MthSc 810: Mathematical Programming Lecture 16 Sensitivity ... MthSc 810: Mathematical Programming Lecture 16 Sensitivity ...

MthSc810: MathematicalProgramming

Lecture16

PietroBelotti

Dept. ofMathematicalSciences

Clemson University

November 3,2011

Reading fortoday: Sections 5.1

ReadingforNov.8: Sections 5.2, 5.3-5.5

Homework#9 duenextTuesday

Homework #10is out! Due next Thursday.

◮ A = (B|N) → Ã = (˜ B| Ñ),where

... Ñ = (N|An+1) and ˜ B = B.

◮ ˜c = (c|cn+1).

◮ Unchanged b.

Forsimplicity,let’s assume the new variablexn+1 ≥ 0.

◮ Primal feasibility: ˜ B−1˜ b = B−1b ≥ 0? Of course.

◮ Dual feasibility: ˜cN − ˜cB ˜ B−1Ñ ≥ 0? Maybenot.

Dualfeasibilityrequires ¯cn+1 = cn+1 −cBB −1 An+1 ≥ 0.

If ¯cn+1 < 0,pivotonxn+1 starting fromthe currentbasisB.

Sensitivityanalysis

AftersolvingP = min{c ⊤ x : Ax = b,x ≥ 0},we usuallyhave

anoptimal basis B and optimal primal-dualsolutionsx ⋆ ,u ⋆ .

Is B stillfeasible/optimalif

◮ A,b,orcchange?

Dependson primal and dualoptimality: doesthe new problem

P ′ = min{˜c ⊤ x : Ãx = ˜ b,x ≥ 0}stillhave

◮ Primal feasibility: B −1 b ≥ 0?

◮ Dual feasibility: cN −cBB −1 N ≥ 0?

Equalityconstrainta ⊤ m+1 x = bm+1 added

≡ Addingacolumn to the dualproblem

◮ If a ⊤ m+1 x⋆ = bm+1, STOP: stilloptimal.

◮ Assume a⊤ m+1 isnotalinearcombination ofrows ofA

A

◮ Otherwise,two casesfor Ã =

a⊤

,

m+1

˜

b

b = :

bm+1

◮ rank( Ã) = rank(Ã|˜ b): the new constraint isredundant

◮ rank( Ã) < rank(Ã|˜ b): the problem isinfeasible.

Equalityconstrainta ⊤ m+1 x = bm+1 added

◮ We may be tempted to check primal feasibility 1 ,if

necessaryperformone iterationof the dualsimplex.

◮ Problem: If the oldprimal-dualpair had solution (x ⋆ ,u ⋆ ),

forthe new primal-dualpair

P : min c⊤x D : max u⊤b +um+1bm+1

s.t. Ax = b s.t. (u⊤

A

,um+1)

a⊤ m+1

a⊤ m+1x = bm+1

x ≥ 0

≤ c ⊤ ,

u u⋆ ũ =

= may not beaBFSfor the dual

um+1 0

i.e. Theremaynot ∃m +1lin.ind.,activedualconstraintsatũ ⋆

1 Dual feasibility check not necessary.

Inequalityconstrainta ⊤ m+1 x ≤ bm+1 3 added

◮ If a ⊤ m+1 x⋆ ≤ bm+1, STOP

◮ Addvariablexn+1,withcn+1 = 0: a⊤ m+1x +xn+1 = bm+1

A 0

◮ Ã =

a⊤ m+1 1

◮ Define new basis ˜

B 0

B =

ǎ⊤

, ǎ

1

⊤ definedas before

◮ Basicsolution ˜x ⋆

x

=

bm+1 −a ⊤ m+1x⋆

≥ 0 ○ ..

◮ However, ˜ B is dualfeasible! Reducedcosts:

¯c = (c ⊤ ,0) − (c ⊤ B ,0)˜ B −1Ã = (c⊤ −c ⊤ B B−1 A,0) ≥ 0

⇒ Apply dual simplex

3 Note: oppositesign totextbook.

⇒Do ittheprimalway!

◮ Otherwise,suppose a ⊤ m+1 x⋆ < bm+1 2 . Thenadd fictitious

variablexn+1 with ludicrouscostM ≫ 0:

P : min c⊤x + Mxn+1

s.t. Ax = b

a⊤ m+1x +

x,

xn+1 = bm+1

xn+1 ≥ 0.

Here,the oldoptimal basis is stillfeasible!

⇒ Apply primal simplexfrombasis ˜

B 0

B =

ǎ⊤

1

... whereǎ ⊤ isthe subvectorof basiccomponents ofa ⊤ m+1

◮ The optimal solutionx ⋆⋆ of this problemhas

◮ x⋆⋆ n+1 > 0: Asin Phase I, the problem became infeasible

◮ x⋆⋆ n+1 = 0: The solution isnot only feasible but also optimal

2 Note: oppositesign totextbook.

Changesinb

b → ˜ b = b + δ

◮ Dual feasibilityis safe

◮ Primal feasibilitymay be not! Suppose xB = B −1 b.

◮ We mustensure B −1˜ b ≥ 0,or

B −1 (b + δ) ≥ 0.

◮ Suppose δ = δej = δ(0,0, . . . ,0,1,0, . . . ,0) ⊤

⇒ If xB + δB −1 ej = xB + δ(B −1 )j ≥ 0, where (B −1 )j is the j-th

columnof B −1 , STOP: primal feasibility

◮ Otherwise,apply dual simplexmethod fromcurrentbasis.

Changesinc

c → ˜c = c + δ

◮ Primal feasibilityis safe

◮ Dual feasibilitymay benot!

◮ Suppose ¯cN = cN −cBB −1 N ≥ 0.

◮ Suppose δ = δej = δ(0,0, . . . ,0,1,0, . . . ,0) ⊤

◮ If j /∈ B,i.e.,xj isnonbasic, then ¯cj → ¯cj + δ

⇒ ifcj + δ ≥ 0,then we’restilloptimal, otherwiseapply

primal simplexfromthe same basis

◮ xj basic: consider columnqofB −1 N associatedwithxj.

cN −cBB −1 N − δq ≥ 0 ⇔ ¯cN ≥ δq

◮ Again,we obtainlowerand upper bound for δ. If δ isout

ofthese bounds,apply primal simplex

ChangesinA

Twocases: Band N.

Case 1,N: aij → aij + δ with j /∈ B.

◮ Primal feasibilityis safe: B −1 b ≥ 0

◮ BecauseAj → Aj + δei,check reducedcostof xj:

¯cj = cj −cBB −1 Aj → cj −cBB −1 (Aj + δei)

◮ Atoptimality, cBB −1 = u ⋆ ,hence

cj −cBB −1 (Aj + δei) = ¯cj − δu ⋆ i ≥ 0

◮ If not,apply primal simplex

Case 2,B: Neither primal nor dualfeasibilityareensured:

B → ˜ B ⇒may imply that

˜B −1 b ≥ 0 and cN −cB ˜ B −1 N ≥ 0

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