IEOR 4004 Midterm review(Part II)

March 12, 2014

Summary

• Matrices, Tableaux, and Dictionaries• Duality– Shadow prices– Complementary slackness

• Sensitivity analysis• Multiple solutions• Goal Programming

Matrices, Tableaux and Dictionaries

x1 − 2x2 + x3 + x4 = 1 − x1 + 3x2 + 2x3 + x4 = 2max z = 2x2 − 3x3

x1, x2, x3, x4 ≥ 0

3 21 1B-1=−B-1N = −5 −8−2 −3B-1b = 73

x1 = 7 − 5x3 − 8x4 x2 = 3 − 2x3 − 3x4 z = 6 − 7x3 − 6x4 cN − cBB-1N = (−7, −6)

Basis: {x1, x2}B NcB cN bxB xN1 −2−1 3B= 1 211

N=

cBB-1b = 6

x1 x2 x3 x4 z 0 0 7 6 6 x1 1 0 5 8 7 x2 0 1 2 3 3

Duality

• Normal formmax 3x1 + 2x2x1 + x2 ≤ 802x1 + x2 ≤ 100x1, x2 ≥ 0

y1 + 2y2 ≥ 3y1 + y2 ≥ 2y1y2 x1x2min 80y1 + 100y2

y1, y2 ≥ 0PRIMAL

DUAL

Primal max Dual min ≤ constraint ≥ 0 variable ≥ constraint ≤ 0 variable= constraint unrestricted var ≥ 0 variable ≥ constraint ≤ 0 variable ≤ constraintunrestricted var = constraint

Duality• (Weak duality) the value

of every dual solution is an upper bound on the value of every primal solution (for maximization)

• (Strong duality) if both primal and dual are feasible, then each have an optimal solution of the same value (no gap)

DUAL(min)

PRIMAL(max)

Simplex

DualSimplex

Optimalsolution

objectivevalue

max 3x1 +2x2x1 + x2 + s1 = 802x1 + x2 + s2 = 100x1, x2, s1, s2 ≥ 0

Duality graphically

20 40 60 800

20

40

60

80

x1 x2s2

s1

max 3x1 +2x2x1 + x2 ≤ 802x1 + x2 ≤ 100x1, x2 ≥ 0 min 80y1 +100y2y1 + 2y2 − t1 = 3y1 + y2 − t2 = 2y1, y2, t1, t2 ≥ 0

min 80y1 + 100y2y1 + 2y2 ≥ 3y1 + y2 ≥ 2y1, y2 ≥ 0

1 2 30

1

2

y1 y2t2t1

100

feasible

optimal

duallyfeasible

complementarysolution

(shadow prices)

Complementary slackness(x1,…,xn) a primal solution, (y1,…,ym) a dual solution are complementary to one another if for every i:a) If i-th constraint in primal is not tight for values (x1,…,xn),

then yi=0b) if xi is positive, then the corresponding dual constraint is tight

for values (y1,…,ym)(Complementary slackness) If both (x1,…,xn) and (y1,…,ym) are optimal, then they are complementary to one another.80y1 + 100y2 = vy1 + 2y2 ≥ 3y1 + y2 ≥ 2y1, y2 ≥ 0

DUAL (min)

3x1 + 2x2 = zx1 + x2 ≤ 802x1 + x2 ≤ 100x1, x2 ≥ 0PRIMAL (max)

x1 : 20x2 : 60y1 : 1y2 : 1

50001.5010002

Complementary sol’s

Shadow prices

• πi = shadow price of i-th constraint– change in the value of z when bi changed to bi+1

• every solution yields different shadow prices• shadow prices = complementary dual solution

PRIMAL(max)

DUAL(min)

x πx π

feasible

infeasible feasible

infeasible

If both feasiblethen both optimal

max 3x1 + 2x2π1: x1 + x2 ≤ 80π2: 2x1 + x2 ≤ 100x1, x2 ≥ 0x1 = 20x2 = 60z = 180

solution

complementary

complementary

π1 = 1π2 = 1

Pricing-out variables

• pricing-out = calculating cost in shadow prices“multiplying the column of xi by shadow prices”

• reduced cost = coefficient ci in z minus cost

• If reduced cost positive, increasing xi increases zmax 3x1 + 2x2 x1 + 1x2 ≤ 80 2x1 + 1x2 ≤ 100x1, x2 ≥ 0

x1 = 50x2 = 0z = 150solutionπ1 = $0π2 = $1.5Price out x2:

Reduced cost of x2:

increasing x2 yields a better solution positive

1 × $0 + 1 × $1.5$2 − $1.5 = $0.5

opportunitycost

x1 = 20 + s1 − s2

x2 = 60 − 2s1 + s2

z = 180 −s1 − s2

Complementary dictionary

x1x2s1s2

t1t2y1y2

values objective

BasicNon-basic

Non-basicBasic

rotate

replacecomplementary

variables

− y1 =− y2 =

t2 t1

−1 − 2t2 +t1−1 +t2 − t1y1 = 1 + 2t2 − t1y2 = 1 − t2 + t1

(min) v = 180+60t2+20t1Complementarydual dictionary

(dual basis)

x1 = 20 + s1 − s2x2 = 60 − 2s1 + s2z = 180 −s1 − s2

Primaldictionary

(basis)(max)primal

vars

primalslack

dualslack

dualvars

Complementarityrules

Complementarity - summary

Max cx Min by Ax + s = b yA − t = c x, s ≥ 0 y, t ≥ 0• primal and dual bases are complementary• If (xB,sB) is a primal basis then (yN,tN) is the

corresponding complementary dual basis.xi × ti = 0 yj × sj = 0(complementarity)• values and reduced costs switch roles

Sensitivity analysis

• Determining the effect of changes in the input (coefficients) on the (optimal) solution

Problem

Modifiedproblem

Optimal(modified)

basis

Optimalsolution

Optimal(modified)

solution

Optimalbasis

Sensitivity analysis

x1 − 2x2 + x3 + x4 = 1 − x1 + 3x2 + 2x3 + x4 = 2max z = 2x2 − 3x3

x1, x2, x3, x4 ≥ 03 21 1B-1= −B-1N = −7 −5−3 −2B-1b = 73

x1 = 7 − 7x3 − 5x4 x2 = 3 − 3x3 − 2x4 z = 2x2 − 3x3= 6 − 9x3 − 4x4

x1 = 7 + 3Δ ≥ 0x2 = 3 + 1Δ ≥ 0Changing rhs b1=1 to (1+Δ)

− 9 − 3Δ ≤ 0− 4 − 2Δ ≤ 0

Changing z from 2x2 to (2+Δ)x2

cN − cBB-1N = (−9, −4)

1 −2−1 3B= Optimal basis:{x1, x2}

− 9 + Δ ≤ 0− 3x3 to (− 3+Δ)x3

Alternative solutions• Every LP has 0, 1, or ∞ optimal solutions• If (x1,x2) and (x1’ , x2’) are optimal solutions

then so is every convex combination(x1’’,x2’’) = λ(x1,x2) + (1 − λ) (x1’ , x2’)where 0 ≤ λ ≤ 1

max 2x1 + 2x2 x1 + x2 ≤ 80 2x1 + x2 ≤ 100x1, x2 ≥ 0

optimal (basic) solutions: (0,80) and (20,60) x1 = 20 + s1 − s2x2 = 60 − 2s1 + s2z = 180 − 2s1

x1 = 20 − Δx2 = 60 + Δ 0 ≤ Δ ≤ 20

(x1, x2) = λ(0,80)+(1 − λ) (20,60)x1 = 20 − 20 λ x2 = 60 + 20 λ

Goal ProgrammingConstraints:x1 + x2 ≤ 802x1 + x2 ≤ 100x1, x2 ≥ 0

Goals#1: 3x1 + 2x2 ≥ 140#2: 3x1 + 4x2 ≤ 130#3: x1 ≤ 40Penalties:

#1: pay $1.2/unit#2: pay $0.8/unit#3: pay $3/unitfor each missing unit

Introduce slack (unrestricted)#1: 3x1 + 2x2 − s1 = 140#2: 3x1 + 4x2 + s2 = 130#3: x1 + s3 = 40

We pay penalty onlyif slack is negative

s1 = s1+ − s1−s2 = s2+ − s2−s3 = s3+ − s3−

Penalties:1.2s1−0.8s2−3s3−Positive part

Negative part

Minimize total penalty!

Minimize 1.2s1− + 0.8s2− + 3s3−3x1 + 2x2 − s1+ + s1− = 1403x1 + 4x2 + s2+ − s2− = 130x1 + s3+ − s3− = 40x1 + x2 ≤ 802x1 + x2 ≤ 100x1, x2, s1+, s1−, s2+, s2−, s3+, s3− ≥ 0

Notes• Substitute variables if it helps• Use variants of Simplex if it helps• Solve the dual if it is easier• Optimal basis is all you need to find out• Shadow prices = dual variables• Pricing out = checking dual constraints• Any feasible dual solution is an upper (lower)

bound if we maximize (minimize) in primal• LP has 0,1,or ∞ solutions• If both primal and dual are feasible then they both

have optimal solutions

Write your answers clearly !!!(for the questions asked)