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1. Dual:
1 2 1 2 3 (^1 2 3 1 ) 1 2 3
min 12 6 8 when (^2 ) , , 0
λ λ λ λ λ λ λ λ (^) λ λ λ λ λ
Initial tableau:
Second tableau:
Final tableau:
The primal solution is given by x ˆ 1 = 0, x ˆ (^) 2 = 2, x ˆ 3 = 8 , the maximum value is 20. The dual solution is given by λ ˆ 1^ = 1, λ ˆ 2^ = 0, λ ˆ 3 = 1 , the minimum value is 20.
2. Using two bottom rows the initial tableau associated with the first phase is given by:
Using three bottom rows the final tableaus of both the first and the second phase are given by: 1 0 1 1 0 1 6 1 1 0 1 1 1 6 0 0 0 0 0 1 0 1 2 0 0 0 0 1 0 0 2 0 12
The feasible point x 1 (^) = 0, x 2 (^) = 6, x 3 = 6 given by the first phase is also a global maximum.
3. The slack s ˆ 2 = 1 > 0 , so λ ˆ 2^ = 0. From x ˆ 1 > 0, x ˆ 3 > 0 conclude that μ ˆ 1 = 0, μ ˆ 3 = 0.
It follows that 1 3 3
λ λ λ
(^) =^ and hence^1 λ ˆ^ = 1.
4. λ ˆ T^ = 1 1 0 ,
c^ T λT^ A cT cT
− = − ^ ^ ^ = − ^ ^ =^
so cT = ^2 3 .
The basic variables are given by x (^) 2 , x 1 (^) , y (^) 3 so cTB = ^3 2 0 . Note that λ ˆ T = c BTB −^1 , 2 1 0 1 1 0 2 3 1
B
= ^
and
c^ TB
= ^ ^
5. (a) The only part of the tableau that changes is the
( c + ∆ c ) T^ − c BTB^ −^1 A = cT − c BTB^ −^1 A + ∆ c T = ^ − 12 0 ^ + ^ ∆ c 1 0 ^ ≤^0 0 so the increase allowed is at most 12.
(b) The basic variables are given by x (^) 2 , y 2 so cTB = ^2 0 . This time
(^12) (^12 2 2 712 12 ) 2
c^ T cT cTB cTB B A cT cTB B A cT c BTB A
c c c
+ ∆ − + ∆ −^ = − −^ + ∆ − ∆ − =
^
−^ +^ ∆^ − ∆ =^ −^ −^ ∆^ ≤
It is also necessary to check
(^16) (^1 12 2 12 16 ) (^16)
cB cB TB − 0 c 0 1 c 0 0 0
^
The decrease allowed is at most 1.
6. (a) 1 ( )^1 1
B −^ b b B −^ b B −^ b B − b
+ ∆ = + ∆ = ^ + ∆ ≥
leads to 907 + 73 ∆ b 1 ≥ 0 ,
(^757) − 37 ∆ b 1 ≥ 0 , 957 − 218 ∆ b 1 ≥ 0 , so − 30 ≤ ∆ b 1 ≤ 25
(b) (^ )^45757 657 157
λ^ T b b λT b λT b
+ ∆ = + ∆ = + ^ ^ =
