Introduction to Operations Research

ANSWERS

Duality and Sensitivity Analysis

Prof. dr. E-H. Aghezzafir. Rodrigo Rezende Amaral

Exercise 1 (items a, c, h)

Exercise 1

0 5 10 15 20 25 30 35 400

5

10

15

20

25

30

35

40

x1

x2

x1 +2x

2 <= 40

x1 +x

2 <= 30

2x1 +x

2 <= 40

z = 40x1 +50x

2

ALE = 40/3BEER = 40/3z = 1200

x1 + 2x2 = 402x1 + x2 = 40

BV = {x1, x2, s2}

Exercise 1

The slope of the isoprofit line has to stay between the slopes of the corn and malt constraints:

z = c1x1 + 50x2 → x2 = (-c1/50) x1 + (1/50) z

x1 + 2x2 = 40 → x2 = (-1/2) x1 + 20

2x1 + x2 = 40 → x2 = (-2) x1 + 40

(-2) ≤ (-c1/50) ≤ (-1/2) → 25 ≤ c1 ≤ 100

Exercise 1

The current basis remains optimal for 20 ≤ b1 ≤ 50

b1 = 40 → z = 1200 (original problem)

b1 = 50 → z = 1400 (increased corn availability)

Shadow Price =(1400 - 1200)/(50 - 40) = $20/lb corn

Check:z = 1200 + 20*(20 - 40) = 800 (decreased corn availability b1 = 20)

Exercise 1

0 10 20 30 40 50 60 70 80 900

500

1000

1500

X: 0

Y: 0

lb corn available

optim

al z-v

alu

e

X: 10

Y: 400

X: 20

Y: 800

X: 30

Y: 1000X: 40

Y: 1200

X: 50

Y: 1400

X: 60

Y: 1500

X: 70

Y: 1500

Exercise 2

check the LINDO output.

Exercise 2

Exercise 1

a).50

b) 1.75

c) Basis would have to change

→ New Price ≥ Current Price + Reduced Cost

→ New Price ≥ 6.00 + .25 → New Price ≥ 6.25

d) Basis would not change (allowable increase = 30)

→ Current Profit + (∆b3) * Shadow Price = 97.50 + (100-90) *.75 = 105

e) Basis would not change (allowable increase = 3 ≤ 15-13 )

→ Current Profit + (∆c3) * Current Value X3 = 97.50 + (15-13) 7.50 = 112.50

Exercise 3

(ci is the coefficient of xi in the objective function)

x1 x2 x3 s1 s2 s3

basis cb 5 1 2 0 0 0

s1 0 0 0,1667 0 1 -0,1667 -0,8333 3

x1 5 1 -0,1667 0 0 0,1667 -0,1667 1

x3 2 0 1 1 0 0 1 2

zj 5 1,1667 2 0 0,8333 1,1667 9

cj-zj 0 -0,1667 0 0 -0,8333 -1,1667

Exercise 3

Primal

Dual: Solution:

x1 x2 x3 s1 s2 s3

basis cb 5 1 2 0 0 0

s1 0 0 0,1667 0 1 -0,1667 -0,8333 3

x1 5 1 -0,1667 0 0 0,1667 -0,1667 1

x3 2 0 1 1 0 0 1 2

zj 5 1,1667 2 0 0,8333 1,1667 9

cj-zj 0 -0,1667 0 0 -0,8333 -1,1667

Exercise 3

Exercise 4

Exercise 4

Exercise 4

a) Basis would not change (allowable increase = 20 > 310-300)

→ Current Profit + (∆c1) * Current Value X1= 32540 + 10 * 88 = 33420

c) $0. Carco does not need more steel

→ Slack > 0, Shadow Price = 0

d) Basis would not change (allowable decrease = 3 > 88-86)

→ Current Profit + (∆b6) * Shadow Price = 32540 - 2* (-20) = 32580

e) Constraints 4 (machine 2) and 5 (steel availability) are not binding (slack > 0,

shadow price = 0). So we only have to check machine 1 constraint:

→ 1.2 * Shadow Priceb1 = 1.2 * 400 < 600 → Carco should produce jeeps

Exercise 5

–1

!x1 x2 x3 s1 e2 a2 a3

basis cb 3 4 1 0 0 -M -M

e2 0 -3 0 0 1 1 -1 -2 15

x3 1 0 0 1 1 0 0 -1 40

x2 4 1 1 0 0 0 0 1 10

zj 4 4 1 1 0 0 3 80

cj-zj -1 0 0 -1 0 -M -M-3

Exercise 5

Exercise 5

BV cB x1 x2 x3 s1 e2 a2 a3 RHS

c1 4 1 0 0 -M -M

e1 0 -3 0 0 1 1 -1 -2 15

x3 1 0 0 1 1 0 0 -1 40

x2 4 1 1 0 0 0 0 1 10

Zj 4 4 1 1 0 0 3 80

Cj-Zj c1-4 0 0 -1 0 -M -M-3

Current basis remains optimal if

c1 - 4 ≤ 0 → c1 ≤ 4

Exercise 5

40+10c2

Exercise 6

a) 12.50

b) 75.00

c) Basis would not change (allowable decrease = 6.66 > 45-40)

→ Current Profit + (∆b1) * Shadow Price = 4250 - 5* (75) = 3875

d) New c1 = 5 * 26 = 130

→ Basis would not change (allowable decrease = 30 > 150-130)

→ Current Profit + (∆c1) * Current Value X1= 4250 - 20 * 25 = 3750

e) Used Area and Used Labor constraints (1 & 2) are binding

→ 1 * Shadow Priceb1 + 3 * Shadow Priceb2 = 1*75 + 3*12.5 = 112.5 < 4*30

→ Farmer Leary should grow barley

Exercise 7

x1 x2 s1 e2 a2 a3

basis cb 3 1 0 0 -M -M

s1 0 0 0 1 0 0 -0,5 0,5

x2 1 0 1 0 -2 2 -1,5 1,5

x1 3 1 0 0 1 -1 1 1

zj 3 1 0 1 -1 1,5 4,5

cj-zj 0 0 0 -1 -M+1 -M-1,5

Exercise 7

Exercise 7

BV cB x1 x2 s1 e2 a2 a3 RHS

3 1 0 0 -M -M

s1 0 0 0 1 0 0 -1/2 1/2-1/2∆b3

x2 1 0 1 0 -2 2 -3/2 3/2-3/2∆b3

x1 3 1 0 0 1 -1 1 1+1∆b3

Zj 3 1 0 1 -1 3/2 9/2+3/2∆b3

Cj-Zj 0 0 0 -1 -M+1 -M-3/2

Current basis remains optimal if

-1 ≤ ∆b3 ≤ 1 → 6 ≤ b3 ≤ 8

New optimal solution if b3 = 15/2 (∆b3 = 1/2)

9/2+3/2*1/2 = 21/4

Exercise 10

Exercise 10

RS LS HB s1 s2 s3 e4 a4

Basis cb 3 5 4 0 0 0 0 -M

HB 4 0 0 1 0,1875 -0,125 0 0,9375 -0,938 187,5

LS 5 0 1 0 -0,05 0,1 0 0,75 -0,75 150

s3 0 0 0 0 -0,175 -0,15 1 -1,875 1,875 5025

RS 3 1 0 0 0 0 0 -1 1 1000

zj 3 5 4 0,5 0 0 4,5 -4,5 4500

cj-zj 0 0 0 -0,5 0 0 -4,5 -M+4,5

Exercise 10

Exercise 10

RS LS HB s1 s2 s3 e4 a4

Basis cb 3 5 4 0 0 0 0 -M

HB 4 0 1,25 1 0,125 0 0 1,875 -1,875 375

s2 0 0 10 0 -0,5 1 0 7,5 -7,5 1500

s3 0 0 1,5 0 -0,25 0 1 -0,75 0,75 5250

RS 3 1 0 0 0 0 0 -1 1 1000

zj 3 5 4 0,5 0 0 4,5 -4,5 4500

cj-zj 0 0 0 -0,5 0 0 -4,5 -M+4,5

This solution uses 16,500 sewing minutes.

Sewing time = 15 * RS + 15* LS + 4 * HB =

15 * 1000 + 15 * 0 + 4 * 375 = 16,500

Exercise 10

$0, since shadow price of sewing time is 0

All available sewing time is used in the

current optimal solution, but if it would be

(slightly) decreased, an alternative solution

with the same revenue can be found

Exercise 10

RS LS HB s1 s2 s3 e4 a4

Basis cb 3 5 4 0 0 0 0 -M

HB 4 0 0 1 0,1875 -0,125 0 0,9375 -0,938 187,5

LS 5 0 1 0 -0,05 0,1 0 0,75 -0,75 150

s3 0 0 0 0 -0,175 -0,15 1 -1,875 1,875 5025

RS 3 1 0 0 0 0 0 -1 1 1000

zj 3 5 4 0,5 0 0 4,5 -4,5 4500

cj-zj 0 0 0 -0,5 0 0 -4,5 -M+4,5