Linear Programming Lecture Notes - Penn State Personal Web Server

simplex tableau). We obtain:
z1 − c1 = wA·1 − c1 = 1 10 0
⎡
⎣ 1
z2 − c2 = wA·2 − c2 =
⎤
0 ⎦ − 0 = 1
3/8
1 10 0
⎡
⎣ 0
z6 − c6 = wA·6 − c6 =
⎤
1 ⎦ − 0 = 10
3/2
1 10 0
⎡
⎣ 0
⎤
0 ⎦ − 5 = −5
−1
By Dantzig’s rule, we enter variable x2. We append B −1 A·2 and the reduced cost to the
revised simplex tableau to obtain:
(8.12)
z
y1
y2
⎡
1
⎢ 1
⎣ 0
10
0
1
0
0
0
⎤ ⎡
100
50 ⎥ ⎢
⎥ ⎢
5 ⎦ ⎣
10
0
1
⎤
⎥
⎦
0 0 1 10 3/2
z1
z1
MRT
−
5
20/3
After pivoting on the indicated element, we obtain the new tableau:
(8.13)
z
y1
x2
⎡
1
⎢ 1
⎣ 0
0
0
1
0
0
0
50
50
5
⎤
⎥
⎦
0 −3/2 1 5/2
We can compute reduced costs for the non-basic variables (except for y2, which we know will
not re-enter the basis on this iteration) to obtain:
z1 − c1 = wA·1 − c1 = 1
z6 − c6 = wA·6 − c6 = −5
In this case, x1 will enter the basis and we augment our revised simplex tableau to obtain:
(8.14)
Note that:
z
y1
x2
z1
⎡
⎢
⎣
B −1 A·1 =
1 0 0 50
1 0 0 50
0 1 0 5
0 −3/2 1 5/2
⎤ ⎡
⎥ ⎢
⎥ ⎢
⎦ ⎣
1
1
0
3/8
⎤
⎥
⎦
MRT
50
−
20/3
⎡
1
⎣0 0
1
⎤ ⎡
0
0⎦
⎣
0 −3/2 1
1
⎤ ⎡
0 ⎦ = ⎣
3/8
1
⎤
0 ⎦
3/8
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