Answers to Homework Chapter 7 (a)
BSBN2120, J. Wang
Page 280 7-14.
Let
X1 = number of air conditioners to be produced;
X2 = number of large fans to be produced.
Max
S.T.
25X1 +15X2
3X1 + 2X2  240
2X1 + X2  140
X1, X2  0
(profit)
(wiring hours)
(drilling hours)
Module: Linear Programming
Problem 7-14
Objective: Maximize
X1
X2
RHS
Dual
----------------------------------------------------------------------Maximize
25
15
Wiring hours
3
2
<=
240
5
Drilling hours
2
1
<=
140
5
----------------------------------------------------------------------Solution
40
60
1,900
The optimal solution from QM is (X1 = 40, X2 = 60), and the optimal objective
function value is $1,900.
That is, the Electrocomp Corporation should produce 40 air conditioners and 60
large fans to maximize the profit.
7-15 (a).
Add two constraints to the LP in Problem 7-14:
X1
 20
(at least 20 air conditioners)
X2  80
(at most 80 large fans)
Module: Linear Programming
Problem 7-15(a)
Objective: Maximize
X1
X2
RHS
Dual
----------------------------------------------------------------------------------------------Maximize
25
15
Wiring hours
3
2
<=
240
5
Drilling hours
2
1
<=
140
5
At least 20 air conditioners
1
0
>=
20
0
At most 80 large fans
0
1
<=
80
0
----------------------------------------------------------------------------------------------Solution
40
60
1,900
1
The optimal solution from QM is (X1 = 40, X2 = 60), and the optimal objective
function value is $1,900, which are same as those in Problem 7-14.
7-16.
Let
X1 = number of radio ads to be used;
X2 = number of television ads to be used.
Max
S.T.
3,000X1 +7,000X2
200X1 + 500X2  40,000
X1
 10
X2  10
X1 
X2  0
X1, X2  0
(total people to be reached)
(budget)
(at least 10 radio ads)
(at least 10 television ads)
(radio ads no less than tele. ads)
Module: Linear Programming
Problem 7-16
Objective: Maximize
X1
X2
RHS
Dual
----------------------------------------------------------------------------------------- ----------Maximize
3,000 7,000
Budget
200
500
<=
40,000
15
At least 10 radio ads
1
0
>=
10
0
At least 10 TV ads
0
1
>=
10
-500
Radio ads no less than TV ads
1
-1
>=
0
0
---------------------------------------------------------------------------------------------------Solution
175
10
595,000
The optimal solution from QM is (X1 = 175, X2 = 10), and the optimal objective
function value is 595,000.
That is, 175 radio ads and 10 television ads should be used so that 595,000 people
will be reached.
7-17.
Let
Max
S.T.
X1 = number of benches to be produced;
X2 = number of picnic tables to be produced.
9X1 +20X2
4X1 + 6X2  1,200
10X1 + 35X2  3,500
X1, X2  0
(profit)
(carpenter hours)
(redwood material)
2
Module: Linear Programming
Problem 7-17
Objective: Maximize
X1
X2
RHS
Dual
---------------------------------------------------------------------------------Maximize
9
20
Carpenter hours
4
6
<=
1,200
1.4375
Redwood material
10
35
<=
3,500
0.325
---------------------------------------------------------------------------------Solution
262.5
25
2,862.5
The optimal solution from QM is (X1 = 262.5, X2 = 25), and the optimal objective
function value is $2,862.50.
That is, the Electrocomp Corporation should produce 262.5 (or 262) benches and
25 picnic tables to maximized the total profit.
7-18.
Let
X1 = number of undergraduate courses to offer;
X2 = number of graduate course to offer.
Min.
S.T.
2,500X1 +3,000X2
X1 +
X2  60
X1
 30
X2  20
X1, X2  0
(total cost of salary)
(at least 60 courses in total)
(at least 30 undergraduate courses)
(at least 20 graduate courses)
Module: Linear Programming
Problem 7-18
Objective: Minimize
X1
X2
RHS
Dual
----------------------------------------------------------------------------------------------Minimize
2,500 3,000
At least 60 courses in total
1
1
>=
60
-2,500
At least 30 undergraduate courses
1
0
>=
30
0
At least 20 graduate courses
0
1
>=
20
-500
----------------------------------------------------------------------------------------------Solution
40
20
160,000
The optimal solution from QM is (X1 = 40, X2 = 20), and the optimal objective
function value is $160,000.
That is, 40 undergraduate courses and 20 graduate courses should be offered, with
the minimized total salary cost $160,000.
3
7-39.
Let X1 = Number of pounds of Stock X feed to buy for a cow per month,
X2 = Number of pounds of Stock Y feed to buy for a cow per month,
X3 = Number of pounds of Stock Z feed to buy for a cow per month.
LP formulation with the standard format:
Min. 2X1 + 4X2 + 2.5X3
S.T. 3X1 + 2X2 + 4X3  64 (4 lbs *16 ounces/lb = 64 ounces)
2X1 + 3X2 + X3  80 (5 lbs *16 ounces/lb = 80 ounces)
X1
+ 2X3  16 (1 lbs *16 ounces/lb = 16 ounces)
6X1 + 8X2 + 4X3  128 (8 lbs *16 ounces/lb = 128 ounces)
X3  5 (lbs) (It is not X3500 because X3 is defined for
one cow rather than for 100 cows.)
X1, X2, X3  0
Module: Linear Programming
Problem 7-39
Objective: Minimize
X1
X2
X3
RHS
Dual
---------------------------------------------------------------------------------------------------Minimize
2
4
2.5
Ingredient A
3
2
4
>=
64
0
Ingredient B
2
3
1
>=
80
-1
Ingredient C
1
0
2
>=
16
0
Ingredient D
6
8
4
>=
128
0
No more than 5 lbs of Z
0
0
1
<=
5
0
--------------------------------------------------------------------------------------------------Solution
40
0
0
80
The optimal solution from QM is: {X1=40, X2=0, X3=0}.
The optimal objective function value is 80.
Therefore, the ranch should buy 40 pounds of Stock X feed for a cow per month,
without needing Stock Y and Stock Z. The minimized feed cost for a cow per month is
$80.
4
7-40.
Let X1 = Number of units of product XJ201 to produce in the next month,
X2 = Number of units of product XM897 to produce in the next month,
X3 = Number of units of product TR29 to produce in the next month,
X4 = Number of units of product BR788 to produce in the next month.
LP formulation with standard format:
Max.
9X1 + 12X2 + 15X3 + 11X4
S.T. 0.5X1 + 1.5X2 + 1.5X3 + X4  15,000
0.3X1 + X2 + 2X3 + 3X4  17,000
0.2X1 + 4X2 + X3 + 2X4  26,000
0.5X1 + X2 + 0.5X3 + 0.5X4  12,000
X1

150
X2

100
X3

300
X4 
400
X1, X2, X3, X4  0
Module/submodule: Linear Programming
Problem title: Problem 7-40
Objective: Maximize
X1
X2
X3
X4
RHS
Dual
---------------------------------------------------------------------------------------------------------------Maximize
9
12
15
11
Wiring hours
0.5
1.5
1.5
1
<= 15,000
6
Drilling hours
0.3
1
2
3
<= 17,000
0
Assembly hours
0.2
4
1
2
<= 26,000
0
Inspection hours
0.5
1
0.5
0.5 <= 12,000
12
X1201 at least 150 units 1
0
0
0
>=
150
0
XM897 at least 100 units 0
1
0
0
>=
100
-9
TR29 at least 300 units
0
0
1
0
>=
300
0
BR788 at least 400 units 0
0
0
1
>=
400
-1
--------------------------------------------------------------------------------------------------------------Solution
20,650 100 2,750
400
232,700
The optimal solution from QM is: {X1=20,650, X2=100, X3=2,750, X4=400}.
The optimal objective function value is 232,700.
Therefore, Weinberger Electronics Corporation should produce 20,650 XJ201’s, 100
XM897’s, 2,750 TY29’s, and 400 BR788’s in the next month. The maximized total
monthly profit is $232,700.
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