OR methods for Profit and Value Engineering
Solution for HW#1
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/
6
Solution for HW#1 of
1.
P21 Problem 12 Solution:
Operations Research Methods for Profit and Value Engineering
The following table provides the basic data of the problem (Suppose time to produce a Type
2 hat is T):
Required Labor Time
Profit($)
Market limit
Type 1 Hat
2T
8
150
Type 2 Hat
T
5
200
Labor time limit
400T
For the Wild West problem we need to determine the daily amounts to be produced of Type 1
and Type 2 hats. Thus the variables of the model are defined as
x
1
=
hats produced daily of Type 1
x
2
=
hats produced daily of Type 2
Since Wild West want to
maximize
the total daily profit. It follows that
Profit from Type 1 hat= 8
x
1
Profit from Type 2 hat=5
x
2
Letting
z
represent the total daily profit, the objective of Wild West is
Maximize
z
=8
x
1
+5
x
2
The constraints restrict labor time and market limit are expressed as:
!
1
"
150
!
2
"
200
2
# $ !
1
+
# $ !
2
"
400T
that is
2
!
1
+
!
2
"
400
Thus, the complete Wild West model is
Maximize
z
=8
x
1
+5
x
2
Subject to
!
1
"
150
!
2
"
200
2
!
1
+
!
2
"
400
!
1
,
!
2
%
0
The solution of the LP problem is:
!
1
= 100;
!
2
= 200; then
&
= 1800
2.
P21 Problem 15 Solution:
The following table provides the basic data of the problem:
Cost
Reached audiences
Allocated limit of
total budget
First ad
Additional ad
Radio
$300
5,000
2,000
80%
TV ad
$2,000
4,500
3,000
80%
Budget limit
$20,000
For the Top Toy problem we need to determine the amount of budget allocated to radio
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OR methods for Profit and Value Engineering
Solution for HW#1
2
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6
commercial and TV ad separately. Thus the variables of the model are defined as
x
1
=
number of radio commercial
x
2
=
number of TV ad
P.S.:
both
x
1
and
x
2
should be integers
Since Top Toy want to
maximize
the total people could be reached by radio commercial and
TV ad. It follows that
People reached by radio commercial= 5,000+2,000
×(
x
1
!
1)
People reached by TV ad=4,500+3,000
×(
x
2
!
1)
Letting
z
represent the total people can be reached, the objective of Top Toy is
Maximize
z
=[5,000+2,000
×(
x
1
!
1)]
"
[
4,500+3,000
×(
x
2
!
1)]
that is,
Maximize
z
=4,5000+2,000
x
1
"
3,000
x
2
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 Spring '09
 Nedich
 Optimization, Value engineering, BMW Sports Activity Series, 2010s automobiles

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