Formulation 3:
Formulation 4 appears to be proper as is. Note that the constraint 4X1 + 6X2
48 is redundant.
7-28. Using the isoprofit line or corner point method, we see that point b (where X = 37.5 and Y
= 75) is optimal if the profit = $3X + $2Y. If the profit changes to $4.50 per unit of X, the optimal
solution shifts to point c. If the objective function becomes P = $3X + $3Y, the corner point b
remains optimal.
7-29. The optimal solution of $26 profit lies at the point X = 2,Y = 3.
If the first constraint is altered to 1X + 3Y 8, the feasible region and optimal solution shift con-
siderably, as shown in the next graph.
7-30.
Using the corner point method, we determine that the optimal solution mix under the new con-
straint yields a $29 profit, or an increase of $3 over the $26 profit calculated. Thus, the firm
should not add the hours because the cost is more than $3.
7-31. a. The corner points and profits are
X = 0, Y = 0, profit = 0
X = 60, Y = 0, profit = 300
X = 30, Y = 60, profit = 510 Optimal solution
c. If profit = 3X + 6Y, a new corner point is optimal.
7-32. The corner points change and the new optimal solution is X = 40, Y = 40, and profit =
440. The corner points are
X = 0, Y = 0, profit = 0
7-33. a. It could increase by 7 (for an upper limit of 12) or decrease by 1 (for a lower limit of 4).
7-34. a. 25 units of product 1 and 0 units of product 2.
b. All of resource 3 is being used (there is no slack for constraint 3). A total of 25 units of
resource 1 are being used since there were 45 units available and there are 20 units of
slack. A total of 75 units of resource 2 are being used since there were 87 units available
7-35.
a. The feasible corner points and their profits are:
Feasible corner points
Profit = 8X1 + 5X2
(0,0)
0
(6,0)
48
(6,4)
68
(0,10)
50
b. The feasible corner points and their profits are:
Feasible corner points
Profit = 8X1 + 5X2
(0,0)
0
(6,0)
(6,5)
c. The feasible corner points and their profits are:
Feasible corner points
Profit = 8X1 + 5X2
(0,0)
0
(6,0)
48
(0,6)
30
As a result of this change, the feasible region got smaller. Profit decreased by $20. The right-
hand side decreased by 4 units, and the profit decreased by 4 x dual price.
d. The feasible corner points and their profits are:
Feasible corner points
Profit = 8X1 + 5X2
(0,0)
0
(5,0)
40
(0,5)
25
e. The computer output indicates that the dual price for constraint 1 is $5, but this is valid up to a
7-36. Let: X1 = number of coconuts carried
X2 = number of skins carried
Maximize profit = 60X1 + 300X2 (in rupees)
subject to 5X1 + 15X2 300 pounds
The three princes should carry 24 coconuts and 12 lions’ skins. This will produce a wealth of
5,040 rupees.
7-37. a. $120,000 in money market fund; $80,000 in stock fund; total risk = 1,560,000
b. Total return = $14,000. Rate of return = 14,000/200,000 = 0.07
7-38. a. $40,000 in money market fund; $160,000 in stock fund; total return = 18,000
b. Total risk = 12(160,000) + 5(40,000) = 2,120,000. Average risk = 2,120,000/200,000 =
10.6.
7-39. a.Let: X1 = number of pounds of stock X purchased per cow each month
X2 = number of pounds of stock Y purchased per cow each month
X3 = number of pounds of stock Z purchased per cow each month
Four pounds of ingredient A per cow can be transformed to:
4 pounds (16 oz/lb) = 64 oz per cow
5 pounds = 80 oz
Minimize cost = $2X1 + $4X2 + $2.50X3
b. Cost = $80
X1 = 40 lbs. of X
7-40. Let: X1 = number units of XJ201 produced
X2 = number units of XM897 produced
X3 = number units of TR29 produced
X4 = number units of BR788 produced
7-41.
Let ED1, ED2, and ED3 represent the ending inventory for the first three months respectively.
Let PD1, PD2, and PD3 represent the production for the first three months respectively. Then
the formulation is:
Minimize cost: 6ED1 + 6ED2 + 6ED3 + 120PD1 + 120PD2 + 120PD3
subject to
7-42.
For the addition of the Family Rolls Tents to the problem above, a few new variables are needed.
Let EF1, EF2, and EF3 represent the ending inventory for the first three months of the Family
Rolls product line respectively. Let PF1, PF2, and PF3 represent the production levels of the
Family Rolls Tent in the first three month respectively. Then we have the following formulation:
PF1 – EF1 = 60 Demand in first month for Family Rolls
EF1 + PF2 – EF2 = 70 Demand in second month for Family Rolls
EF2 + PF3 – EF3 = 65 Demand in third month for Family Rolls
The optimal production schedule is to produce {220, 180, 215} for the Double Inn and {60, 70,
65} for the Family Rolls respectively for a total of $108,620 in production and inventory costs.
7-43. a. Let: X1 = number of MCA regular modems made and sold in November
X2 = number of MCA intelligent modems made and sold in November
Data needed for variable costs and contribution margin are in the table.
Hours needed to produce each modem:
Table for Problem 7-43(a)
MCA REGULAR MODEM
MCA INTELLIGENT MODEM
Total
Per Unit
Total
Per Unit
Net sales
$424,000
$47.11
$613,000
$58.94
Variable costsa
Direct labor
60,000
6.67
76,800
7.38
Indirect labor
9,000
1.00
11,520
1.11
Materials
90,000
Sales commissions
Total variable costs
$220,000
$24.44
$311,320
$29.93
Contribution margin
$204,000
$22.67
$301,680
$29.01
b.
c. The optimal solution suggests making all MCA regular modems. Students should dis-
cuss the implications of shipping no MCA intelligent modems.
7-44. Minimize cost = 0.12X1 + 0.09X2 + 0.11X3 + 0.04X4
subject to X1 + X2 + X3 + X4 = 50
X4 7.5
X1 + X2 22.5
X2 + X3 15.0
Solution:
7-45. Let A1 = gallons of crude A used in Regular
A2 = gallons of crude A used in Premium
Minimize cost = 0.42A1 + 0.42A2 + 0.42A3 + 0.47B1 + 0.47B2 + 0.47B3
Subject to
0.40A1 + 0.52B1 0.41(A1 + B1)
0.40A2 + 0.52B2 0.44(A2 + B2)
0.40A3 + 0.52B3 0.48(A3 + B3)
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
7-46.
X1 = number of model A tubs produced
X2 = number of model B tubs produced
7-47. Let: X1 = number of pounds of compost in each bag
X2 = number of pounds of sewage waste in each bag
Minimize cost = 5X1 + 4X2 (in cents)
subject to X1 + X2 60 (pounds per bag)
X1 30 (pounds compost per bag)
X2 40 (pounds sewage per bag)
7-48.
X1 = $ invested in Treasury notes
X2 = $ invested in bonds
7-49.
Let: X1 = number of TV spots
X2 = number of newspaper ads
Maximize exposures = 35,000X1 + 20,000X2
subject to 3000X1 + 1,250X2 $100,000
X1 5
X1 25
X2 10
7-50. Maximize Z = [220 − (0.45)(220) − 44 − 20]X1 + [175 − (0.40)(175) − 30 − 20]X2
= 57X1 + 55X2
Students should point out that those three options are so close in profit that production desires
and sensitivity of the RHS and cost coefficient are important issues. This is a good lead-in to the
discussion of sensitivity analysis. As a matter of reference, the right-hand side ranging for the
first constraint is a production limit from 384 to 400 units. For the second constraint, the hours
may range only from 936 to 975 without affecting the solution.
SOLUTION TO MEXICANA WIRE WORKS CASE
1. Maximize P = 34 W75C + 30 W33C + 60 W5X + 25 W7X
subject to:
1 W75C 1,400
1 W33C 250
Solution: Produce:
1,100 units of W75C—backorder 300 units
2. Bringing in temporary workers in the Drawing Department would not help. Drawing is not a
binding constraint. However, if these former employees could do rework, we could reduce our
3. The plant layout is not optimum. When we install the new equipment, an opportunity for im-
proving the layout could arise. Exchanging the locations for packaging and extrusion would cre-
ate a better flow of our main product. Also, as we improve our quality and reduce our rework
INTERNET CASE STUDY:
AGRI-CHEM CORPORATION
This case demonstrates an interesting use of linear programming in a production setting.
Let X1 = ammonia
X2 = ammonium phosphate
Objective function:
Subject to the following constraints:
X1 1,200 X5 560
Current natural gas usage = 85,680 cu. ft. 103/day
20 percent curtailment = 68,554 cu. ft. 103/day
Hence, the ninth constraint is:
8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 68,544
The following is the production schedule (tons/day);
X1 = 1,200 X5 = 560
For a 40 percent natural gas curtailment, the ninth constraint is:
8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8
51,408
The optimal solution results in the following production schedule:
X1 = 1200 X5 = 560