Answers to Homework Problems, Cht. 3
BSNS5230, J. Wang
P. 99 -#22.
X1 = number of sausage biscuits to make each morning
X2 = number of ham biscuits to make each morning
Max
s.t.
0.6X1 + 0.5X2
0.01X1 + 0.024X2  6
0.1X1
 30
0.15X2  30
0.04X1 + 0.04X2  16
X1, X2  0
(Total profit of a morning)
(labor hour)
(sausage)
(ham)
(flour)
Solved by QM, the optimal solution is : {X1*= 300, X2*= 100}. That is, the manager
should prepare 300 sausage biscuits and 100 ham biscuits each morning. The best objective
function value is $230.
Problem #22. Objective: Maximize
Results ---------X1
X2
RHS
Dual
-------------------------------------------------------------------------Maximize
0.6
0.5
Labor hour
0.01
0.024 <=
6
0
Sausage
0.1
0
<=
30
1
Ham
0
0.15
<=
30
0
Flour
0.04
0.04
<=
16
12.5
-------------------------------------------------------------------------Solution
300
100
230
Ranging ---------Variable
X1
X2
Constraint
Labor hour
Sausage
Ham
Flour
Value
300
100
Dual
Value
0
1
0
12.5
Reduced
Cost
0
0
Original
Value
0.6
0.5
Lower
Bound
0.5
0
Upper
Bound
Infinity
0.6
Slack/
Surplus
0.6
0
15
0
Original
Value
6
30
30
16
Lower
Bound
5.4
25.7143
15
12
Upper
Bound
Infinity
40
Infinity
17
#23.
(a) For “sausage: Plug the optimal solution in the constraint of “sausage”: 0.1*300=30 which is
equal to the RHS. So, there is no sausage left over.
For “ham”: Plug the optimal solution in the constraint of “ham”: 0.15*100=15 lbs to be
used. Comparing to 30 lbs of ham available, there are 15 lbs of ham left over.
1
For “labor hours”: Plug the optimal solution in the constraint of “labor hour”:
0.01*300+0.024*100=5.4, which is 0.6 less than the RHS=6. So, there is 6-5.4=0.6 labor hour
left over.
(b) If the unit profit for a ham biscuit were increased from $0.5 to $0.6, the current optimal
solution would not change, because 0.6 is not beyond the sensitivity range for X2’s coefficient in
objective function (0~0.6).
(c) Change the RHS of constraint of “flour” from 16 to 18 and resolve the LP by QM, the new
optimal solution is (X1=300, X2=125). (Attach QM solution)
#25.
X1 = number of telephone interviewers to hire
X2 = number of personal interviewers to hire
Min
s.t.
50X1 + 70X2
80X1 + 40X2  3,000
80X1
 1,000
40X2  800
X1, X2  0
(Total cost of interviewers per day)
(total interviews per day)
(telephone interviews per day)
(personal interviews per day)
By QM, the optimal solution is: {X1*= 27.5, X2*= 20}. That is, the firm should hire 27.5
telephone interviewers and 20 personal interviewers. The best objective function value is $2,775
of total cost per day.
Problem #25. Objective: Minimize
Results ---------X1
X2
RHS
Dual
-----------------------------------------------------------------------------------------------------Minimize
50
70
Total interviews / day
80
40
>=
3,000 -0.625
Telephone interviews / day
80
0
>=
1,000
0
Personal interviews / day
0
40
>=
800
-1.125
--------------------------------------------------------------------------------------------------------Solution
27.5
20
2,775
Ranging ---------Variable
X1
X2
Value
27.5
20
Constraint
Total interviews / day
Telephone interviews / day
Personal interviews / day
Reduced
Original
Lower
Cost
Value
Bound
0
50
0
0
70
25
Dual
Value
-0.625
0
-1.125
Slack/
Surplus
0
1,200
0
Original
Value
3,000
1,000
800
Upper
Bound
140
Infinity
Lower
Bound
1,800
-Infinity
0
Upper
Bound
Infinity
2,200
2,000
2
#26
(a) As given in the results of QM in Problem #25:
Sensitivity range for daily cost of a telephone interviewer which is the coefficient of X1
in the objective function, is (0~140).
Sensitivity range for number of personal interviews required which is RHS of the third
constraint, is (0~2000).
(b) Plug the optimal solution in the second constraint (telephone interviews per day):
80*27.5=2200 which is larger than its RHS=1000. So, the firm conducts 2200-1000=1200 more
telephone interviews per day than the required.
Plug the optimal solution in the third constraint (personal interviews per day): 40*20=800
which is equal to its RHS. So, the firm conducts no more personal interviews than the required.
(c) If total personal interview must be at least 1,200, we need to change RHS of the third
constraint from 800 to 1200 and resolve it by QM. The new optimal solution would be
(X1=22.5, X2=30). (QM result should be attached)
#27.
(a) The dual value for “personal interviews” 1.125. That means that the total cost will be
worse off by 1.125 (i.e., increase by 1.125) if the required personal interviews increases by 1
from 800 to 801. That is, the total cost will be better off by 1.125 (i.e., decrease by 1.125) if the
required personal interviews is reduced by 1 to 799.
The shadow price for “telephone interviews” is 0, which tells that there is no effect on the
total cost by changing the required telephone interviews that is currently 1,000.
Therefore, daily personal interview requirement should be reduced because it would bring
the total cost down more.
After reducing personal interview requirement by 1 from 800 to 799, QM results tell that
the total optimal cost is reduced by 1.125 to 2773.875 with the new optimal solution {X1*=
27.5125, X2*= 19.975}. (Attach QM solution)
(b) The sensitivity range for the cost of a personal interviewer, which is the coefficient of X2 in
the objective function, is (25   ).
The sensitivity range for number of total interviews that must be done per day, which is the
RHS of the first constraint, is (1,800  ).
#46.
X1 = number of cartons of “own brand” milk to stock
X2 = number of cartons of “local brand” milk to stock
X3 = number of cartons of “national brand” milk to stock
Max
s.t.
0.97X1 + 0.83X2 + 0.69X3
16X1 + 16X2 + 16X3  5184
(Total profit)
(space in inch2)
3
X1 
3X1
X2 + X3 
0
 X3 
0
X2
 120
X1, X2, X3  0
(local brand supply)
Note: (i) 1 foot = 12 inches; 1 foot2 = 144 inches2. 5184 (inches2) of RHS of constraint 1 is
obtained from 36*144.
(ii) 120 in constraint four is obtained from 10*12. (1 dozen = 12)
By QM, the optimal solution is : {X1*= 54, X2*= 108, X3*= 162}. That is, store should
stock 54 cartons of “own brand” milk, 108 cartons of “local brand” milk, and 162 cartons of
“national brand” milk. The best objective function value is $253.8 of total profit.
Problem #42. Objective: Maximize
Results ---------X1
X2
X3
RHS
Dual
-----------------------------------------------------------------------------------------Maximize
0.97
0.83
0.69
Space in inches2
16
16
16
<=
5,184
0.049
Constraint 2
-1
-1
1
>=
0
-0.0467
Constraint 3
3
0
-1
<=
0
0.0467
Local brand
0
1
0
<=
120
0
-----------------------------------------------------------------------------------------Solution
54
108
162
253.8
Ranging ---------Variable
X1
X2
X3
Constraint
Space in inches2
Constraint 2
Constraint 3
Local brand
Value
54
108
162
Dual
Value
0.049
-0.0467
0.0467
0
Reduced
Cost
0
0
0
Slack/
Surplus
0
0
0
12
Original
Lower
Value
Bound
0.97
0.83
0.83
0.76
0.69
-0.8767
Original
Value
5,184
0
0
120
Lower
Bound
0
-18
-36
108
Upper
Bound
1.25
0.97
0.7833
Upper
Bound
5,760
162
324
Infinity
#47.
(a) The dual value for “space” is 0.049, which means that for each additional inch2 the total profit
will increase by $0.049. Each carton takes 16 inches2. Each additional carton implies additional
16 inches2. So, each additional carton space would make additional $0.049*16=$0.784 in profit.
But the above dual value is true only when the total space does not exceed 5760 inches2
since the upper limit for the RHS of “space” constraint is 5760.
(b) Getting more supply of the local brand milk will not increase the total profit, since the
shadow price for “local brand supply” is 0.
(c) The discount would change the objective function to
4
Max 0.86X1 + 0.83X2 + 0.69X3
and change the third constraint to
1.5X1
 X3 
0.
The new optimal solution by QM is {X1*= 108, X2*= 54, X3*= 162}. (QM solution should
be attached)
The best objective function value is 249.48. Since the total profit is lower than that
without discount, the discount should not be implemented.
5