Finding Feasible Solutions To A Linear Programming Worksheet - Columbia University In The City Of New York Page 11
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14Iteration 2
2M
3
M
3
3
4M
60 10M
z =
+
x
+
e
+
a
(46)
1
1
1
3
3
3
3
x
e
a
1
1
1
20
x
=
+
(47)
2
3
3
3
3
5
e
a
1
1
7
s
=
x
+
(48)
1
1
3
12
12
12
2x
e
a
1
1
1
10
a
=
+
(49)
2
3
3
3
3
Now we pivot in x
and pivot out a
and obtain:
1
2
e
1
2M
3
2M
1
z = 25
+
a
+
a
(50)
1
2
2
2
2
e
a
3a
1
1
2
x
= 5
+
(51)
1
2
2
2
e
a
a
1
1
2
x
= 5 +
+
(52)
2
2
2
2
e
a
5a
1
1
2
1
s
=
+
+
(53)
1
4
8
8
8
Now all the coefficients in the objective row are negative, so we have an
optimal solution. Also, a
and a
are both non-basic, so the problem is
1
2
feasible. The optimal solution, in the original variables, is x
= 5, x
= 5
1
2
with objective value 25.
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