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10-26. a. Let S = dollars invested in stocks; B = dollars invested in bonds;
R = dollars invested in real estate
Minimize d1– + d2– + d3+
Subject to
0.13S + 0.08B + 0.10R + Return is at least 10%
d1– – d1+ = 25,000
S + B + R = 250,000
S 150,000
150,000
P 150,000
10-27. Maximize profit = X1(1,800 – 50X1)
+ X2(2,400 – 70X2)
10-28. Let X1 = no. of XJ6s and X2 = no. of XJ8s
10-29. The optimal solution found using Solver in Excel is X = 62.73, Y = 8.64, Profit = 720.41.
10-30. The optimal solution found using Solver in Excel is X = 0.333, Y = 0.667, with a variance
of 0.102 and a return of 0.09.
17.33, profit = $4,946.67.
10-32. a. Z = $665,000
Variable
Value
X1
0
X2
0
X3
0
b. The expected return drops to $625,000. Osceola opens and Cocoa Beach closes.
c. As seen below, with Apopka corrected, the new solution has a return of $635,000 but the
same locations as part a.
Solution:
Z = $635,000
Variable
Value
Variable
Value
X1
0
X6
0
10-33. Solving this with Excel after changing the objective to maximization, we get a smoothing
constant of 0. The maximum MAD is 28.17 (ignoring the error in time period 1).
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
10-35. Maximize return = 50X1 + 100X2 + 30X3 + 45X4 + 65X5 + 20X6 + 90X7 + 35X8
subject to 500X1 + 1,000X2 + 350X3 + 490X4
+ 700X5 + 270X6 + 800X7 + 400X8 3,000
10-36. Define the variables Ai, Bi, Ci, Di, to be 1 if job A, B, C, or D respectively is assigned to
machine i, where i = 1, 2, 3, and 4. The integer program is:
Minimize cost = 85A1 + 70A2 + 60A3 + 10A4 + 6B1 + 15B2 + 90B3 + 76B4 + 50C1 + 80C2 + 5C3 +
75C4 + 75D1 + 84D2 + 82D3 + 25D4
Subject to:
A1 + A2 + A3 + A4 = 1
B1 + B2 + B3 + B4 = 1
Using Excel, we get two optimal solutions:
Assignment
Cost
Assignment
Cost
A4
$ 10
A4
$ 10
B1
6
B2
15
10-37. X1 = number of TV spots
X2 = number of newspaper ads
d1+ = deviation above budget funds of $120,000
10-38. The first two priorities are fully satisfied. The best solution is
X1 = 10 TV spots
10-39. Let S = number of Standard blenders produced each week
D = number of Deluxe blenders produced each week
(1) use 240 hours per week
1.5S + 2D + 2.5C –
1
d+
+
d1
−
= 240
(2) produce 60 of the Chef’s Delight blenders
C – d2+ + d2– = 60
(4) produce 60 of the Standard blenders
S – d4+ + d4– = 60
(5) generate profit of at least $3,500
SOLUTION TO SCHANK MARKETING RESEARCH CASE
1. The first part of this case is an assignment problem that can be formulated with LP. A dummy
project manager can be added to create a balanced 4 4 cost matrix.
2. This part is a goal programming formulation with five goals, ranked from P1 (highest) to P5
(lowest):
P1: assign a manager to the NASA account.
P2: do not assign Gardener to CBT Television account.
Constraints
For client’s demand:
X11 + X21 + X31 + d1– = 1 Hines
These constraints assume no more than one assignment per manager.
For project managers:
X11 + X12 + X13 + X14 + d5– – d5+ = 1 Gardener
X21 + X22 + X23 + X24 + d6– – d6+ = 1 Ruth
X31 + X32 + X33 + X34 + d7– – d7+ = 1 Hardgraves
These constraints permit assigning three managers to four clients while minimizing positive and
negative deviational variables (d5, d6, d7).
( )
43
10
11
0
ij ij
ji
C X d +
==
−=
This attempts to minimize total cost, bringing it as close to zero as possible; d10+ is the deviation
from the goal.
Objective function:
SOLUTION TO THE OAKTON RIVER BRIDGE CASE
For a given set of requirements, the smallest number of toll collectors that will meet them can be
obtained from the following integer linear programming problem:
minimize Z = X1 + X2 + X3 + X4 + X5 + X6 + X7
subject to
X3 + X4 + X5 + X6 + X7 R7
X1 + X4 + X5 + X6 + X7 R1
X1 + X2 + X5 + X6 + X7 R2
1. The following table summarizes the requirements for shifts A, B, and C for each of the seven
days of the week along with the allocations that yield the minimum numbers of collectors start-
ing each shift: 18 for shift A, 16 for shift B, and 18 for shift C.
Toll Collector Requirements for Oakton River Case
Shift
A
B
C
Mix
DAY
Req.
Start
Req.
Start
Req.
Start
Req.
Start
Sun.
8
0
10
0
15
5
33
3
Mon.
13
3
10
1
13
2
36
9
Tue.
12
5
10
5
13
1
35
8
2. If mixing of shifts is allowed, the daily requirements become the sum of the daily shift re-
quirements, as shown in the second part of the table. The minimum number of collectors starting
each day is shown in the last Start column. The total 50 is a reduction of two from the total re-
quired without allowing for the mixing of shifts.
SOLUTION TO PUYALLUP MALL CASE
The problem can be expressed as the following integer linear programming problem with Xi be-
ing a 0–1 variable, 1 if store i is to be included and 0 if not:
Maximize
28.1X1 + 34.6X2 + 50.0X3 + 162.0X4 + 77.8X5
subject to the space constraint
1.0X1 + 1.6X2 + 2.0X3 + 3.2X4 + 1.8X5 + 2.1X6
the construction cost constraint
24.6X1 + 32.0X2 + 41.4X3 + 124.4X4 + 64.8X5
+ 79.8X6 + 38.6X7 + 66.8X8 + 45.1X9 + 54.3X10
+ 15.0X11 + 13.4X12 + 42.0X13 + 63.7X14
+ 40.0X15 700
at least one clothing store
X1 + X2 + X3 1
