DOC PREVIEW
Clemson MTHSC 440 - lecture 23

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

MthSc 440/H440/640: Linear ProgrammingLecture 23Pietro BelottiDept. of Mathematical SciencesClemson UniversityNovember 18, 2010Reading for today: pages 158-166.Sensitivity AnalysisAfter solving an LP problem and getting an optimum x⋆,sometimes we need to re-solve the LP, due to changes in◮variable bounds [ℓ, u] → [˜ℓ,˜u]◮right-hand side b →˜b◮objective function coefficients c →˜c◮coefficient matrix A →˜AWe could simply solve the new LPmax˜c⊤xs.t.˜Ax ≤˜bx ∈ [˜ℓ,˜u]from scratch, and find the (possibly different)¯x⋆.Or we could simply use x⋆and fewer steps to find¯x⋆.How to deal with:◮changes in c◮changes in b◮added variable◮added constraint◮all of the aboveUsing the right toolChanges to A, b, c, ℓ, or u may have different consequences.The old solution x⋆, associated with a basis B, might be(A) feasible (B−1˜b ≥ 0), but not optimal (˜cN−˜cBB−1N 6≤ 0)i.e. primal feasible, but not dual feasible(B) infeasible (B−1˜b 6≥ 0), but dual feasible (˜cN−˜cBB−1N ≤ 0)(C) neither primal nor dual feasible ..◦We know that if a basic solution is◮primal feasible but dual infeasible1⇒ useprimal simplex◮primal infeasible but dual feasible ⇒ use dual simplexCase (C) may require running some iterations of both.1i.e. not optimalNew objective function: c →˜c◮Constraints don’t change, so x⋆(old optimum) is stillfeasible: Ax⋆≤ b and x⋆∈ [ℓ, u]◮Basis B associated with x⋆might be dual infeasible⇒ Check˜cN−˜cBB−1N:◮solve yB =˜cB◮compute˜cN− y⊤N◮if˜cN− y⊤N ≤ 0 ⇒ still dual feasible ⇒ optimum!◮otherwise, apply primal simplexNew right-hand side: b →˜bWe could write the dual problem and apply the same as theprevious slide . . . But let’s stay on the primal side.◮Constraints do change, so x⋆could now be infeasiblei.e. it could be Ax⋆6≤˜b◮However, B is still a (possibly infeasible) basis!◮Nonbasic variables are still nonbasic (no change in ℓ or u):∀i ∈ N , x⋆i∈ {ℓi, ui}.◮Though primal feasibility might be gone, dual feasibilityis not: we still have cN− cBB−1N ≤ 0.⇒ Check B−1˜b:◮solve¯xB =˜b◮if¯x ≥ 0 ⇒ still primal feasible ⇒ optimum!◮otherwise, apply dual simplexAdding a variable3xn+m+1Objective coefficient: cn+m+1; associated column in A:An+m+1= (a1,n+m+1, a2,n+m+1. . . , am,n+m+1). Suppose it has zerolower or upper bound:0 ≤ xn+m+1≤ un+m+1orℓn+m+1≤ xn+m+1≤ 0.Take the same basis B and define¯x = (x⋆1, x⋆2. . . , x⋆n+m,0), andmake xn+m+1nonbasic. B still feasible: B−1b ≥ 0.◮Consider˜c = (c1, c2. . . , cn+m, cn+m+1)⇒˜cB= cBand˜cN= (cN,cn+m+1)◮Solve yB = cBif necessary2◮Compute˜cN− yN; if ≤ 0, ⇒ optimum!◮Otherwise, apply primal simplex.2We may have y from the last iteration if we used primal simplex.3We call it xn+m+1because we had n + m variables (original + slack).Adding a xn+m+1with ℓn+m+16= 0 6= un+m+1This is a little more complicated:◮We cannot set xn+m+1to zero and treat it as nonbasic.◮We could make it nonbasic and set it to one of its boundsℓn+m+1or un+m+1, but then the basis would readxB= B−1b − B−1NxN6= x⋆unlike the case where ℓn+m+1= 0 ∨ un+m+1= 0.◮The basis may be neither primal nor dual feasibleAdding a constraintFrom a dual standpoint, this is like adding a variable.◮Suppose it is a constraintPni=1am+1,ixi≤ bm+1◮Add a slack variable ⇒Pni=1am+1,ixi+xn+m+1= bm+1◮Make xn+m+1basic: B ← B ∪ {n + m + 1}◮This basis may be infeasible, i.e.,Pni=1am+1,ix⋆i6≤ bm+1◮set x⋆n+m+1= bm+1−Pni=1am+1,ix⋆i◮if x⋆n+m+1< 0, primal infeasible◮however, it is dual feasible⇒ apply the dual simplex◮if x⋆n+m+1≥ 0, primal feasible ⇒ optimalChanging A, b, c, ℓ, and u◮x⋆is primal and dual feasible.◮convert it to a new basic solution¯x◮In the conversion,¯x may turn out◮dual feasible (but primal inf.) ⇒ apply dual simplex◮primal feasible (but dual inf.) ⇒ apply primal simplexChanging A, b, c, ℓ, and u (cont.)Computing¯x is not trivial.◮If B did not change (i.e. changes A →˜A = (B|˜N) affect onlyN) then B is still nonsingular⇒ Keep B as the basis, then◮For each i ∈ N , set¯xias follows:◮if˜ℓi= −∞ and˜uifinite, set¯xi=˜ui◮if˜ui= +∞ and˜ℓifinite, set¯xi=˜ℓi◮if˜ℓi= −∞,˜ui= + ∞ (i.e. xiu.r.s.), set¯xi= x⋆i◮if both finite, set xito either◮Then set¯xBto B−1˜b − B−1N¯xNi.e. solve the system B¯xB=˜b −˜N¯xnThe resulting¯x may be primal feasible, dual feasible, both,


View Full Document

Clemson MTHSC 440 - lecture 23

Documents in this Course
Load more
Download lecture 23
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view lecture 23 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view lecture 23 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?