MODULE B
END-OF-MODULE PROBLEMS
B.1
Let x = number of standard model to produce
y = number of deluxe model to produce
Maximize 40x + 60y
Subject to 30 x  30 y  450
10 x  15y  180
x6
x, y  0
B.2
y
20
18
16
14
12
Optimal
 x  0, y  10 
10
8
6
4
2
0
2
4
6
8
10
12
14
16
18
20 x
Feasible corner points (x, y): (0, 3), (0, 10), (2.4, 8.8), (6.75, 3). Maximum profit is 100 at (0, 10).
Quantitative Module B: Linear Programming
1
B.3
y
20
18
16
14
12
10
Optimal
 x  4, y  8
8
6
4
2
0
2
4
6
8
10
12
14
16
18
20 x
Feasible corner points (x, y): (0, 2), (0, 10), (4, 8), (10, 2). Maximum profit is 52 at (4, 8).
B.4
(a)
(b)
Corner points (0, 50), (50, 50), (0, 200), (75, 75), (50, 150).
Optimal solutions: (75, 75) and (50, 150). Both yield profit of $3,000.
x2
400
350
300
250
200
150
Feasible Region
100
50
0
2
50
100 150 200 250 300 350 400 450
x1
Instructor’s Solutions Manual t/a Operations Management
B.5
(a)
(b)
B.6
(a)
Adding a new constraint will reduce the size of the feasible region unless it is a redundant
constraint. It can never make the feasible region any larger.
A new constraint can only reduce the size of the feasible region; therefore the value of the
objective function will either decrease or remain the same. If the original solution is still
feasible, it will remain the optimal solution.
Let x1  number of liver flavored biscuits in a package
x2  number of chicken flavored biscuits in a package
Minimize x1  2 x2
Subject to x1  x2  40
2 x1  4 x2  60
x1  15
(b)
(c)
B.7
x1, x2  0
Corner points are (0, 40) and (15, 25). Optimal solution is (15, 25) with cost of 65.
Minimum cost = 65 cents.
160
140
120
Drilling Constraint
b
100
80
c
60
40
Wiring Constraint
Feasible Region
20
0
20
d
40
60
80
100
Number of Air Conditioners: x1
120
Let x1  number of air conditioners to be produced
x2  number of fans to be produced
Maximize 25x1  15x2
Subject to 3x1  2 x2  240 wiring
2 x1  1x2  140 drilling
x1, x2  0 nonnegativity
Profit:
@ a: ( x1  0 , x2  0 )
@ b: ( x1  0 , x2  120 )
@ c: ( x1  40 , x2  60 )
@ d: ( x1  70 , x2  0 )
 Obj  $0
 Obj  25  0  15  120  $1,800
 Obj  25  40  15  60  $1,900 *
 Obj  25  70  15  0  $1,750
The optimal solution is to produce 40 air conditioners and 60 fans each period. Profit will be $1,900.
Quantitative Module B: Linear Programming
3
300
B.8
250
a
200
x2
b
150
100
Feasible Region
50
0
50
100
c
200
150
x1
250
300
Let x1  number of Model A tubs produced
x2  number of Model B tubs produced
Maximize 90 x1  70 x2
Subject to 125x1  100 x2  25,000 steel 
20 x1  30 x2  6,000 zinc
x1, x2  0 nonnegativity
Profit:
 Obj  90  0  70  200  $14,000.00
@ a: ( x1  0 , x2  200 )
 Obj  90  85.71  70  142.86  $17,71410
@ b: ( x1  85.71, x2  142.86 )
.

@ c: ( x1  200 , x2  0 )
Obj  90  200  70  0  $18,000.00 *
The optimal solution is to produce 200 Model A tubs, and 0 Model B tubs. Profit will be $18,000.
300
B.9
250
200
x2
150
100
a
Optimal Solution
50
b
Feasible Region
0
50
100
150
200
250
c
300
350
x1
Let x1  number of benches produced
x2  number of tables produced
Maximize 9x1  20 x2
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Instructor’s Solutions Manual t/a Operations Management
Subject to 4 x1  6 x2  1,200  hours
10 x1  35x2  3,500 board-feet
x1, x2  0 nonnegativity
Profit:
@ a: ( x1  0 , x2  100 )
@ b: ( x1  262.5, x2  25 )
@ c: ( x1  300 , x2  0 )
 Obj  9  0  20  100  $2,000.00
 Obj  9  262.5  20  25  $2,862.50 *
 Obj  9  300  20  0  $2,700.00
The optimal solution is to make 262.5 benches and 25 tables per period. Profit will be $2,862.50. Because
benches and tables may be matched (two benches per table), it may not be reasonable to maximize profit in
this manner. Also, this problem brings up the proper interpretation of the statement that “One should make
262.5 (a fractional quantity) benches per period.”
B.10
40
Optimal Solution
30
Feasible Region is
Heavily Shaded Line
a
x2 20
b
10
0
10
20
30
40
50
x1
Let x1  number of Alpha-4 computers
x2  number of Beta-5 computers
Maximize: 1200 x1  1800 x2
Subject to 20 x1  25x2  800  total hours
x1  10 Alpha-4s
x2  15 Beta-5s
x1, x2  0 nonnegativity
Profit:
 Obj  1200  10  1800  24  $55,200 *
@ a: ( x1  10 , x2  24 )
 Obj  1200  2125
. , x2  15 )
.  1800  15  $52,500
@ b: ( x1  2125
The optimal solution is to produce 10 Alpha-4 and 24 Beta-5 computers per period. Profit is $55,200.
Quantitative Module B: Linear Programming
5
50
B.11
5x1  3x2  150
40
30
x1  2 x2  10
x2
20
a
10
0
3x1  5 x2  150
Feasible Region
10
20
30
40
50
x1
Note that this problem has one constraint with a negative sign.
The optimal point, a, lies at the intersection of the constraints:
3x1  5x2  150
5x1  3x2  150
To solve these equations simultaneously, begin by writing them in the form shown below:
3x1  5x2  150
5x1  3x2  150
Multiply the first equation by 5, the second by –3, and add the two equations and solve for x2:
x2  300 16  18.75 . Given: 3x1  5x2  150 then
3x1  150  5x2  150  5  18.75
and
56.25
 18.75
3
Thus, the optimal solution is: x1  18.75, x2  18.75
The profit is given by:
x1 
C  4 x1  4 x2  4  18.75  4  18.75  $150
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Instructor’s Solutions Manual t/a Operations Management
100
B.12
8 x1  2 x2  160
x2  70
80
60
3x1  2 x2  120
x2
40
Feasible Region
20
0
a
20
40
x1  3x2  80
60
x1
80
100
The optimal point, a, lies at the intersection of the constraints:
3x1  2 x2  120
x1  3x2  90
To solve these equations simultaneously, begin by writing them in the form shown below:
3x1  2 x2  120
x1  3x2  90
. . Given:
Multiply the second equation by –3, and add it to the first and solve for x2: x2  150 7  2143
3x1  2 x2  120 then
3x1  120  2 x2  120  2  2143
.
and
x1 
7714
.
 25.71
3
.
Thus, the optimal solution is: x1  25.71 , x2  2143
The cost is given by:
C  x1  2 x2  25.71  2  21.43  $68.57
B.13 The fifth constraint is not linear because it contains the square root of x and the objective function
and first constraint are not because of the x1x2 term.
B.14 (a)
Using software, we find that the optimal solution is:
x1  7.95
x2  5.95
x3  12.60
Profit  $143.76
(b)
There is no unused time available on any of the three machines.
Quantitative Module B: Linear Programming
7
B.15 (a)
(b)
An additional hour of time on the third machine would be worth $0.26.
Additional time on the second machine would be worth $0.786 per hour for a total of $7.86
for the additional 10 hours.
B.16 (a)
Let Xij  number of students bused from sector i to school j. Objective:
minimize total travel miles  5 X AB  8 X AC  6 X AE  0 XBB  4 XBC  12 XBE  4 XCB  0 XCC
7 XCE  7 XDB  2 XDC  5 XDE  12 XEB  7 XEC  0 XEE
subject to
X AB  X AC  X AE  700  number students in sector A
XBB  XBC  XBE  500  number students in sector B
XCB  XCC  XCE  100  number students in sector C 
XDB  XDC  XDE  800  number students in sector D
XEB  XEC  XEE  400  number students in sector E 
X AB  XBB  XCB  XDB  XEB  900 school B capacity
X AC  XBC  XCC  XDC  XEC  900 school C capacity
X AE  XBE  XCE  XDE  XEE  900 school E capacity
(b)
Solution: X AB  400
X AE  300
XBB  500
XCC  100
XDC  800
XEE  400
Distance  5,400 “student miles”
B.17 Because the decision centers about the production of the two different cabinet models, let:
x1  number of French Provincial cabinets produced per day
x2  number of Danish Modern cabinets produced each day
The equations become:
Objective 28x1  25x2 (Maximize revenue)
Subject to 3x1  2 x2  360 hours, carpentry
1.5 x1  1x2  200 hours, painting
0.75 x1  0.75 x2  125 hours, finishing
x1  60  units, contract 
x2  60  units, contract 
x1, x2  0 nonnegativity
The solution is:
x1  60 , x2  90 , Revenue = $3930/day
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Instructor’s Solutions Manual t/a Operations Management
B.18
Problem Data
Workers
Time Period
Required
3 AM–7 AM
3
7 AM–11 AM
12
11 AM–3 PM
16
3 PM–7 PM
9
7 PM–11 PM
11
11 PM–7 AM
4
Period
1
2
3
4
5
6
Hire
Solution 1
0
16
0
9
2
3
  30
Solution
Hire
Solution 2
3
9
7
2
9
0
  30
Hire
Solution 3
3
14
2
7
4
0
  30
Let xi  number of workers reporting for the start of work in period i, where i  1 , 2, 3, 4, 5, or 6.
The equations become:
Objective:
x1  x2  x3  x4  x5  x6 (Minimize staff size)
Subject to:
x1  x2
 12
x2

 16
x3
x3

x4
x4

x5
x5


x1
x6
x6
 9
 11
 4
 3
x1, x2 , x3, x4 , x5 , x6  0
Note that three alternate optimal solutions are provided to this problem. Either solution could be
implemented using only 30 staff members.
20
B.19
15
Feasible Region
x2
10
5
2 x1  1x2  20
a
x2  5
0
2
4
6
8
10
12
14
16
18
20 x
x1
Quantitative Module B: Linear Programming
9
Define the following variables:
x1  thousands of round tables produced per month
x2  thousands of square tables produced per month
The appropriate equations then become:
Objective: 10 x1  8x2 (minimize handling and storage costs)
Subject to x2  5 square tabletop contract 
2 x1  1x2  20 total labor capacity
x1, x2  0 nonnegativity
Cost:
 Obj  10  75
@ a: ( x1  7.5, x2  5 )
.  8  5  $115*
 Obj  10  0  8  20  $160
@ b: ( x1  0 , x2  20 )
The optimal solution is to produce 7500 round tables and 5000 square tables, for a cost of $115,000.
B.20 The original equations are:
Objective: 9x1  12 x2 (maximize)
Subject to: x1  x2  10 gallons, varnish
x1  2 x2  12 lengths, redwood
where: x1  number of coffee tables/week
x2  number of bookcases/week
Optimal: x1  8 , x2  2 , Profit = $96
10
B.21
2nd corner point
8
Optimal corner
6 b
c
x2
4
1st corner point
2
a
0
4 thcorner point
d
2
4
6
8
10
x1
The original equations are:
Objective: 3x1  5x2 (maximize)
Subject to: x2  6
3x1  2 x2  18
x1, x2  0 nonnegativity
The optimal solution is found at the intersection of the two constraints. Solving for the values of x1
and x2 at the intersection, we have:
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Instructor’s Solutions Manual t/a Operations Management
x2  6
18  2 x2 18  2  6 6

 2
3
3
3
Profit  3x1  5x2  3  2  5  6  6  30  $36
x1 
B.23 Let x1  the number of class A containers to be used
x2  the number of class K containers to be used
x3  the number of class T containers to be used
The appropriate equations are:
Maximize: 8x1  6 x2  14 x3
Subject to: 2 x1  x2  3x3  120  Material 
2 x1  6 x2  4 x3  240  Time
x1, x2 , x3  0 nonnegativity
Using software we find that the optimal solution is:
2
1
x1  0 , x2  17 , x3  34
7
7
or
x1  0 , x2  17143
. , x3  34.286
and
Profit  582
B.24 (a)
(b)
6
 $582.86
7
The unit profit of the air conditioner must fall in the range $22.50–$30.00
The shadow price for the wiring constraint is $5.00, and it holds within the range 210–280 hours.
B.25 Let x1  number of newspaper ads placed
x2  number of TV spots purchased
Given that we are to minimize cost, we may develop the following set of equations:
Minimize: 925x1  2000 x2
Subject to: 0.04 x1  0.05x2  0.40 city exposure
0.03x1  0.03x2  0.60 exposure in NW suburbs
Note that the problem is not limited to unduplicated exposure (for example, one person seeing the
Sunday newspaper three weeks in a row counts for three exposures).
Solution:
x1  20 ads, x2  0 TV spots, cost = $18,500
B.26 (a)
Minimize: 6x1a  5x1b  3x1c  8x2a  10 x2b  8x2c  11x3a  14 x3b  18x3c
Quantitative Module B: Linear Programming
11
Subject to: x1a  x2 a  x3a  7
x1b  x2b  x3b  12
x1c  x2c  x3c  5
x1a  x1b  x1c  6
x2 a  x 2 b  x 2 c  8
x3a  x3b  x3c  10
All variables  0
(b)
Solution:
x1b  6
x2 b  3
x2 c  5
x3a  7
x3b  3
Cost  $219
B.27
Labor Hours
400
300
Optimal Solution
200
Feasible Region
Production Limit
100
0
50
100 150
200 250 300 350
Boy’s Bicycles
400 450
Maximize: 57x1  55x2
Subject to: x1  x2  390
2.5x1  2.4 x2  960
@ ( x1  384 , x2  0 )
@ ( x1  0 , x2  390 )
@ ( x1  240 , x2  150 )
 Obj  57  384  55  0  $21,888
 Obj  57  0  55  390  $21,450
 Obj  57  240  55  150  $21,930 *
The optimal solution occurs at x1  240 boy’s bikes, x2  150 girl’s bikes, producing a profit of $21,930
However, the possible optimal solutions are so close in profit that sensitivity analysis is important here.
B.28 Let x1  number of medical patients
x2  number of surgical patients
The appropriate equations are:
Maximize 2280 x1  1515x2
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Instructor’s Solutions Manual t/a Operations Management
Subject to: 8 x1  5 x2  32,850 patient days available
31
. x1  2.6 x2  15,000  lab tests
1x1  2 x2  7,000 X rays
x2  2,800 operations
x1, x2  0 nonnegativity
Optimal: x1  2790 , x2  2104 , Profit = $9,551,659
or: 2790 medical patients, 2104 surgical patients, Profit $9,551,659
Beds required:
Use: Medical: 8  2790  22,320
Surgical: 5  2104  10,520
32,840
22,320
Medical uses:
 68%  61 beds
32,840
10,520
Surgical uses:
 32%  29 beds
32,840
Here is an alternative approach that solves directly for the number of beds:
Maximize revenues  104,025x1  110,595x2
Subject to: x1  x2  90 beds
8x1  5x2  32,850 patients yr
141.44 x1  189.8x2  15,000 lab tests
45.63x1  146 x2  7,000 x-rays
73x2  2,800 operations
where x1  no. of medical beds = 61.17
x2  no. of surgical beds = 28.83
Revenue is $9,551,659, as before
Quantitative Module B: Linear Programming
13