X 2
0 S1
0 S2
S 1 S 2
1 1
4 2
1 0
0 1
24 16
Z j C j Z j
0 3
0 9
0 0
0 0
0
Solution Mix
3 X 1
9 X 2
0 S1
0 S2
X 2 S 2
⁄ 4 1 ⁄ 2
1 0
1 ⁄ 2
Z j C j Z j
9 ⁄ 4 3 ⁄ 4
9 0
9 ⁄ 4
Quantity
2nd Iteration C j l b 9 0
1
⁄ 4 1
9 ⁄ 4
Quantity
0 1
6 4
0 0
54
This is not an optimum solution since the X 1 column contains a positive value. More profit remains ($ C\v per #1).
3rd/Final Iteration C j l Solution b Mix
3 X 1
9 X 2
9 3
X 2 X 1
0 1
1 0
1 ⁄ 2 3 1 ⁄ 2
Z j C j Z j
3 0
9 0
0 S1
0 S2 1 ⁄ 2
Quantity
3 2 ⁄ 2
4 8
3 ⁄ 2
3 ⁄ 2
60
3 ⁄ 2
3 ⁄ 2
This is an optimum solution since there are no positive values in the C j Z j row. This says to make 4 of item #2 and 8 of item #1 to get a profit of $60. Set up an initial simplex tableau, given the following two constraints and objective function: Alternative Example 9.2:
Minimize Z 8 X 1 6 X 2 Subject to:
2 X 1 4 X 2 8
3 X 1 2 X 2 6 The constraints and objective function may be rewritten as: Minimize 8 X 1 6 X 2 0S 1 0S 2 MA1 MA2 2 X 1 4 X 2 1S 1 0S 2 1 A1 0 A2 8 3 X 1 2 X 2 0S 1 1S 2 0 A1 1 A2 6 115
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The first tableau would be: C j l b
Solution Mix
8 X 1
6 X 2
0 S1
0 S2
M A1
M A2
0
Quantity
A1 A2
2 3
4 2
1
0
1
1 0
0 1
8 6
Z j
5M
6M
M
M
M
M
14M
C j Z j
8 5M
6 6M
M
M
0
0
M M
The second tableau: C j l b
Solution Mix
8 X 1
6 X 2
0 S1
X 2 A2
⁄ 2 2
1 0
Z j
3 2M
6
3 1 ⁄ 2 ⁄ 2M
C j Z j
5 2M
0
⁄ 2 ⁄ 2M
6 M
1
0 S2
M A1
⁄ 4
0
⁄ 4
1 ⁄ 2
1
1 ⁄ 2
0 1
2 2
M
3 1 ⁄ 2 ⁄ 2M
M
12 2M
⁄ 2 ⁄ 2M
0
1
3
1
M A2
1
3
M
3
Quantity
The third and final tableau: C j l Solution b Mix
8 X 1
6 X 2
0 S1
6 8
X 2 X 1
0 1
1 0
⁄ 8
⁄ 4
⁄ 8
⁄ 4
1 ⁄ 4
1 ⁄ 2
1 ⁄ 4
1 ⁄ 2
Z j
8
6
1 ⁄ 4
5 ⁄ 2
1 ⁄ 4
5 ⁄ 2
C j Z j
0
0
⁄ 4
⁄ 2
3
1
0 S2
M A1
1
M A2
3
M ⁄ 4
5
1
Quantity 3 ⁄ 2 1
1
17
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 3.000
9.000
0.000
0.000
Cb\
\Cj Basis
Bi
x1
x2
s1
s2
9.000
x2
4.000
0.000
1.000
0.500
0.500
3.000
x1
8.000
1.000
0.000
1.000
60.000
3.000
9.000
0.000
0.000
Zj Cj
Zj
1.500
2.000
1.500
Value
x 1
8.000
0.000
4.000
0.000
C on st ra in t
S la ck /S ur pl us
S ha do w P ri ce
C1
0.000
1.500
C2
0.000
1.500
1 X 1 2 X 2 16 glaze where X 1 small vases made
Levine Micros assembles both laptop and desktop personal computers. Each laptop yields $160 in profit; each desktop $200. The firm’s LP primal is: Alternative Example 9.4:
Maximize profit $160 X 1 $200 X 2 subject to:
Objective Coefficient Ranges
1 X 1 2 X 2 20 labor hours
Lower
Current
Upper
Allowable
Allowable
9 X 1 9 X 2 108 RAM chips
Limit
Values
Limit
Increase
Decrease
12 X 1 6 X 2 $120 royalty fees
x 1
2.250
3.000
4.500
1.500
0.750
x 2
6.000
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
1 X 1 4 X 2 24 clay
The optimal solution was X 1 8, X 2 4. Profit $60. Using software (see the printout to the left), we can perform a variety of sensitivity analyses on this solution.
Reduced Cost
x 2
Variables
Subject to:
X 2 large vases made
60.000
Variable
Maximize Profit $3 X 1 $9 X 2
1.500
1.500
Final Optimal Solution Z
Referring back to Hal, in Alternative Example 7.1, we had a formulation of: Alternative Example 9.3:
where X 1 no. laptops assembled daily X 2 no. desktops assembled daily
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Here is the primal optimal solution and final simplex tableau. C j l b
Solution Mix
$160 X 1
$200 X 2
0 S1
200 160 0
X 2 X 1 S 3 Z j C j Z j
0 1 0 160 0
1 0 0 200 0
1 1 6 40 40
0 S2
0 S3 Quantity
1 ⁄ 9 2 ⁄ 9
2 1 13 ⁄ 3 1 13 ⁄ 3
0 0 1 0 0
8 4 24 $2,240
or X 1 4, X 2 8, S 3 $24 in slack royalty fees paid Profit $2,240/day Here is the dual formulation: 1 y1 9 y2 12 y3 160 2 y1 9 y2 6 y3 200 Here is the dual optimal solution and final tableau. C j l b
Solution Mix
20 y 1
108 y 2
108 20
y 2 y 1 Z j
0 1 20 0
1 0 108 0
C j Z j
120 y 3
0 S1
2
2 ⁄ 9
6
2 1 ⁄ 9 2 4 ⁄ 9 2 4 ⁄ 9
96 24
0 S2 Quantity 1 ⁄ 9
1 8
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 C j Z j value (in a maximization problem) or smallest negative C j Z j 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 20 y1 108 y2 120 y3 subject to:
117
1 13 ⁄ 3 40 $ 2, 24 0
8
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 C j Z j row is the main difference. In maximization problems, the greatest positive C j Z j indicates the new pivot column; in minimization problems, it’s the smallest negative C j Z j. The Z j entry in the “quantity” column stands for profit contribution or cost, in maximization and minimization problems, respectively. 9-7. The Z j values indicate the opportunity cost of bringing one unit of a variable into the solution mix. 9-8. The C j Z j value is the net change in the value of the ob jective 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.
The variable with the largest objective function coefficient should enter as the first decision variable into the second tableau for a maximization problem. Hence X 3 (with a value of $12) will enter first. In the minimization problem, the least-cost coefficient is X 1, with a $2.5 objective coefficient. X 1 will enter first. 9-10.
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. 9-11.
9-12. An optimal solution will still be reached if any positive C j Z j 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. A shadow price is the value of one additional unit of a scarce resource. The solutions to the U i dual variables are the primal’s shadow prices. In the primal, the negatives of the C j Z j values in the slack variable columns are the shadow prices. 9-13.
9-14.
The dual will have 8 constraints and 12 variables.
The right-hand-side values in the primal become the dual’s objective function coefficients. 9-15.
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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. The student is to write his or her own LP primal problem of the form: 9-16.
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. a. See the table below.
9-18.
b. 14 X 1 4 X 2 3,360
maximize profit C 1 X 1 C 2 X 2
10 X 1 12 X 2 9,600
A11 X 1 A12 X 2 B1
subject to
X 1, X 2 0
A21 X 1 A22 X 2 B2
c. Maximize profit 900 X 1 1,500 X 2
and for a dual of the nature:
d. Basis is S 1 3,360, S 2 9,600.
minimize cost B1U 1 B2U 2 A11U 1 A21U 2 C 1
subject to
A12U 1 A22U 2 C 2 9-17.
e.
X 2 should enter basis next.
f.
S 2 will leave next.
g. 800 units of X 2 will be in the solution at the second tableau.
a. X2
h. Profit will increase by (C j Z j)(units of variable entering the solution)
60
(1,500)(800)
1,200,000
Table for Problem 9-18 C j l b
20
Solution Mix
$900 X 1
0
S 1
14
4
1
0
3,360
0
S 2
10
12
0
1
9,600
Z j
0
0
0
0
0
900
1,500
0
0
120
X1
C j Z j
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)
(7,200
4,800/160
30
2,400)/(240
$1,500 X 2
$0 S1
$0 S2 Quantity
9-19. a. Maximize earnings 0.8 X 1 0.4 X 2 1.2 X 3 0.1 X 4 0S 1 0S 2 MA1 MA2
subject to X 1 2 X 2 X 3 5 X 4 S 1 150
80)
X 2 4 X 3 8 X 4 A1 70
6 X 1 7 X 2 2 X 3 X 4 S 2 A2 120 c.
S 1 150, A1 70, A2 120, all other variables 0
Table for Problem 9-19b C j l b
Solution Mix
0.8 X 1
0.4 X 2
1.2 X 3
0.1
0 S1
0 S2
M
M
X 4
A1
A2
0
S 1
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
0
M
M
M
190M
0
M
0
0
6M
Z j C j Z j
0.8 6M
8M
0.4 8M
2M
1.2 2M
7M 0.1
9-20.
7M
Quantity
First tableau:
C j l Solution b Mix
$3 X 1
$5 X 2
$0 S1
$0 S2 Quantity
$0
S 1
0
1
1
0
6
$0
S 2
3
2
0
1
18
Z j
$0
$0
$0
$0
$0
C j Z j
$3
$5
$0
$0
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Second tableau:
b.
C j l Solution b Mix
$3 X 1
$5 X 2
$0 S1
$0 S2 Quantity
C j l Solution b Mix
10 X 1
8 X 2
0 S1
0 S2 Quantity
$5
X 2
0
1
1
0
6
0
S 1
4
2
1
0
80
$0
S 2
3
0
2
1
6
0
S 2
1
2
0
1
50
Z j
$0
$5
$5
$0
$30
Z j
0
0
0
0
0
C j Z j
$3
$0
$5
$0
C j Z j
10
8
0
0
This represents the corner point (0,0).
Third and optimal tableau: C j l Solution b Mix
$3 X 1
$5 X 2
$0 S1
c. The pivot column is the X 1 column. The entering variable is X 1.
$0 S2 Quantity
d. Ratios: Row 1: 80/4 20
$5
X 2
0
1
1
0
6
Row 2: 50/1 50
$3
X 1
1
0
2 ⁄ 3
1 ⁄ 3
2
These represent the points (20,0) and (50,0) on the graph.
Z j
$3
$5
$3
$1
$36
$0
$0
$3
$1
C j Z j
e.
The smallest ratio is 20, so 20 units of the entering variable ( X 1) 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.
f.
The leaving variable is the solution mix variable in row with the smallest ratio. Thus, S 1 is the leaving variable. The value of this will be 0 in the next tableau.
X 1 2, X 2 6, S 1 0, S 2 0, and profit $36
Graphical solution to Problem 9-20: 9 Second Corner Point of Simplex
g. Second iteration
(Optimal Corner Point of Simplex) (X 1 = 2, X 2 = 6; Profit = $36) 6
b
C j l Solution b Mix
c
X 2
1
0.5
0.25
0
20
0
S 2
0
1.5
0.25
1
30
Z j
10
5
2.5
0
200
C j Z j
0
3
2.5
0
a
3
6
9
8 X 2
0 S1
0 S2 Quantity
X 1
1
0
0.3333
0.3333
10
8
X 2
0
1
0.1667
0.6667
20
Z j
10
8
2
2
260
C j Z j
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 X 1
10
X2 40
25
0 S2 Quantity
X 1
C j l Solution b Mix
X 1
9-21.
0 S1
Third iteration
First Corner Point of Simplex
0
8 X 2
10
3
0
10 X 1
Basis for first tableau: A1 80 A2 75
Isoprofit line
( X 1 0, X 2 0, S 1 0, S 2 0) Second tableau:
A1 55 X 1 25
20
X1
50
( X 2 0, S 1 0, S 2 0, A2 0)
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Graphical solution to Problem 9-22:
b. The variable X 2 has a C j Z j value of $0, indicating an alternative optimal solution exists by inserting X 2 into the basis. c. The alternative optimal solution is found in the tableau in the next column to be X 1 C\m 0.42, X 2 ZX\m 1.7, ROI $6. Tableau for Problem 9-25c
80 (X 1 = 0, X 2 = 75)
60
C j l b X 2
40
(X 1 = 14, X 2 = 33) (Optimal Solution)
20
Solution Mix
2 X 1
3 X 2
X 2
0
1
1 ⁄ 7
2 ⁄ 21
2 ⁄ 7
X 1
1
0
⁄ 21
⁄ 7
⁄ 7
Z j
2
3
⁄ 3
0
0
C j Z j
0
0
⁄ 3
0
M
3 2
(X 1 = 80, X 2 = 0)
d. 0
1
0 S2
M
A1 Quantity
1
1
1
20
40
60
$6
9X 1 + 3X 2 ≥ 9
X 1 14 X 2 33
(
2
X 1 = 3 / 7, X 2 = 12 / 7
)
a
Cost 221 at optimal solution This problem is infeasible. All C j Z j are zero or negative, but an artificial variable remains in the basis. 9-23.
X 2
At the second iteration, the following simplex tableau is
6X 1 + 9X 2 ≤ 18 1
found: C j l Solution b Mix
6 X 1
6
X 1
3 X 2
0 S1
0 S2
1
1
⁄ 2
0
S 2
0
Z j C j Z j
1
0
⁄ 2
1
2
6
6
3
0
6
0
9
3
0
Tableau for Problem 9-25a C j l Solution b Mix
2 X 1
3 X 2
0 S1
0 S2
M
0
S 1
0
7
⁄ 2
3
⁄ 2
1
1
⁄ 2
A1 Quantity 6
X 1
1
⁄ 2
0
0
3
Z j
2
3
3 1 ⁄ 2
0
0
$6
C j Z j
0
0
3 1 ⁄ 2
0
1
0
c
0
b
1
2
3
X 1
Alternative optimum at a and b, Z $6.
a. The optimal solution using simplex is X 1 3, X 2 0. ROI $6. This is illustrated in the problem’s final simplex tableau:
3
(X 1 = 1, X 2 = 0)
1
1
(X 1 = 3, X 2 = 0)
Feasible Region
Quantity
At this point, X 2 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.
2
⁄ 7 3
80
(S 1 0, S 2 0, A1 0, A2 0)
9-25.
⁄ 7
12
3 0
Third tableau:
0
3
The graphical solution is shown below.
X 1
9-24.
0 S1
M
This problem is degenerate. Variable X 2 should enter the solution next. But the ratios are as follows: 9-26.
X 3 row
X 1
row
S 2
row
5 1
12 3
5
unacceptable
10 2
5
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Since X 3 and S 2 are tied, we can select one at random, in this case S 2. The optimal solution is shown below. It is X 1 27, X 2 5, X 3 0, profit $177. C j l Solution b Mix
6 X 1
3 X 2
5 X 3
0 S1
$5
X 3
0
0
1
1 ⁄ 2
1 ⁄ 2
7 ⁄ 2
0
$6
X 1
1
0
0
3
⁄ 2
3
⁄ 2
⁄ 2
27
$3
X 2
0
1
0
1 ⁄ 2
1 ⁄ 2
1 ⁄ 2
Z j
6
3
5
C j Z j
0
0
0
9-27. MA2
Minimum cost
0 S2
13 ⁄ 2 3
0 S3 Quantity 1
8 ⁄ 2 3
5
13 ⁄ 2 $177 3
3 13 ⁄ 8 ⁄ 2 2 13 ⁄ 2 3
3
50 X 1 10 X 2 75 X 3 0S 1 MA1
subject to 1 X 1 1 X 2 0 X 3 0S 1 1 A1 0 A2 1,000 0 X 1 2 X 2 2 X 3 0S 1 0 A1 1 A2 2,000 1 X 1 0 X 2 0 X 3 1S 1 0 A1 0 A2 1,500 First iteration: C j l b
Solution Mix
50 X 1
10 X 2
75 X 3
0 S1
M A1
M A2
M
A1
1
1
0
0
1
0
1,000
M
A2
0
2
2
0
0
1
2,000
0
S 1
1
0
0
1
0
0
1,500
0
M
M
0
0
0
Z j C j Z j
M M
2M
M 50
M
10
2M
75
Quantity
3,000M
Second iteration: C j l b
Solution Mix
50 X 1
10 X 2
75 X 3
0 S1
M A1
M A2
A1
1
1
0
0
1
0
75
X 3
0
1
1
0
0
0
S 1
1
0
0
1
0
0
75
0
M
37 ⁄ 2
0
0
0
1 M 37 ⁄ 2
M
Z j C j Z j
M M
M
50
75
M 65
Quantity 1,000
⁄ 2
1
1,000 1,500
1
1,000M 75,000
Third iteration: C j l b
Solution Mix
50 X 1
10 X 2
75 X 3
0 S1
M A1
X 1
1
1
0
0
1
75
X 3
0
1
1
0
0
0
S 1
0
1
0
1
1
0
Z j
50
25
75
0
50
37 ⁄ 2
C j Z j
0
15
0
0
M 50
M 37 ⁄ 2
50
M A2
Quantity
0
1,000
⁄ 2
1
1,000 500
1
1
$125,000
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Fourth and final iteration: C j l b
Solution Mix
50 X 1
10 X 2
75 X 3
0 S1
M A1
M A2 0
50
X 1
1
0
0
1
0
75
X 3
0
0
1
1
1
10
X 2
0
1
0
1
1
0
Z j
50
10
75
15
65
37 ⁄ 2
C j Z j
0
0
0
15
M 65
M 37 ⁄ 2
Quantity 1,500
1 ⁄ 2
500 500
1
$117,500
1
X 1 1,500, X 2 500, X 3 500, Z $117,500 9-28.
X 1 number of kilograms of brand A added to each batch X 2 number of kilograms of brand B added to each batch
Minimize costs 9 X 1 15 X 2 0S 1 0S 2 MA1 MA2 subject to
X 1 2 X 2 S 1 A1 30 X 1 4 X 2 S 2 A2 80
C j l b
Solution Mix
$9 X 1
$15 X 2
$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
Z j
2M
6M
M
M
M
M
110M
M
M
0
0
C j Z j
2M
9
6M
15
Quantity
First iteration: Solution Mix
$9 X 1
$15 X 2
15
X 2
⁄ 2
1
M
A2
C j l b
Z j C j Z j
1
1
15
3 ⁄ 2 M
0
$0 S2
M A1
M A2
⁄ 2
0
⁄ 2
0
15
1
20
1
0
⁄ 2 M
15
$0 S1 2 15 ⁄ 2
1
1
2M
2
⁄ 2 2M
15
M
Quantity
⁄ 2 2M
M 225 20M
3M 15 ⁄ 2
0
15
M
Second iteration: C j l b
Solution Mix
$9 X 1
$15 X 2
$0 S1
$0 S2
M A1
M A2
15
X 2
0
S 1
⁄ 4
1
0
⁄ 4
1 ⁄ 2
0
1
1 ⁄ 2
Z j
4 15 ⁄
15
4
15 ⁄ 4
⁄ 4
0
0
1
C j Z j
21
Third and final iteration: X 1 0 kg, X 2 20 kg, cost $300 9-29.
X 1 number of mattresses X 2 number of box springs
Minimize cost 20 X 1 24 X 2 subject to
X 1 X 2 30 X 1 2 X 3 40 X 1, X 2 0
1
⁄ 4
15
Quantity
0
1
⁄ 4
20
1
1 ⁄ 2
10
0
15
M
⁄ 4
M ⁄ 4 15
$300
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Initial tableau: C j l b
Solution Mix
$20 X 1
$24 X 2
$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
M
M
M
M
70M
M
M
0
0
2M
Z j
2M
C j Z j
3M 20
3M
24
Second tableau: C j l b
Solution Mix
$20 X 1
$24 X 2
A1
1
⁄ 2
0
X 2
⁄ 2
1
Z j
⁄ 2M 12
M $24
$0 S2
M A1
M A2
1
⁄ 2
0
24
M
0
M
1
1
1 ⁄ 2M
C j Z j
12
$0 S1
Quantity
1
1
⁄ 2
10
⁄ 2
0
⁄ 2
20
⁄ 2M 12
0
1
1
1
1 ⁄ 2M
12
1
⁄ 2M 1
0
12
10M 480
3 ⁄ 2M 12
Final tableau: C j l b
Solution Mix
$20
X 1
1
0
$24
X 2
0
1
Z j
20
C j Z j
0
$20 X 1
$24 X 2
$0 S1
$0 S2
M A1
M A2
2
1
2
1
20
1
1
1
1
10
24
16
4
16
4
$640
0
16
4
M 16
M 4
Quantity
X 1 20, X 2 10, cost $640 9-30.
Maximize profit 9 X 1 12 X 2 subject to
X 1 X 2 10 X 1 2 X 2 12 X 1, X 2 0
Initial tableau: C j l Solution b Mix $0 $0
$9 X 1
$12 X 2
$0 S1
$0 S2
S 1
1
1
1
0
10
S 2
1
2
0
1
12
Z j
0
0
0
0
$0
9
12
0
0
C j Z j
Quantity
Final tableau: C j l Solution b Mix
$9 X 1
$12 X 2
$0 S1
$0 S2 Quantity
$0
S 1
1 ⁄ 2
0
1
1 ⁄ 2
4
$12
X 2
1 ⁄ 2
1
0
1 ⁄ 2
6
Z j
6
12
0
6
$72
C j Z j
3
0
0
6
$12 X 2
$0 S1
$0 S2
Quantity
$4
X 1
1
0
2
1
8
$12
X 2
0
1
1
1
2
Z j
9
12
6
3
$96
C j Z j
0
0
6
3
Second tableau: C j l Solution Mix b
$9 X 1
X 1 8, X 2 2, profit $96 9-31.
Maximize profit 8 X 1 6 X 2 14 X 3 subject to
2 X 1 X 2 3 X 3 120 2 X 1 6 X 2 4 X 3 240 X 1, X 2 0
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Initial tableau: C j l b
Solution Mix
$8 X 1
$6 X 2
$14 X 3
0 S1
M
S 1
2
1
3
1
0
0
A1
A2
2
6
4
0
1
Z j
2M
6M
4M
0
M
C j Z j
8 2M
6 6M
14 4M
0
0
M
Quantity 120 240 240M
Second tableau: C j l b
Solution Mix
$8 X 1
$6 X 2
$14 X 3
0 S1
M
$0
S 1
5
⁄ 3
0
7
⁄ 3
1
⁄ 6
80
$6
X 2
1 ⁄ 3
1
2 ⁄ 3
0
1 ⁄ 6
40
Z j
2
6
4
0
C j Z j
6
0
10
0
A1
Quantity
1
1 M
$240 1
Final tableau: C j l b
Solution Mix
$8 X 1
$6 X 2
$14 X 3
0 S1
M
$14
X 3
5 ⁄ 7
0
1
3 ⁄ 7
1 ⁄ 14
240
X 2
1
⁄ 7
1
0
2
⁄ 7
⁄ 14
120
⁄ 7
6
14
⁄ 7
⁄ 7
1.1
0
0
30 ⁄ 7
M
$6
64
Z j
C j Z j
X 1
0,
X2
120 , 7
X 3
A1
Quantity ⁄ 7 ⁄ 7
3
30
2
240 , profit = $582 7
2 ⁄ 7
N\m
(which is X 1 0, X 2 17.14, X 3 34.29, profit $582.86) 9-32.
6 $582 ⁄ 7
b. Maximize profit 8,000 X 1 6,000 X 2 5,000 X 3 3,500 X 4 0S 1 0S 2 0S 3 0S 4 0S 5 0S 6 0S 7 0S 8 MA1 MA2 subject to 1,100 X 1 1,000 X 2 600 X 30, 500 X 4 S 1
35,000
a.
700 X 10, 600 X 20, 400 X 30, 300 X 4 S 2
28,000
X 1 number of deluxe one-bedroom units converted
2,000 X 1 1,600 X 2 1,200 X 3 900 X 4 S 3
45,000
X 2 number of regular one-bedroom units converted
1,000 X 1 400 X 20, 900 X 30, 200 X 4 S 4
19,000
X 3 number of deluxe studios converted
X 1 X 2
X 3
X 4
S 5
50
X 4 number of efficiencies converted
X 1 X 2
X 3
X 4
S 6 A1
25
Objective: maximum profit 8,000 X 1 6,000 X 2 5,000 X 3 3,500 X 4 subject to 1,100 X 1 1,000 X 2 600 X 30, 500 X 4 $35,000 700 X 10, 600 X 20, 400 X 30, 300 X 4 $28,000 2,000 X 1 1,600 X 2 1,200 X 3 900 X 4 $45,000 1,000 X 1 400 X 20, 900 X 30, 200 X 4 $19,000 X 1 X 2
X 3
X 4
X 1 X 2
X 3
X 4
50 25
X 1 X 2 0.40( X 1 X 2 X 3 X 4) X 1 X 2 0.70( X 1 X 2 X 3 X 4)
The above constraints can be rewritten as: 0.6 X 1 0.6 X 2 0.4 X 3 0.4 X 4 0 0.3 X 1 0.3 X 2 0.7 X 3 0.7 X 4 0
a
9-33.
0.6 X 1 0.6 X 2 0.4 X 3 0.4 X 4 S 7 A2
0
0.3 X 1 0.3 X 2 0.7 X 3 0.7 X 4 S 8 A2
0
a.
The initial formulation is
minimize cost $12 X 1 18 X 2 10 X 3 20 X 4 7 X 5 8 X 6 subject to
X 1
25 X 2 2 X 1
3 X 3 X 3 2 X 4 8 X 5
X 2
18 X 1 15 X 2
2 X 3
4 X 4
X 4
15 X 5 6 X 5
100
900
X 6 250
25 X 6 2 X 4
150
300
70
b. Variable X 5 will enter the basis next. (Its C j Z j value is the smallest negative number, that is, 21 M 7.) Variable A3 will leave the basis because its ratio (150/15) is the smallest of the three positive ratios.
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a. We change $10 (the C j coefficient for X 1) to $10 and note the effect on the C j Z j row in the table below. 9-34.
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Simplex table for Problem 9-34 C j l b
Solution Mix
$10 X 1
$30 X 2
$0 S1
$0 S2
$10
X 1
1
4
2
0
160
$0
S 2
0
6
7
1
200
Z j
10
40
4
20
2
0
$1,600 160
C j Z j
0
10
4
20
2
0
From the X 2 column, we require for optimality that 10 4
0
or
0
or
10
Since the 2 ⁄ 2 is more binding, the range of optimality is 1
1 2 C $7 ⁄ j (for X 1)
b.
The range of insignificance is
9-35.
9-36.
9-37.
C j (for X 2) $40
c. One more unit of the first scarce resource is worth $20, which is the shadow price in the S 1 column. d. Another unit of the second resource is worth $0 because there are still 200 unused units ( S 2 200). e. This change is within the range of insignificance, so the optimal solution would not change. If the 30 in the C j row were changed to 35, the C j Z j 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. 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 S 1 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 S 3. Thus, the right-hand-side can be decreased by this amount without changing the solution mix. a. Produce 18 of model 102 and four of model H23. b. S 1 represents slack time on the soldering machine; S 2 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.
Quantity
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.
1 2 ⁄ 2
From the S 1 column, we require that 20 2
a. The first shadow price (in the S 1 column) is $5.00. The second shadow price (in the S 2 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 ( X 1) is established from Table 9-37c on the next page. 5
15 0
15 5
0
or
3.333
or
30
from S 1 column
from S 2 column
Hence the C j for X 1 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 C j $100.00 for X 1. d.
The range of optimality for X 2. See Table 9-37d. 5 2 15
0
0
or or
2.5
5
from S 1 column from S 2 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
S1
Ratio
30
3 ⁄ 2
20
40
2
20
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 60).
40 to 240
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Table for Problem 9-37c C j l b
Solution Mix
70 X 1
50 X 2
0 S1
0 S2
X 1
1
0
⁄ 2
⁄ 2
70 50
3
Quantity 30
1
X 2
0
1
2
1
Z j
70
50
3 5 ⁄ 2
1 15 ⁄ 2
C j Z j
0
0
Solution Mix
70 X 1
50 X 2
0 S1
0 S2
70
X 1
1
0
3 ⁄ 2
1 ⁄ 2
30
50
X 2
0
1
2
1
40
Z j
70
50
5 2
15
C j Z j
0
0
5
3 ⁄ 2
40
$4,100 30
1 15 ⁄ 2
Table for Problem 9-37d C j l b
5
2
Quantity
$4,100 40
15
Note that artificial variables may be omitted from the sensitivity analysis since they have no physical meaning. a. Range of optimality for X 1 (phosphate): 9-38.
C j l b
Solution Mix
$5 X 1
$6 X 2
$0 S1
$0 S2
$0
S 2
0
0
1
1
550
$5
X 1
1
0
1
0
300
$6
X 2
0
1
1
0
700
Z j
5
6
1
0
$5,700 300
C j Z j
0
0
1
0
10
or
Quantity
1
If the C j value for X 1 increases by $1, the basis will change. Hence C j (for X 1) $6.
Range of optimality for X 2 (potassium): C j l b
Solution Mix
5 X 1
6 X 2
0 S1
0 S2
0
S 2
0
0
1
1
550
5
X 1
1
0
1
0
300
6
10
Quantity
X 2
0
1
1
0
700
Z j
5
6
1
0
$5,700 700
C j Z j
0
0
1
0
or
1
If the C j value for X 2 decreases by $1, the basis will change. The range is thus $5 C j (for X 2) . b. This involves right-hand-side ranging on the slack variables S 1 (which represents number of pounds of phosphate under the 300-pound limit).
Quantity
S2
550
1
550
Ratio
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 X 1, X 2, and S 2 would change. X 1 would now be 400, X 2 would be 600, and S 2 would be 450. This is best seen graphically in Figure 9.3.
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127
9-41.
U 1 $80, U 2 $40, cost $1,000
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. 8U 1 4U 2 9U 3 is the value of 8, 4, a nd 9 of tests 1, 2, and 3, respectively, performed per hour by a biochemist. This means that the prices U 1, U 2, and U 3 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, 4U 1 6U 2 4U 3 is the value of 4, 6, and 4 of tests 1, 2, and 3, respectively, performed per hour by a biophysicist. Again, the prices U 1, U 2, and U 3 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.
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.
9-47.
a. X 1 27.38 tables, X 2 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 ( X 4) 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: 19 Z\x to 41. g. Table profit range: $41.67 to $160 Chair profit range: $21.87 to $84.
9-39.
Minimize cost 4U 1 8U 2 subject to
1U 1 2U 2 80 3U 1 5U 2 75 U 1, U 2 0
The dual of the dual is the original primal. 9-40.
Maximize profit 50U 1 4U 2 subject to
12U 1 1U 2 120 20U 1 3U 2 250 U 1, U 2 0
maximize profit 0.5 X 1 0.4 X 2 primal constraints: 2 X 1 1 X 2 120 2 X 1 3 X 2 240 X 1, X 2 0
primal solution: X 1 30, X 2 60, profit $39 9-43.
Maximize profit 10 X 1
5 X 2 31 X 3 28 X 4 17 X 5
X 1 X 2
subject to
12 X 5
2 X 5
2 X 2 2 X 3 5 X 4
X 2
X 1
X 5
5 X 3
28
53
70
18
X 1, X 2, X 3, X 4, X 5 0 9-44.
9-45.
a. Machine 3, as represented by slack variable S 3, 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. The dual is maximize Z 120U 1 115U 2 116U 3 subject to
8U 1
4U 2 9U 3 23
4U 1
6U 2 4U 3 18 U 1, U 2, U 3 0
U 1 $2.07 is the price of each test 1 U 2 $1.63 is the price of each test 2 U 3 $0
is the price of each test 3
Using the dual objective function: Z 120U 1 115U 2 116U 3
120(2.07)
$248.4
$435.85
115(1.63)
$187.45
$0
116(0)
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. 9-48.
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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 ***** 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
Z
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS 9-49.
Maximize 20 X 1 10 X 2 0S 1 0S 2 Subject to: 5 X 1 4 X 2 S 1 250 2 X 1 5 X 2 S 2 150 X 1, X 2 0
$9,683.228
Variable
Value
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 1X 11 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.
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
C j l Solution b Mix
20 X 1
10 X 2
0 S1
0 S2
Quantity
0
S 1
5
4
1
0
250
0
S 2
2
5
0
1
150
0
0
0
0
0
10
0
0
Z j C j Z j
20
REVISED M09_REND6289_10_IM_C09.QXD
5/12/08
12:01 PM
Page 129
CHAPTER 9
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
L I N E A R P R O G R A M M I N G : T HE S I M P L E X M E T H O D
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 ( S 2 425), so additional units of this resource would simply increase the slack. 9-50.
9-51.
Maximize 10 X 1 8 X 2
a.
Subject to: 2 X 1 1 X 2 24 2 X 1 4 X 2 36 X 1, X 2 0 S 1 24; S 2 26; X 1 0; X 2 0. Profit 0.
b. c.
***** Program Output ***** Final Optimal Solution At Simplex Tableau Z
C j l Solution b Mix
$8865.500
Variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
0 S1
0 S2
Quantity
0
S 1
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
S 2
2
4
0
1
36
Z j
0
0
0
0
0
C j Z j
10
8
0
0
The pivot column is the X 1 column. d. Variable X 1 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 S 1. g. The ratio for the pivot row is 12, so 12 units of X 1 will be in the next solution. h. The total profit will increase by ($10 per unit) (12 units) $120. 9-52. a. Maximizeprofit 20 X 1 30 X 2 15 X 3 0S 1 0S 2 MA2 MA3
Subject to: 3 X 1 5 X 2 2 X 3 S 1 120
Final Optimal Solution at Simplex Tableau : 21
8 X 2
Value
Printout 4 for Problem 9-48
Z
10 X 1
2 X 1 X 2 2 X 3 S 2 A2 250
$9,380.234
X 1 X 2 X 3 A3 180
Variable
Value
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15
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
X 1, X 2, X 3 0 b. S 1 120; A2 250; A3 180; all Profit 430 M . 9-53.
others 0.
a. S 1 12; X 2 16; X 1 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 C j Z j value).
