This preview shows pages 1–3. Sign up to view the full content.
CO 350 Assignment 2 – Winter 2010
Solutions
Question #
Max. marks
Part marks
1
8
3
2
2
1
2
6
3
3
3
6
2
2
2
4
6
3
3
5
4
Total
30
1. Consider the following linear programming problem (
P
):
min
2
x
1
+
x
2
+ 3
x
3
subj. to 3
x
1
+ 2
x
2
= 10
x
2
+ 2
x
3
≥
8
2
x
1
+
x
2

x
3
≤
8
x
1
≥
0
x
2
≤
0
(a) Convert (
P
) into standard inequality form. Denote the LP problem after the transfor
mation by (
P
0
).
(b) Convert (
P
) into standard equality form. Denote the LP problem after the transformation
by (
P
00
).
(c) Let
x
= (
x
1
,x
2
,x
3
)
T
be a feasible solution of (
P
). Determine feasible solutions
x
0
of (
P
0
)
and
x
00
of (
P
00
) that correspond to
x
.
(d) Apply your results from (c) to the feasible solution
x
= (4
,

1
,
5)
T
of (
P
) and determine
corresponding feasible solutions of (
P
0
) and (
P
00
).
Solution:
(a) Write
x
2
=

x
0
2
and
x
3
=
u
3

v
3
. The problem in SIF is
max

2
x
1
+
x
0
2

3
u
3
+ 3
v
3
subj. to
3
x
1

2
x
0
2
≤
10

3
x
1
+ 2
x
0
2
≤ 
10
x
0
2

2
u
3
+ 2
v
3
≤

8
2
x
1

x
0
2

u
3
+
v
3
≤
8
x
1
,
x
0
2
,
u
3
,
v
3
≥
0
.
(b) As in (a), write
x
2
=

x
0
2
and
x
3
=
u
3

v
3
. The problem in SEF is
max

2
x
1
+
x
0
2

3
u
3
+ 3
v
3
subj. to
3
x
1

2
x
0
2
=
10
x
0
2

2
u
3
+ 2
v
3
+
x
4
=

8
2
x
1

x
0
2

u
3
+
v
3
+
x
5
=
8
x
1
,
x
0
2
,
u
3
,
v
3
,
x
4
,
x
5
≥
0
where
x
4
and
x
5
are slack variables.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document (c) If
x
3
≥
0, we let
u
3
=
x
3
and
v
3
= 0. If
x
3
<
0, we let
u
3
= 0 and
v
3
=

x
3
. We obtain
x
0
= (
x
1
,

x
2
,x
3
,
0)
T
if
x
3
≥
0
x
0
= (
x
1
,

x
2
,
0
,

x
3
)
T
if
x
3
<
0
and
x
00
= (
x
1
,

x
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/28/2011 for the course CO 250 taught by Professor Guenin during the Fall '10 term at Waterloo.
 Fall '10
 GUENIN

Click to edit the document details