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300 BUSINESS ANALYTICS MODULE B LIN E A R PR O G R A M M I N G
Solution Value = 9380.234
Shadow
Slack or
Original
Lower
Upper
Prices
Surplus
RHS
Limit
Limit
const 1
1.494825
0.00
980.00
969.9414
1,202.002
const 2
0.00
120.755
400.00
279.2448
Infinity
const 3
0.7247843
0.00
600.00
598.0171
811.0541
const 4
0.8810187
0.00
2,500.00
2492.973
2,917.931
const 5
0.0234673
0.00
1,800.00
1530.888
1,805.481
const 6
6.716568
0.00
1,000.00
918.2866
1,002.674
const 7
0.00
0.00
0.00
−Infinity
0.00
const 8
0.00
0.00
0.00
−Infinity
0.00
const 9
0.00
0.00
0.00
−Infinity
0.00
const 10
0.00
0.00
0.00
−Infinity
0.00
const 11
0.00
28.7226
0.00
−Infinity
28.72255
const 12
−8.85034
0.00
20.00
17.19764
40.10845
const 13
−5.71531
0.00
10.00
0.00
25.09986
const 14
0.00
17.5172
20.00
−Infinity
37.51723
const 15
−10.3985
0.00
50.00
42.69018
75.98374
const 16
−4.85238
0.00
20.00
0.00
38.00887
const 17
0.00
13.94
20.00
−Infinity
33.94098
const 18
0.00
27.4846
10.00
−Infinity
37.485
const 19
−0.494562
0.00
20.00
1.392963
21.02138
const 20
−61.9896
0.00
10.00
0.7036638
10.96196
const 21
0.00
0.2774
10.00
−Infinity
10.27741
B.25 We solve this problem by the isocost line method:
B.24* (cont.)
BUSINESS ANALYTICS MODULE B LI N EA R PR O G R A M M I N G 301
B.28 Let: X1 = number of pounds of compost in each bag
(X1 = 60, X2 = 0) cost = 5(60) + (4)(0) = $3.00
B.29
The optimal point, a, lies at the intersection of the constraints:
3x1 + 2x2 120
x1 + 3x2 90
To solve these equations simultaneously, begin by writing
them in the form shown below:
3x1 + 2x2 = 120
x1 + 3x2 = 90
Multiply the second equation by –3, and add it to the first:
3x1 + 2x2 = 120 → 3x1 + 2x2 = 120
–3(x1 + 3x2 = 90) → −3x1 – 9x2 = –270
–7x2 = –150
Therefore, x2 = 150/7 = 21.43. Given: 3x1 + 2x2 = 120, then
The original equations are:
Objective: 4x1 + 5x2 (minimize)
Subject to: x1 + 2x2 80
3x1 + x2 75
The optimal solution is found at the intersection of the two
constraints:
x1 + 2x2 80
3x1 + x2 75
1 2 1 2
1
5 70
x
− = −
Thus, x1 = 70/5 = 14. Given: x1 + 2x2 = 80,
2x2 = 80 – x1 = 80 – 14, or x2 = 66/2 = 33.
The cost is given by:
C = 4x1 + 5x2 = (4 14) + (5 33) = 221
302 BUSINESS ANALYTICS MODULE B LIN E A R PR O G R A M M I N G
B.31
Note that the problem is not limited to unduplicated expo-
sure (for example, one person seeing the Sunday newspa-
per three weeks in a row counts for three exposures).
Solution:
x1 = 20 ads, x2 = 0 TV spots, cost = $18,500
B.34 (a) Let Xij = number of students bused from sector i to
school j. Objective:
X X X
++
Minimize total travel miles = 5 8 6
BUSINESS ANALYTICS MODULE B LI N EA R PR O G R A M M I N G 303
B.36
Problem Data
Solution
Period
Time Period
Workers
Required
Hire
Solution 1
Hire
Solution 2
Hire
Solution 3
5
7 P.M.–11 P.M.
11
2
9
4
6
11 P.M.–3 A.M.
4
3
0
0
= 30
= 30
= 30
B.35
X1 + X2 = $250,000
Point b (X1 = 250,000, X2 = 0), ROI = $20,000
Let xi = number of workers reporting for the start of working peri-
Subject to: 2x1 + x2 + 3x3 130 (material)
B.38 (a) Minimize:
6 5 3 8 10 8 11 14 18
x x x x x x x x x+ + + + + + + +
1 1 1
8
All variables 0
a b c
x x x
+ +
B.38 (cont.)
$I$22
Fruit + Veg I
200
$I$22 $J$2
Binding
0
BUSINESS ANALYTICS MODULE B LI N EA R PR O G R A M M I N G 305
Adjustable cells
Sensitivity Report (Relevant Section)
Cell
Name
Final
Value
Reduced
Cost
Objective
Coefficient
Allowable
Increase
Allowable
Decrease
$B$14
serving AS
0
0.1726
0.3
1E + 30
0.1726
$C$14
serving CC
1.333
0
0.4
0.2589
0.2256
$D$14
serving FC
0.457
0
0.9
0.1051
0.1006
$E$14
serving FF
0
0.1527
0.2
1E + 30
0.1529
$F$14
serving M
1.130
0
0.5
0.0629
0.7078
$G$14
serving TB
0
0.1693
1.5
1E + 30
0.1694
$H$14
serving GS
0
0.6661
0.9
1E + 30
0.6882
Constraints
Cell
Name
Final
Value
Shadow
Price
Constraint
R.H. Side
Allowable
Increase
Allowable
Decrease
$I$17
Cals min LI
800
0
500
300
1E + 30
$I$18
Cals max L
800
– 0.00023
800
200
251.6129
$I$19
Protein min
200
0.008983
200
155
40
$I$20
Carb min L
311.43
0
200
111.4285
1E + 30
$I$21
Fat max LI
288.57
0
400
1E + 30
111.4286
$I$22
Fruit + Veg I
200
0.0015
200
485.7143
200
B.41 (a) For i = 1,…,5, we define the following decision
variables:
– 10,000Y4 – 18,000Y5
Subject to: X1 + X2 + X3 + X4 + X5 20,000
Profit = $487,000
Let i =
1 if investment is chosen
0 otherwise
i
+ 500D + 750E + 640F
+ 300G
A + B = 1
A, B, C, D, E, F, G = 0 or 1
CASE STUDY
following set of equations:
Minimize 12x1 + 9x2 + 11x3 + 4x4 (in cents/lb.)
A more general set of equations can be developed:
Minimize 12x1 + 9x2 + 11x3 + 4x4
+ + + =
+ + + + − −
+ + + + − −
1 2 3 4
1 2 1 2 3 4
2 3 1 2 3 4
1 2 3 4
Subject to 50 (50 pounds)
0.45 ( )(C 92,C 30)
0.30 ( )(D 21,C 92)
, , , 0 (non-negative)
x x x x
x x x x x x
x x x x x x
x x x x
These equations can be rewritten as:
Minimize:
1 2 3 4
12 9 11 4x x x x+ + +
1 2 3 4
Subject to: 50.0 (50 pounds)x x x x+ + + =
1 2 3 4
1 2 3 4
, , , 0 (non-negativity)
x x x x
Now, if the size of the bag is changed, only the first constraint
306 BUSINESS ANALYTICS MODULE B LIN E A R PR O G R A M M I N G
1
=−
=
4
27.5 (E 11)
Cost $3.35
x
LO B.1: Formulate linear programming models, including an
objective function and constraints
AACSB: Analytical thinking
VIDEO CASE STUDY
SCHEDULING CHALLENGES AT ALASKA
AIRLINES
1. Scheduling is more complex for Alaska Air than for other
airlines for several reasons: (1) Its route system is far-flung and
problems
AACSB: Reflective thinking
2. Operational considerations in network expansion include: (1)
shortage of trained pilots/crew, (2) limited number of planes
AACSB: Reflective thinking
3. The risks on the quick turnaround situation route of Seattle–
Chicago–Seattle are that if the pilots “time out,” there is no re-
placement crew in Chicago. This can be a very expensive propo-
4. The decision by Alaska to risk a crew “time out” is a good one
if you consider the costs and probabilities involved. Students can
estimate the financial downsides of a cancellation, which would
4 hours, including taxi time on both ends, and turnaround can be
accomplished in 45 minutes, the 98.7% success rate is under-
standable. Here are some cost estimates:
Parking/ramp fees for overnight of plane: $1,000
Hotel/meals for 5 crew @ $200/each: $1,000
This would occur about 4 times a year (equals (1 – .987)
365 days)) implying an annual cost of $88,000. But the alterna-
tive, keeping a spare crew in Chicago at $1,000/day, is $365,000.
A good business decision by the airline.
LO B.6: Formulate production-mix, diet, and labor scheduling
problems
AACSB: Analytical thinking
ADDITIONAL CASE STUDIES (available in MyOM-
Lab)
CHASE MANHATTAN BANK
hours, none can start after 3 P.M. because the entire operation
ends at 7 P.M. Similarly, some number of part-time employees, Qj,
leave at the end of hour j, j = 4, 5, . . . , 9.
The workforce requirements for the first two hours, 9 A.M.
0.5F + P1 + P2 + P3 26
0.5F + P1 + P2 + P3 + P4 38
Starting at 1 P.M., some of the part-time employees begin to
leave. For the remainder of the straight-time day:
+ − − − − −
+ + + + + +
+ − − − − − −
7 4 5 6 7 8 14
1 2 3 4 5 6
7 4 5 6 7 8 9 9
P Q Q Q Q Q
F P P P P P P
P Q Q Q Q Q Q
(although 8F are paid for), F1 + F2 full-time labor hours are used
and paid for at overtime rates, and the total number of part-time
hours is:
308 BUSINESS ANALYTICS MODULE B LIN E A R PR O G R A M M I N G
(b) Supply constraints (40 percent gas curtailment):
5.5x1 + 7x2 + 8x3 + 10x4 + 15x5 + 16x6 + 12x7 + 11x8 21,600
(Note: 36,000 cu. ft. 103 0.60 = 21,600 cu. ft. 103)
2. Obviously, those products that have high energy consumption
factors must undergo extensive scrutiny to conserve energy. These
products include chlorine (15.0) and caustic soda (16.0). Energy
