Solutions 249
w=[0,2,2]Ei^1 B-^1 = [2,2,0].
w = cTBB~l = [0,2,2]
rXl = cXl - wNxi = 1 - [2,2,0]
rS2 = cS2 - wNS2 = 0 - [2,2,0]
rSl =cSl -wNs> =0-[0,2,2]
= -3 < 0.
= -2 < 0.
= -2<0.
Optimal.
- revised simplex with B = LU decomposition:
Let XB = (si,X3,X2)T, XN = (x\,S2,sz)T. Then, B =
triangular, L = I3. Solve BXB — bhy back substitution.
X2 = 2, x 3 = 10, si = 8 — X2 = 6.
Solve wB = CB by back substitution.
iyi=0, W2 = 2, W3 = 2 — tui = 2.
101
010
001
is upper
The rest is the same, S3 enters and s± leaves.
001"
010
-10 1
New basis is B =
"00 r
010
100
001"
010
-101
=
"-10 r
010
001
PB = LU &
1
Solve BxB - Pb= (2,10, 8)T by substitution.
x 2 = 2, £3 = 10, s 3 = x 2 - 2 = 6.
Solve wB = PcB = (2,2,0)r by substitution.
wi = 0, w 2 = 2, w 3 = 2 - wi = 2.
The rest is the same.
= hU.
