Answers to Homework Chapter 7 (b)
BSBN2120, J. Wang
Page 295 7-33. According to the QM output on the top of page 287:
(a) Unit profit of X, which is currently 5, could decrease by 1 (to 4) or increase by 7 (to
12) without changing the current optimal solution (since the sensitivity range for X’s
coefficient is [4, 12]).
(b) The total profit in the objective function would increase by 0.75 (the dual price), if the
RHS of constraint 1 were increased by 1 unit.
(c) If RHS of constraint 1 were increased by 10 units, the RHS would become 130 that is
still within the lower bound (80) and upper bound (240) of constraint 1’s RHS range,
within which the current dual price, 0.75, would be still valid.
Therefore, the total profit in the objective function would increase by 10*0.75=7.5.
7-34. According to the QM output at the bottom of page 287:
(a) 25 units of product 1 and 0 unit of product 2 should be produced.
(b) Plug the optimal solution into constraint 1’s left-hand-side. We have 1*25+2*0=25.
That is, 25 units of the resource in constraint 1 are used. There are 45-25=20 units
surplus for this resource.
Plug the optimal solution into constraint 2’s left-hand-side. We have 3*25+3*0=75.
That is, 75 units of the resource in constraint 2 are used. There are 87-75=12 units
surplus for this resource.
Plug the optimal solution into constraint 3’s left-hand-side. We have 2*25+1*0=50.
That is, 50 units of the resource in constraint 3 are used. There is 50-50=0 unit surplus
for this resource.
(c) The dual prices are 0, 0, and 25, respectively, for constraints 1, 2, and 3.
(d) More resource 3 should be obtained because it has the largest dual price, 25. Each
additional unit of resource 3 can make the total profit increase by $25. I would be willing
to pay no more than $25 per unit for this resource.
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.
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.