Initial tableau:
Cj→
Solution
Mix
$20
$24
$0
$0
M
M
Quantity
X1
X2
S1
S2
A1
A2
M
A1
1
1
–1
0
1
0
30
M
A2
1
2
0
0
1
40
–M
0
0
Second tableau:
Solution
Mix
$20
$24
$0
$0
M
M
Quantity
X1
X2
S1
S2
A1
A2
A1
X2
1
0
0
20
0
0
0
0
–1
1
10
Final tableau:
Cj→
Solution
Mix
$20
$24
$0
$0
M
M
Quantity
X1
X2
S1
S2
A1
A2
$20
X1
1
0
–2
1
2
–1
20
$24
X2
0
1
–1
–1
10
–4
$640
0
0
4
M – 16
M7-30. Maximize profit = 9X1 + 12X2
subject to X1 + X2 10
X1 + 2X2 12
X1, X2 0
Initial tableau:
Cj→
Solution
Mix
$9
$12
$0
$0
Quantity
X1
X2
S1
S2
$0
$0
Second tableau:
Cj→
Solution
Mix
$9
$12
$0
$0
Quantity
X1
X2
S1
S2
$0
S1
$12
1
0
6
6
0
6
3
0
0
–6
0
1
4
Final tableau:
Cj→
Solution
Mix
$9
$12
$0
$0
Quantity
X1
X2
S1
S2
$4
X1
1
0
2
–1
8
$12
X2
0
1
–1
1
2
9
6
3
0
0
–6
–3
X1 = 8, X2 = 2, profit = $96
M7-31. Maximize profit = 8X1 + 6X2 + 14X3
subject to 2X1 + X2 + 3X3 120
2X1 + 6X2 + 4X3 = 240
X1, X2 0
Initial tableau:
Cj→
Solution
Mix
$8
$6
$14
0
−M
Quantity
X1
X2
X3
S1
A1
0
Second tableau:
Cj→
Solution
Mix
$8
$6
$14
0
−M
Quantity
X1
X2
X3
S1
A1
$0
S1
$6
1
0
40
0
1
80
Final tableau:
Cj→
Solution
Mix
$8
$6
$14
0
−M
Quantity
X1
X2
X3
S1
A1
$14
X3
$6
X2
1
0
6
0
0
0
1
M7-32. a.
X1 = number of deluxe one-bedroom units converted
subject to
1,100X1 + 1,000X2 + 600X3 + 500X4 $35,000
700X1 + 600X2 + 400X3 + 300X4 $28,000
2,000X1 + 1,600X2 + 1,200X3 + 900X4 $45,000
b. Maximize profit = 8,000X1 + 6,000X2 + 5,000X3 + 3,500X4 + 0S1 + 0S2 + 0S3 + 0S4 + 0S5
+ 0S6 + 0S7 + 0S8 – MA1 – MA2
subject to
1,100X1 + 1,000X2 + 600X3 + 500X4 + S1 = 35,000
M7-33. a. The initial formulation is
minimize cost = $12X1 + 18X2 + 10X3 + 20X4 + 7X5 + 8X6
subject to
X1 – 3X3 = 100
25X2 + X3 + 2X4 + 8X5 900
b. Variable X5 will enter the basis next. (Its Cj – Zj value indicates the most improvement,
that is, 7 – 21M ) Variable A3 will leave the basis because its ratio (150/15) is the smallest of
the three positive ratios.
M7-34. 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 M7-34
Cj→
Solution
Mix
$10 +
$30
$0
$0
Quantity
X1
X2
S1
S2
$10 +
X1
1
4
2
0
160
0
From the S1 column, we require that
–20 – 2 0 or –10
b. The range of insignificance is
– 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.
M7-35. a. The shadow prices are: 3.75 for constraint 1; 22.5 for constraint 2; and 0 for con-
straint 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.
M7-36. a. Produce 18 of model 102 and 4 of model H23.
b. S1 represents unused or slack time on the soldering machine; S2 represents unused or
slack 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.
M7-37. 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.
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
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
Range is thus from 200 hours to 300 hours (or 240 – 40 to 240 + 60).
Table for Problem M7-37c
Cj→
Solution
Mix
70 +
50
0
0
Quantity
X1
X2
S1
S2
70 +
X1
1
0
$4,100 + 30
30
Table for Problem M7-37d
Cj→
Solution
Mix
70
50 +
0
0
Quantity
X1
X2
S1
S2
70
X1
1
0
32
−12
30
M7-38. Note that artificial variables may be omitted from the sensitivity analysis since they
have no physical meaning.
a. Range of optimality for X1 (phosphate):
Cj→
Solution
Mix
$5 +
$6
$0
$0
Quantity
X1
X2
S1
S2
$0
S2
0
0
–1
1
550
$6
0
1
–1
0
700
X1
1
0
1
0
300
If the Cj value for X1 increases by $1, the basis will change. Hence – Cj (for X1) $6.
Range of optimality for X2 (potassium):
Cj→
Solution
Mix
5
6 +
0
0
Quantity
X1
X2
S1
S2
0
S2
0
0
–1
1
550
5
1
0
0
300
If the Cj value for X2 decreases by $1, the basis will change. The range is thus $5 Cj (for X2)
.
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
700
–1
–700
subject to 1U1 + 2U2 80
3U1 + 5U2 75
U1, U2 0
The dual of the dual is the original primal.
M7-40. Maximize profit = 50U1 + 4U2
M7-41. U1 = $80, U2 = $40, cost = $1,000
M7-42. Primal objective function:
maximize profit = 0.5X1 + 0.4X2
M7-43. Maximize profit =
10X1 +5X2 + 31X3 + 28X4 + 17X5
M7-44. 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.
M7-45. The dual is
maximize Z = 120U1 + 115U2 + 116U3
subject to 8U1 + 4U2 + 9U3 23
4U1 + 6U2 + 4U3 18
U1, U2, U3 0
Using the dual objective function:
Z = 120U1 + 115U2 + 116U3
= 120(2.07) + 115(1.63) + 116(0)
= $248.4 + $187.45 + $0
= $435.85
M7-46. a. There are 8 variables (2 decision variables, 3 surplus variables, and 3 artificial varia-
bles) and 3 constraints.
M7-47. a. Rounded to two decimals, the solution is X1 = 27.38 tables, X2 = 37.18 chairs daily,
profit = $3775.60.
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.
M7-48. Printout 1 illustrates the model formulation.
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.
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
M7-49. Maximize 20X1 + 10X2 + 0S1+ 0S2
Subject to: 5X1 + 4X2 + S1 = 250
2X1 + 5X2 + S2 = 150
X1, X2 0
Cj→
Solution
Mix
20
10
0
0
Quantity
X1
X2
S1
S2
0
S1
5
4
1
0
250
0
0
0
0
0
0
M7-50. 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.
M7-51. a. Maximize 10X1 + 8X2
Subject to: 2X1 + 1X2 24
Solution
Mix
10
8
0
0
X1
X2
S1
S2
0
S1
2
1
1
0
0
S2
2
4
0
1
0
0
0
0
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).
M7-52. a. Maximize profit = 20X1 + 30X2 + 15X3 + 0S1 + 0S2 – MA2 – MA3
Subject to: 3X1 + 5X2 – 2X3 + S1 = 120
M7-53. 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