Stochastic Programming - Index of

RECOURSE PROBLEMS 199
The first idea we wish to test is based on comparing pairs of extreme points,
to see how well the optimal dual solution (which is dual feasible for all righthand
sides) at one extreme-point works at a neighbouring extreme point. We
use the indexing L and U to indicate Low and Up of the support.
LL:UL We first must test the optimal dual solution π LL together with the
right-hand side b UL .Weget
α =(π LL ) T b UL − φ(U, L)
=(0, 0, 0, 0, 0, 1, 2)(6, 21, 49, 120, 45, 20, 0) T − φ(U, L)
=20− 10.5 =9.5.
We then do the opposite, to find
β =(π UL ) T b LL − φ(L, L)
=(0, 1 , 0, 0, 0, 0, 7)(6, 21, 49, 120, 45, 0, 2 2 0)T − φ(L, L)
=10.5 − 0=10.5.
The minimum is therefore 9.5 for the pair LL:UL.
LL:LU Following a similar logic, we get the following:
α =(π LL ) T b LU − φ(L, U)
=(0, 0, 0, 0, 0, 1, 2)(6, 21, 49, 120, 45, 0, 10) T − φ(L, U)
=20− 12 = 8,
β =(π LU ) T b LL − φ(L, L)
=(2, 0, 0, 0, 0, 3, 0)(6, 21, 49, 120, 45, 0, 0) T − φ(L, L)
=12− 0=12.
The minimal value for the pair LL:LU is therefore 8.
LU:UU Forthispairwegetthefollowing:
α =(π UU ) T b LU − φ(L, U)
=(0, 0, 0, 0.0476, 0.476, 0, 0)(6, 21, 49, 120, 45, 0, 10) T − φ(L, U)
=27.143 − 12 = 15.143
β =(π LU ) T b UU − φ(L, L)
=(2, 0, 0, 0, 0, 3, 0)(6, 21, 49, 120, 45, 20, 10) T − φ(U, U)
=72− 27.143 = 44.857.
The minimal value for the pair LU:UU is therefore 15.143.
UL:UU For the final pair the results are given by
α =(π UU ) T b UL − φ(U, L)
=(0, 0, 0, 0.0476, 0.476, 0, 0)(6, 21, 49, 120, 45, 20, 0) T − φ(U, L)
=27.143 − 10.5 =16.643,
β =(π UL ) T b UU − φ(U, U)
=(0, 1 2 , 0, 0, 0, 0, 7 2 )(6, 21, 49, 120, 45, 20, 10)T − φ(U, U)
=46.5 − 27.143 = 18.357.