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Homework 2 Solutions
Due:
February 3, 2009
Solution to Exercise 4, page 83
:
Let
x
11
and
x
12
denote the production level (units) of product
P
1
on machines
M
1 and
M
2, respectively. Similarly, let
x
21
and
x
22
denote the production level of
product
P
2
on machines
M
1 and
M
2, respectively.
Note that
x
11
+
x
21
represents the total production of machine
M
1, and
x
12
+
x
22
represents the production of machine
M
2. The capacity constraints for machines
M
1 and
M
2 are, respectively, given by:
x
11
+
x
21
≤
200
,
x
12
+
x
22
≤
250
.
The constraint balancing the production schedule on the two machines is given
by:

(
x
11
+
x
21
)

(
x
12
+
x
22
)
 ≤
5
.
This constraint can be represented equivalently by two inequalities, as follows:
x
11
+
x
21

x
12

x
22
≤
5
,

x
11

x
21
+
x
12
+
x
22
≤
5
.
The total production of
P
1 is
x
11
+
x
12
, and that of
P
2 is
x
21
+
x
22
. Given the
proﬁt of $10 per unit of
P
1 and $15 per unit of
P
2, the total proﬁt is given by
10(
x
11
+
x
12
) + 15(
x
21
+
x
22
).
Thus, the LP formulation of the problem is
maximize
10
x
11
+ 10
x
12
+ 15
x
21
+ 15
x
22
subject to
x
11
+
x
21
≤
200
x
12
+
x
22
≤
250
x
11
+
x
21

x
12

x
22
≤
5

x
11

x
21
+
x
12
+
x
22
≤
5
x
11
,x
12
,x
21
,x
22
≥
0
,
where
x
11
,x
12
,x
21
,x
22
≥
0 is a short hand notation for denoting that all listed
variables are nonnegative.
1
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View Full DocumentTo transform this LP to a standard form, we need to the transform the inequality
constraints into equalities by adding slack variables. Thus, the standard LP is
maximize
10
x
11
+ 10
x
12
+ 15
x
21
+ 15
x
22
subject to
x
11
+
x
21
+
s
1
= 200
x
12
+
x
22
+
s
2
= 250
x
11
+
x
21

x
12

x
22
+
s
3
= 5

x
11

x
21
+
x
12
+
x
22
+
s
4
= 5
x
11
,x
12
,x
21
,x
22
,s
1
,s
2
,s
3
,s
4
≥
0
.
Solution to the modiﬁcation of Exercise 5, page 83:
The given problem is not LP, so at ﬁrst, we provide its LP formulation. The ﬁrst
step is to eliminate the objective function given by maximization by introducing a
new variable
y
. The equivalent formulation is
maximize
y
subject to
y
≤ 
x
1

x
2
+ 3
x
3

y
≤  
x
1
+ 3
x
2

x
3

x
1
,x
2
,x
3
≥
0
,
y
is unrestricted
.
Note that
y
is a variable in the preceding problem.
The next step toward LP formulation is to eliminate nonlinear constraints involv
ing absolute value. By doing so, we obtain the following LP formulation:
maximize
y
subject to
x
1

x
2
+ 3
x
3

y
≥
0

x
1
+
x
2

3
x
3

y
≥
0

x
1
+ 3
x
2

x
3

y
≥
0
x
1

3
x
2
+
x
3

y
≥
0
x
1
,x
2
,x
3
≥
0
,
y
is unrestricted
.
Now, we provide the standard form of the preceding LP. To do so, we need to
introduce a slack variable for each of the inequality constraints, and we need to write
y
as
y
=
x
+
4

x

4
with
x
+
4
≥
0 and
x

4
≥
0 (since
y
is unrestricted). Thus, the
standard form of the preceding LP is:
maximize
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 Spring '09
 Nedich

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