CSA 273
Solutions to Homework Set #9
Problem #1
Adding artificial variables a
1
and a
2
to each of the constraints and a surplus
variable e
2
to the second constraint, the "artificial" problem in standard form
becomes:
Minimize Z=3x
1
+ 2x
2
+ 4x
3
+ Ma
1
+ Ma
2
such that
2x
1
+ x
2
+ 3x
3
+ a
1
= 10
3x
1
+3x
2
+ 5x
3
e
2
+
a
2
= 20
x
1
,x
2
,x
3
,a
1
,a
2
,e
2
≥
0
Start out with the tableau:
Z
x
1
x
2
x
3
e
2
a
1
a
2
RHS
Row 0
1
3
2
4
0
M
M
0
Row 1
0
2
1
3
0
1
0
10
Row 2
0
3
3
5
1
0
1
20
To get an initial feasible solution with basic variables z,a
1
and a
2
, the
coefficients of a
1
and a
2
must be cleared in row 0. This can be done by applying
the eros:
New Row 0 = Old Row 0 + M*Old Row1 + M*Old Row 2
The other rows would remain unchanged.
Our Initial Simplex Tableau:
Basic
Z
x
1
x
2
x
3
e
2
a
1
a
2
RHS
Solution
Ratio
Row 0
1
5M3
4M2
8M4
M
0
0
30M
Z=30M
Row 1
0
2
1
3
0
1
0
10
a
1
=10
10/3=3.33
Row 2
0
3
3
5
1
0
1
20
a
2
=20
20/5=4
has an initial basic feasible solution of a
1
=10, a
2
=20, x
i
=0 for i=1,2,3, e
2
=0 ,
and Z=30M to the artificial problem. The entering basic variable should be x
3
since it has the largest coefficient in Row 0. The leaving basic variable should
be a
1
by the ratio test.
Problem #2
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 Spring '08
 Patton
 Cost Accounting, Optimization, current basis

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