aaoczc2252

Sensitivity Analysis

The optimal solution of a LPP is based on the conditions that prevailed at the time the LP model was formulated and solved. In the real world, the decision environment rarely remains static and it is essential to determine how the optimal solution changes when the parameters of the model are changed. That is what sensitivity analysis does. It provides efficient computational techniques to study the dynamic behavior of the optimal solution resulting from making changes in the parameters of the model.

In studying the sensitivity analysis, we should be familiar with the lingo that is being used in LPP situations. A general LPP is of the form

Maximize (or Minimize) nn xcxcxcz ...2211

subject to the constraints

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 2

...

...

.

...

, ,..., 0

n n

n n

m m mn n m

n

a x a x a x b

a x a x a x b

a x a x a x b

x x x

The RHS constants of the constraints, mbbb ,...,, 21

are referred to resources or availabilities of the problem.

The objective coefficients, nccc ,...,, 21

are referred to as unit profits (or unit costs).

The decision variables, nxxx ,...,, 21

are referred to as units of activities 1, 2, …, n.

Dual Price of a constraint

This measure actually represents the unit worth of a resource - that is it gives the contribution to the objective function resulting from a unit increase or decrease in the availability of a resource. In terms of duality theory, the dual price of a resource (=constraint) i, is precisely the value of the optimal dual variable yi associated with the constraint i. (Did you understand why it is called “dual price”?). Other non-suggestive names include shadow prices and simplex multipliers.

Reduced cost of a variable xj (=activity j)

is defined as

Cost of consumed resources per unit of activity xj

- profit per unit of activity xj

= zj - cj

Changes affecting feasibility

The feasibility of the current optimum solution may be affected only if

(1) The RHS of the constraints are changed

OR

(2) A new constraint is added to the model.

In both cases, infeasibility occurs when at least one element of the RHS of the optimal tableau becomes negative – that is, one or more of the current basic variables become negative.

is changed to

First we consider the changes in the optimal solution due to changes in the RHS bi. We note that the optimal solution is given by

1B

B bxwhere b is the old RHS and B is the basic matrix. Remember B-1 is found in the optimal tableau below the entries which initially had an identity submatrix.

The optimal value is given by bB 1BC z

b bWhen , the correspondingnew solution and new objective value are got by replacing with .bb

Example 4.3-2 (Pages 135-136)

TOYCO assembles three types of toys: trains, trucks and cars using three operations. The daily limits on the available times for the three operations are 430, 460, and 420 minutes respectively. The profits per toy train, truck and car are $3, $2, and $5 respectively. The assembly times per train at the three operations are 1, 3, and 1 minute respectively. The corresponding times per truck and per car are (2, 0, 4) and (1, 2, 0) minutes respectively. A zero indicates that the operation is not used.

Letting x1, x2, and x3 represent the daily number of units assembled of trains, trucks and cars, the LPP model is:

Maximize 321 523 xxxz

subject to

0,,

4204

46023

4302

321

21

31

321

xxx

xx

xx

xxx

The optimal tableau is given in the next slide. (Note: x4, x5, and x6 are slack variables there.)

x6 0 2 0 0 -2 1 1 20

x3 0 3/2 0 1 0 1/2 0 230

z 1 4 0 0 1 2 0 1350 Basic z x1 x2 x3 x4 x5 x6 Sol

x2 0 -1/4 1 0 1/2 -1/4 0 100

This is the optimal tableau

Suppose that TOYCO wants to change the capacities of the three operations according to the following cases:

460 500 300 450

(a) 500 (b) 400 (c) 800 (d) 700

400 600 200 350

Use Sensitivity analysis to determine the optimal solution in each case.

112

02/10

04/12/1

;052 1BCB

Solution: We note

(b) New Solution is

6

3

2

x

x

x

600

400

500

112

02/10

04/12/11bB

0

200

150

Since this is feasible, it is optimal and the new optimal value =

(c) New Solution is

6

3

2

x

x

x

200

800

300

112

02/10

04/12/11bB

400

400

50

13002005150203

This is not feasible. So we apply dual simplex method to restore feasibility. We note

1900400550203 new z =

x6 0 2 0 0 -2 1 1 400

x3 0 3/2 0 1 0 1/2 0 400

z 1 4 0 0 1 2 0 1900 Basic z x1 x2 x3 x4 x5 x6 Sol

x2 0 -1/4 1 0 1/2 -1/4 0 -50

x6 0 1 4 0 0 0 1 200

x5 0 1 -4 0 -2 1 0 200

z 1 2 8 0 5 0 0 1500

x3 0 1 2 1 1 0 0 300

This is the new optimal tableau.

Feasibility Range of the Elements of the RHS

Another way of looking at the effect of changing the availabilities of the resources, bi, is to determine the range for which the current solution remains feasible.

For example if, in the TOYCO model, b2 is changed to b2+D2= 460+D2, we want to find the range of D2 so that the current solution remains optimal.

When b2 is changed to b2+D2= 460+D2, the new solution is

6

3

2

x

x

x

420

460

430

112

02/10

04/12/1

21 DbB

2

2

2

2/1

4/1

20

230

100

D

D

D (current optimal solution + D2 times the 2nd column of B-1.)

2

2

2

20

2/1230

4/1100

D

D

D

is feasible if

020

02/1230

04/1100

2

2

2

D

D

D

20

460

400

2

2

2

Dor

Dor

Dor

40020 2 DOr

Thus current solution remains optimal if RHS of the 2nd constraint lies between 440 and 860 (the other RHSs being the same).

Problem 5 Problem Set 4.5B Page 151

HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation:

Unit Resources RequirementsResource Model 1 Model2 Maximum Availability

(units) (units) (units)

Resistor 2 3 1200

Capacitor 2 1 1000

Chips 0 4 800

Unit Profit($) 3 4

Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP

Maximize 21 43 xxz

subject to

0,

8004

10002

120032

21

2

21

21

xx

x

xx

xx (Resistors)

(Capacitors)

(Chips)

Taking the slack variables as s1, s2, s3, the optimal tableau is:

x2 0 0 1 1/2 -1/2 0 100

s3 0 0 0 - 2 2 1 400

z 1 0 0 5/4 1/4 0 1750 Basic z x1 x2 s1 s2 s3 Sol

x1 0 1 0 -1/4 3/4 0 450

This is the optimal tableau

(a) Determine the status of each resource

Answer: Since s1 = 0 = s2, the resistor and the capacitor resources are scarce.

Since s3 > 0, the chips resource is abundant.

(b) In terms of the optimal profit, determine the worth of one resistor, one capacitor and one chip.

Answer: They are respectively y1, y2, y3 the dual optimal solution and hence are

5/4, 1/4, 0 respectively.

(c) Determine the range of applicability of the dual prices for each resource.

Resistor: If D1 is the increase in the resource 1,

the new optimal solution is given by

2

3

1

x

s

x

800

1000

1200

02/12/1

122

04/34/1 11

D

bB

1

1

1

2/1

2

4/1

100

400

450

D

D

D

0 gives 200200 1 D

Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given by

200200 2 D

And for a change D3 in the chips, the range of feasibility is given by 4003 D

(d) If the available number of resistors is increased to 1300 units, find the new optimum solution.

The new solution is:

2

3

1

x

s

x 450 25

400 200

100 50

150

200

425 This is feasible and hence optimal.And z = 1875.

Yes, as D1 = 100

(g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer?

Addition of a new constraint

The addition of a new constraint to an existing model can lead to one of two cases:

1. The new constraint is redundant, meaning that it is satisfied by the current optimal solution and hence can be dropped altogether from the model altogether.

2. The current solution violates the new constraint in which case the dual simplex method can be used to recover feasibility.

We introduce this constraint in the final simplex tableau as an additional row where the usual additional (slack or surplus) variable is taken as the basic variable for this new row. Because this new row will probably have non-zero coefficients for some of the other basic variables, the conversion to proper form for Gaussian elimination is done and then reoptimization is done in the usual way. The following example illustrates this.

Example: Consider the LPP

Maximize 3212 xxxz subject to

0,,

20

102

603

321

321

321

321

xxx

xxx

xxx

xxx

Applying the Simplex method, the optimal tableau is given in the next slide. Now a new constraint

1 2 33 2 28x x x solution.

is added. Find the new optimal

x2 0 0 1 - 3/2 0 - 1/2 1/2 5

x1 0 1 0 1/2 0 1/2 1/2 15

z 1 0 0 3/2 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol

s1 0 0 0 1 1 - 1 - 2 10

s4 0 3 -2 1 0 0 0 1 28

s4

0

0

0

0

z 1 0 0 0 0 3/7 2/7 3/7 22

x2 0 0 1 0 0 4/7 5/7 -3/7 8

x1 0 1 0 0 0 1/7 3/7 1/7 14

s1 0 0 0 0 1 - 12/7 -15/7 2/7 8

x3 0 0 0 1 0 5/7 1/7 -2/7 2

0 0 0 -7/2 0 - 5/2 -1/2 1 -7

Changes affecting Optimality

1. Changes in the objective coefficients cj

The changes in the objective coefficients, cj , affect "zj - cj" and hence the optimality. We recompute the z-Row by replacing cj with c'j . We note that CB will also change to C'B.

In the previous problem, we replace the objective function with

321 323 xxxz .

Find the new optimal solution.

Thus c1 is changed from 2 to 3; c2 is changed from -1 to -2; and c3 is changed from 1 to 3;

We note that in z-Row, the coefficients of basic variables x1 , x2 , and x4 will remain zero now also. We have only to calculate the new coefficients of the non-basic variables only and the new z.

Coefficient of x3 = z3 – c3 33

1 cABCB

02

33

2/3

2/1

1

230

55

1 cABCB

02

50

2/1

2/1

1

230

Coefficient of x5 = z5– c5

Coefficient of x6 = z6– c6 66

1 cABCB

02

10

2/1

2/1

2

230

Since all zj– cj are 0, the current solution remains optimal. Of course the new z is

350352153

If in the previous problem, we replace the objective function with 321 332 xxxz

You can verify that z6– c6 = -1/2 < 0. Thus optimality is spoiled. By applying the regular Simplex method, we can show that the new optimum solution is x1 = 10, x2 = 0, x3 = 0 with new z = 20.

Addition of a new activity

(= Changes in the coefficients of an existing activity)

Suppose a new activity n+1 is added with coefficients

1,

1,2

1,1

1

.

nm

n

n

n

a

a

a

c

To find the effect of this on the current optimal solution, we pretend this activity was present initially with all coefficients zero. Hence the new coefficients are calculated using the formulae:

In z-row, the coefficient of xn+1 is

111

11

nnBnn cABCcz

And in the constraint matrix its coefficients are

11

nABIf zn+1 – cn+1 satisfies the optimality condition, the current solution is optimal. Else we apply regular simplex method to restore optimality.

The same procedure is adopted when the coefficients of an existing variable are changed. If the variable ia basic variable, we should see that the new tableau is in proper form (i.e. coefficients of other basic variables in that column should be made zero).

In the previous problem the coefficients of the (non-basic) variable x3 are changed from

1

2

1

1

33

23

13

3

a

a

a

c

to

2

1

3

2

33

23

13

3

a

a

a

c

Using sensitivity analysis, find the new optimal solution and value.

The only change in the optimal tableau will be the x3 column. We calculate the new x3 column and the coefficient of x3 in the z-row.

We first note that

2

1

2

10

2

1

2

10

211

;120 1BCB

Hence the new x3 column is 31AB

2

1

3

2

1

2

10

2

1

2

10

211

2

32

16

New coefficient of x3 in z-row =

331

33 cABCcz B

2

2

32

16

120

02

3

Thus optimality is disturbed. We replace the x3 column in the original optimal tableau by the new values and then find the new optimal solution and optimal value by regular simplex method.

x2 0 0 1 0 - 1/2 1/2 5

x1 0 1 0 0 1/2 1/2 15

z 1 0 0 0 3/2 1/2 25 Basic z x1 x2 x3 s1 s2 s3 Sol

s1 0 0 0 1 - 1 - 2 10

z 1 0 0 0 1/4 5/4 0 55/2

x2 0 0 1 0 1/4 -3/4 0 15/2

x1 0 1 0 0 1/12 5/12 1/3 95/6

x3 0 0 0 1 1/6 - 1/6 -1/3 5/3

-3/2

6

-1/2

-3/2

In the previous problem a new variable x7 is introduced with coefficients

2

1

2

1

37

27

17

7

a

a

a

c


This is like the previous case. We assume the variable x7 was already present with all coefficients 0.

We first note that

2

1

2

10

2

1

2

10

211

;120 1BCB


2

1

2

2

1

2

10

2

1

2

10

211

2

12

37


771

77 cABCcz B

)1(

2

12

37

120

02

7

Hence the original solution remains optimal with the same objective value and it does not help by introducing this new variable.

In the previous problem the coefficients of the (basic) variable x1 are changed from

1

1

3

2

31

21

11

1

a

a

a

c

to


The only change in the optimal tableau will be the x1 column. We calculate the new x1 column and the coefficient of x1 in the z-row.

0

2

2

1

31

21

11

1

a

a

a

c

We first note that

2

1

2

10

2

1

2

10

211

;120 1BCB


0

2

2

2

1

2

10

2

1

2

10

211

1

1

0


111

11 cABCcz B

1

1

1

0

120

2

We replace the x1 column in the original optimal tableau by the new values.

x2 0 1 -3/2 0 - 1/2 1/2 5

x1 0 0 1/2 0 1/2 1/2 15

z 1 0 3/2 0 3/2 1/2 25

Basic z x1 x2 x3 s1 s2 s3 Sol

s1 0 0 1 1 - 1 - 2 10

z 1 0 1/2 0 0 1/2 0 5

s3 0 0 1 -1 0 0 1 20x1 0 1 -1/2 1 0 1/2 0 5

s1 0 0 2 -1 1 - 1 0 50

2

0

1

-1

1 0 0 1/2 0 1/2 -1/2 -5

0 0 1 -1 0 0 1 20

Problem 5 Problem Set 4.5B Page 151

HiDec produces two models of electronic gadgets that use resistors, capacitors, and chips. The following table summarizes the data of the situation:

Unit Resources RequirementsResource Model 1 Model2 Maximum Availability

(units) (units) (units)

Resistor 2 3 1200

Capacitor 2 1 1000

Chips 0 4 800

Unit Profit($) 3 4

Let x1, x2 be the amounts produced of Models 1 and 2 respectively. Then the above model becomes the LPP

Maximize 21 43 xxz

subject to

0,

8004

10002

120032

21

2

21

21

xx

x

xx

xx (Resistors)

(Capacitors)

(Chips)

Taking the slack variables as x3, x4, x5, the optimal tableau is:

(c) Determine the range of applicability of the dual prices for each resource.

Resistors: If D1 is the increase in the resource 1,

the new optimal solution is given by

2

3

1

x

s

x

800

1000

1200

02/12/1

122

04/34/1 11

D

bB

1

1

1

2/1

2

4/1

100

400

450

D

D

D

0 gives 200200 1 D

Similar calculations show that, for a change D2 in the capacitor, the range of feasibility is given by

200200 2 D

And for a change D3 in the chips, the range of feasibility is given by 4003 D

(d) If the available number of resistors is increased to 1300 units, find the new optimum solution.

The new solution is:

2

3

1

x

s

x

50

200

25

100

400

450

150

200

425 This is feasible and hence optimal.And z = 1875.

(e) If the available number of chips is reduced to 350 units, will you be able to determine the new optimal solution directly from the given information? Explain.

Answer: NO, as the feasibility is not spoilt only if the number of chips is 400400800

(f) If the availability of capacitors is limited by the range of applicability computed in (c), determine the corresponding range of the optimal profit and the corresponding ranges for the number of units to be produced of Models 1 and 2.

If the number of capacitors is changed from b2= 400 to b2 = 400+D2, where 200200 2 D

then 4

3450 2

1

Dx

2100 2

2

Dx

41750 2D

z

Hence600300 1 x

2000 2 x

18001700 z

(g) A new contractor is offering to sell HiDec additional resistors at 40 cents each but only if HiDec would purchase at least 500 units. Should HiDec accept the offer?

Answer: Since the new value of b1, the number of resistors is (at least) 1700, the feasibility is spoilt (as D1 > 200). The new solution is (taking D1 =500)

2

3

1

x

s

x

250

1000

125

100

400

450

350

600

325

And new z = 2375. We restore Optimality by Dual Simplex.

Problem 9 Problem Set 4.5E Page 159

Consider the HiDec model in Problem 5, Problem Set 4.5B

(a) Determine the conditions that will keep the current solution optimum if the unit profits of Models 1 and 2 are changed simultaneously.

(b) Suppose that the objective function is changed to

21 25 xxz

Determine the associated optimum solution.

Solution. (a) Let c1 = 3 be changed to c1 =3+ d1. Let c2 = 4 be changed to c2 =4+ d2.

Thus the new coefficient of the non-basic variable s1 in the z-row is z3– c3 33

1 cABCB

0

2/1

2

4/1

403 21

dd

4

5

2421

dd0 for retaining optimality

Thus the new coefficient of the non-basic variable s2 in the z-row is z4– c4 44

1 cABCB

0

2/1

2

4/3

403 21

dd

4

1

24

3 21 dd

0 for retaining optimality

These are the two conditions on d1 and d2 .

Solution (b): Here new objective function is

21 25 xxz

Thus d1 = 5 - 3 =2 and d2 = 2 - 4 = -2.

New z3– c3 = 04

1 New z4– c4 = 0

4

11

Thus optimality is spoiled. And we reoptimize using regular Simplex method.

Note also new z

100

400

450

205 =2450

Problem 3 Problem Set 4.5F Page 161

In the TOYCO model, suppose that a new toy (fire engine) requires 3, 2, and 4 minutes respectively, on operations 1, 2, and 3. Determine the new optimal solution when the profit per unit is given by

(a) $5 (b) $10 (Optimal tableau in next page)

Solution: This amounts to introducing a new variable x7 with coefficients as follows.

37

27

17

7

)(

a

a

a

c

a ;

4

2

3

5

37

27

17

7

)(

a

a

a

c

b

4

2

3

10

We first note that

112

02

10

04

1

2

1

;052 1BCB

Hence the new x7 column is7

1AB

4

2

3

112

02

10

04

1

2

1

0

1

1


771

77 cABCcz B

5

0

1

1

052

2

Thus in case (a) the current solution is optimal and it does not help to introduce the new toy.

in case (a)

And = -3 in case (b)

In case (b), optimality is spoiled and we restore it using Regular simplex method. (see the next slide)

Consider the LPP

Maximize 3212 xxxz subject to

0,,

4

3

15223

321

321

321

321

xxx

xxx

xxx

xxx

The optimal tableau is given in the next slide.

We first note that

021

110

031

;201 1BCB


3

2

1

021

110

031

3

1

5


111

11 cABCcz B

1

3

1

5

201

2

Thus optimality is spoiled. We try to restore it.

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