03.1 Chapter3_SensitivityAnalysis.pdf

8/10/2019 03.1 Chapter3_SensitivityAnalysis.pdf

1/35

Sensitivity Analysis

based on:Bradley, Hax, and Magnanti; Applied

Mathematical Programming, Addison-Wesley,1977Chapter 3, sections 3.1, 3.2, 3.3, 3.4 and 3.5

(FEUP | DEGI) Operations Research March 19, 2014 1 / 35
http://find/http://goback/


2/35


We have already been introduced to sensitivity analysis via the geometry of a simple example.We saw that the values of the decision variables and those of the slack and surplus variablesremain unchanged even though some coefficients in the objective function are varied.We also saw that varying the righthand-side value for a particular constraint alters the optimalvalue of the objective function in a way that allows us to impute a per-unit value, or shadowprice, to that constraint. These shadow prices and the shadow prices on the implicit nonnegativityconstraints, called reduced costs, remain unchanged even though some of the righthand-side

values are varied.Since there is always some uncertainty in the data, it is useful to know over what range and underwhat conditions the components of a particular solution remain unchanged.Further, the sensitivity of a solution to changes in the data gives us insight into possibletechnological improvements in the process being modeled. For instance, it might be that theavailable resources are not balanced properly and the primary issue is not to resolve the mosteffective allocation of these resources, but to investigate what additional resources should be

acquired to eliminate possible bottlenecks. Sensitivity analysis provides an invaluable tool foraddressing such issues.

http://find/


3/35


Learning Objectives

Define shadow price of a constraint.

Calculate shadow prices using only the initial and final simplex tableaux.

Define reduced cost of a decision variable (activity).

Calculate reduced costs using only the initial and final simplex tableaux.

Price out an activity.

Calculate the reduced cost of a variable using the shadow prices and the problem data.

Calculate shadow prices using the problem data and the optimal basis.

Determine how much the objective function coefficients can vary without changing the values of thedecision variables in the optimal solution.

Determine which variables will enter and leave the basis when the new cost coefficient reaches either ofthe extreme values of the range.

Determine how much the right-hand-side values can vary such that the variables that constitute the basis

remain the same.Determine which variables will enter and leave the basis when the new right-hand-side reaches either ofthe extreme values of the range.

Identify alternative optimal solutions when they exist.

Identify alternative optimal shadow prices when they exist.

Analyze Solver (Microsoft Excel) sensitivity report and identify there shadow prices, reduced costs, and

variation ranges for objective coefficients and right-hand-sides.

http://find/


4/35

Production Planning at BA with Favaios Bottles

The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mouldinjection.BA recently won two new clients and intends to assign this production to one of its furnaces. Theorders were of bottles of Sandeman Ruby Port (75cl), one liter bottles of olive oil Oliveira da Serraand one liter bottles of Favaios Moscatel. The customers buy all the bottles that BA can produce.Due to differences in the production (number of cavities and different cycle times), each batch ofportwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30hours and a batch of favaios takes 50 hours to produce. There are a total of 2000 hours availableat the oven.Moreover, there are restrictions in the storage capacity of the bottles prior to their dispatch. Thewarehouse has 300m3 available and each batch of portwine bottles occupies 6m3 of warehousespace, each batch of bottles of olive oil holds 5m3 and each batch of favaios holds 6m3.

Finally we must also consider the capacity available in the decoration sector (e.g. labels,packaging), which is 200 hours. The portwine bottles spend 3 hours per batch, the olive oilbottles 5 hours per batch and the favaios bottles also 5 hours per batch.Barbosa & Almeida wants to maximize the profit from these two orders, knowing that the profitper batch is 50e 60e and 20e, respectively.

http://find/


5/35

Production Planning at BA with Favaios Bottles

Decision variables

xVP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce;

xA Number of batches of Azeite Oliveira da Serra bottles to produce;

xF Number of batches of Favaios bottles to produce.

Objective:

max Z= 50xVP+ 60xA+ 20xF

Subject to:

50xVP +30xA +50xF 2000 (time in the oven)

6xVP +5xA +6xF 300 (space in the warehouse)3xVP +5xA +5xF 200 (capacity in the decoration sector)

xVP, xA, xF 0

In this model xVP, xA and xFmay not be integer.

http://find/


6/35

Production Planning at BAProducing Favaios Bottles

xVP xA xF s1 s2 s3s1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300

s3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0

xVP xA xF s1 s2 s3

s1 32 0 20 1 0 6 800

s2 3 0 1 0 1 1 100xA

35 1 1 0 0

15 40

Z 14 0 40 0 0 12 2400

xVP xA xF s1 s2 s3xVP 1 0

58

132 0

316 25

s2 0 0 78

332 1

716 25

xA 0 1 5

8 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

xA enters the basis because:max(50, 60, 20) = 60s3 leaves the basis because:min( 200030 ,

3005 ,

2005 ) =

2005

xVPenters the basis because:max(14) = 14s1 leaves the basis because:min( 80032 ,

1003 ,

4035

) = 80032 = 25

All the coefficients of the objectivefunction are 0.Optimal solution:(xVP, xA, xF)

= (25, 25, 0) andZ = 2750

http://find/


7/35

Some remarks about the sensitivity analysis

We wish to analyze the effect on the optimal solution of changing various elements of theproblem data without re-solving the linear program or having to remember any of theintermediate tableaux generated in solving the problem by the simplex method.

The type of results that can be derived in this way are conservative, in the sense that theyprovide sensitivity analysis for changes in the problem data small enough so that the samedecision variables remain basic, but not for larger changes in the data.

http://find/


8/35

Shadow PricesReduced Costs

http://find/


9/35


xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 25

s2 0 0 78 3

32 1 7

16 25

xA 0 1 5

8 3

160 0 5

16 25

Z 0 0 48 34 7

16 0 9 3

8 2750

The complete optimal solution for the problem is

(xVP, xA, xF, s1, s2, s3) = (25, 25, 0, 0, 25, 0) and Z = 2750

The binding constraints in this optimal solution are therefore:

the first one (time in the oven, s1 = 0);

the third one (capacity in the decoration sector, s3 = 0);

and xF = 0.

50xVP +30xA +50xF = 20003xVP +5xA +5xF = 200

xF = 0

(xVP, xA, xF) = (25, 25, 0)

Z = 2750

http://find/


10/35

Shadow PriceDefinition

The shadow price associated with a particular constraint is the

change in the optimal value of the objective function per unitincrease in the righthand-side value of that constraint, all otherproblem data remaining unchanged.

http://find/


11/35

Production Planning at BAProducing Favaios Bottles Shadow Prices

What will be the increase of the profit per unit increase in the oven time?

50xVP +30xA +50xF = 20013xVP +5xA +5xF = 200

xF = 0

50xVP +30xA = 20013xVP +5xA = 200

xF = 0

50xVP +30xA = 200118xVP +30xA = 1200

xF = 0

xVP = 25 132

xA = 24157160

xF = 0

(xVP, xA, xF) = (25 1

32, 24 157

160, 0)

Z = 50 25 132

+ 60 24 157160

+ 20 0 = 2750 716

Profit increase: 2750 716 2750 = 7

16

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 25

s2 0 0 78

332

1 716

25

xA 0 1 5

8

3160

0 516

25

Z 0 0 4834

716

0 938

2750

http://find/


12/35

Production Planning at BAProducing Favaios Bottles Shadow Prices

What will be the increase of the profit per unit increase in the decoration capacity?

50xVP +30xA +50xF = 20003xVP +5xA +5xF = 201

xF = 0

50xVP +30xA = 20003xVP +5xA = 201xF = 0

50xVP +30xA = 200018xVP +30xA = 1206

xF = 0

xVP = 241316

xA = 25 83240

xF = 0

(xVP, xA, xF) = (241316, 25 5

16, 0)

Z = 50 24 13

16

+ 60 5

16

+ 20 0 = 2759 3

8Profit increase: 2759 3

8 2750 = 9 3

8

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 25

s2 0 0 7

8

3

32

1 7

16

25

xA 0 1 5

8

3160

0 516

25

Z 0 0 4834

716

0 938

2750

http://find/


13/35

Each constraint of a problem has as shadowprice.

The shadow price for a constraint is the negative of the coefficientof the appropriate slack (or artificial) variable in the objectivefunction of the final tableau.

http://find/


14/35

Reduced CostDefinition

The reduced cost associated with the nonnegativity constraint for

each variable is the shadow price of that constraint (i.e., thecorresponding change in the objective function per unit increase inthe lower bound of the variable).

http://find/


15/35

Production Planning at BAProducing Favaios Bottles Reduced costs

What will be the reduction of the profit if we must produce at least one batch of favaios bottles?

50xVP +30xA +50xF = 20003xVP +5xA +5xF = 200

xF = 1

50xVP +30xA +50 = 20003xVP +5xA +5 = 200

xF = 1

50xVP +30xA = 19503xVP +5xA = 195

xF = 1

xVP = 195/8xA = 195/8

xF = 1

(xVP, xA, xF) = (195

8 , 195

8 , 1)

Z = 50 1958

+ 60 1958

+ 20 1 = 2701, 25

Profit reduction:2750 2701, 25 = 48, 75 = 48 3

4

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 25

s2

0 0 7

8

3

32 1 7

16 25

xA 0 1 5

8

3160

0 516

25

Z 0 0 4834

716

0 938

2750

http://find/


16/35

Production Planning at BAProducing Favaios Bottles Reduced costs

What will be the reduction of the profit if we must produce at least one batch of favaios bottles?

xVP xA xF s1 s2 s3s

1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300s3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0


58

132 0

316 25

s2 0 0 78

332 1

716 25

xA 0 1 5

8 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

We can compute the reduced cost ofxF in a different way by computing theoportunity cost for producing 1 batch offavaios bottles:

50 716

+ 60 + 59 38

= 21 78

+ 46 78

=

68 34

As the revenue of 1 batch of favaiosbottles is 20, the profit will change by20 68 3

4 = 48 3

4


G D
http://find/


17/35

General DiscussionFundamental relationship between shadow prices, reducedcosts, and the problem data. I

Shadow

x1 . . . xn s1 . . . sm price

s1 a11 . . . a1n 1 . . . 0 b1 y1...

......

......

......

sm am1 . . . amn 0 . . . 1 bm ymZ c1 . . . cn 0 . . . 0 0

At the final tableau:

Z c1 . . . cn cn+1 . . . cn+m z0= . . . =

y1 . . . ym


G a D sc ss o
http://find/


18/35

General DiscussionFundamental relationship between shadow prices, reducedcosts, and the problem data. II

At each iteration of the simplex method, the objective function is transformed by subtractingfrom it a multiple of the row in which the pivot was performed.Consequently, the final form of the objective function could be obtained by subtracting multiplesof the original constraints from the original objective function.Consider first the final objective coefficients associated with the original basic variables: s1, . . .,

sm. Let 1, . . ., m be the multiples of each row that are subtracted from the original objectivefunction to obtain its final form. Since siappears only in the ith constraint and has a +1coefficient we should have cn+i = 0 1i = i = yi. Thus the shadow prices are themultiplesi.Since these multiples can be used to obtain every objective coefficient in the final form, thereduced cost cj of variable xj is given by cj =cj

mi=1aijyi (j= 1 . . .n).

Since cj= 0 for the m basic variables: 0 = cjmi=1aijyi (for jbasic). This is a system ofmequations in m unknowns that uniquely determines the values of the shadow prices yi.The current value of the objective function is z0 =

mi=1biyi, z0 =

mi=1biyi

http://find/


19/35

Variation in the objective coefficients

How much the objective-function coefficients can vary withoutchanging the values of the decision variables in the optimal solution?

We will make the changes one at a time, holding all othercoefficients and righthand-side values constant.

http://find/


20/35


Variation in the objective coefficient of xFSuppose that we increase the objective function coefficient ofxF in the original problemformulation by cF:In applying the simplex method, multiples of the rows were subtracted from the objective functionto yield the final system of equations. Therefore, the objective function in the final tableau willremain unchanged except for the addition of cF xF.

xVP xA xF s1 s2 s3s1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300s3 3 5 5 0 0 1 200Z 50 60 20 + cF 0 0 0 0


58

132 0

316 25

s2 0 0 78

332 1

716 25

xA 0 1 5

8 3160 0

516 25

Z 0 0 48 34 + cF 716 0 9

38 2750

xF will become a candidate toenter the basis, only when itsobjective-function coefficient ispositive.

The optimal solution remains un-changed as long as:

48 34

+ cF 0

cF 4834

http://goforward/http://find/http://goback/


21/35


Variation in the objective coefficient of xVP

xVP xA xF s1 s2 s3s1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300s3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0+cVP


58

132

0 316

25

s2 0 0 78

332

1 716

25

xA 0 1 5

8

3160

0 516

25

Z cVP 0 4834

716

0 938

2750


58

132

0 316

25

s2 0 0 78

332

1 716

25

xA 0 1 5

8

3160

0 516

25

Z 0 0 4834

716

0 938

2750

0 0 58

cVP 132

cVP 0 + 316

cVP 25cVP

The current solution re-mains unchanged while:

48 34 58 cVP 0

(cVP 78) 716

132 cVP 0

(cVP 14)9 38 +

316 cVP 0

(cVP 50)

Taking the most limiting in-equalities, the bounds on

cVP are:

14 cVP 5014+50 cnewVP 50+50

36 cnewVP 100


General Discussion
http://find/


22/35

General DiscussionRanges of the cost coefficients in the optimal solution.

Shadowx1 . . . xn s1 . . . sm price

s1 a11 . . . a1n 1 . . . 0 b1 y1

..

....

..

....

..

....

..

.sm am1 . . . amn 0 . . . 1 bm ymZ c1 . . . cn 0 . . . 0 0

At the final tableau:Z c1 . . . cn cn+1 . . . cn+m z0

If xj is a non-basic variable at the final tableau with objective function coefficient cj that waschanged to cnewj = cj+ cj, then the current solution remains unchanged so long as the new reduced

cost cnewj remains nonnegative, that is, cnew

j = cj+ cjm

i=1aijyi = cj+ cj 0.

The range on the variation of the objective-function coefficient of a nonbasic variable is then given by: < cj < cj; ( < cj+ cj = c

newj < cj cj).

If xr is a basic variable at the final tableau with objective function coefficient cr that was changed tocnewr = cr + cr, then c

newr = cr+ cr

mi=1aijyi = cr+ cr.

Since xr is a basic variable, cr= 0. So, to recover a canonical form with cnew

r = 0, we subtract crtimes the rth constraint in the final tableau from the final form of the objective function, obtaining newreduced costs for all nonbasic variables: cnewj = cj arjcr, where arj is the coefficient of variable xj inthe rth constraint in the final tableau.

For all basic variables cnewj = 0 and the current basis remains optimal ifcnew

j 0.

The range on the variation of the objective-function coefficient is:

Maxj

cjarj| arj > 0

cr Minj

cjarj| arj


23/35

Variations on the righthand-side values

http://find/


24/35



Varying the righthand-side value of a constraint that is non-binding in the optimal solution:the warehouse constraint.

xVP xA xF s1 s

2 s3

s1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300 + b2s3 3 5 5 0 0 1 200Z 50 60 20 0 0 0 0

xVP xA xF s1 s

2 s3xVP 1 0

58

132 0

316 25 + 0 b2

s2 0 0 78

332 1

716 25 + 1 b2

xA 0 1 5

8 3160 0

516 25 + 0 b2

Z 0 0 48 34 716 0 9

38 2750

b2 is added to the righthand side ofthe warehouse constraint.This is equivalent to substituting s2 bys2 = s2 + b2 in the original problemformulation.

In order to keep the solution feasible:b2 25

http://find/


25/35



Varying the righthand-side value of a constraint that is binding in the optimal solution:the decoration constraint.

xVP xA xF s1 s2 s3s1 50 30 50 1 0 0 2000s2 6 5 6 0 1 0 300

s3 3 5 5 0 0 1 200 + b3Z 50 60 20 0 0 0 0

xVP xA xF s1 s2 s

3xVP 1 0

58

132

0 316

25 316

b3

s2 0 0 78

332

1 716

25 716

b3

xA 0 1 5

8

3160

0 516

25 + 516

b3

Z 0 0 4834

716

0 938

2750 938

b3

b3 is added to the righthand side ofthe decoration constraint.This is equivalent to substituting s3 bys3 = s3 + b3 in the original problemformulation.

xVP 0 b3 2516

3

s2 0 b3 2516

7

xA 0 b3 5 16

25167 b3 80

200 + 25167 bnew3 200 80

http://find/


26/35


58

132

0 316

25 316

b3

s2 0 0 1 58 332 1 716 25 716 b3

xA 0 1 5

8 3

160 0 5

16 25 + 5

16b3

Z 0 0 48 34 7

16 0 9 3

8 2750 9 3

8b3

The final tableau forbnew

3 = 120 (b

3 = 80) will be:

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 40

s2 0 0 78 3

32 1 7

16 60

xA 0 1 58 3160 0 516 0

Z 0 0 48 34 7

16 0 9 3

8 2480


For bnew3


27/35

What variable should enter the basis to take its place?In order for the new basis to be an optimal solution, the entering variable must be chosen so thatthe reduced costs are not allowed to become positive.Regardless of which variable enters the basis, the entering variable will be isolated in row 3 of thefinal tableau to replace xA, which leaves the basis.

To isolate the entering variable, we must perform a pivot operation, and a multiple, say t, of row3 in the final tableau will be subtracted from the objective-function row.

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 40

s2 0 0 78

332

1 716

60

xA 0 1 5

8 3

160 0 5

16 0

Z 0 0 48 34

716

0 9 38 2480

t t 58

+t 3160

t 516

t 0

Coefficient forXA: t 0 t 0

Coefficient forXF: 4834 t

58 0 t

2885

Coefficient for s1: 716 +t

3160 0 t

703

Coefficient for s3: 938 t

516 0 t 30

0 t 703

Since the coefficient ofs1 is most constraining on t, s1 will enter the basis.

Note that the pivot is made on a negative coefficient.(FEUP | DEGI) Operations Research March 19, 2014 27 / 35

General Discussion
http://find/


28/35

Variations in the Righthand-Side values. I

x1 . . . xn s1 . . . sk . . . sm

s1 a11 . . . a1n 1 . . . 0 . . . 0 b1...

.

.

....

.

.

....

.

.

....

sk ak1 . . . akn 0 . . . 1 . . . 0 bk...

.

.

....

.

.

....

.

.

....

sm am1 . . . amn 0 . . . 0 . . . 1 bmZ c1 . . . cn 0 . . . 0 . . . 0 0

In the final tableau (canonical form):x1 . . . xn s1 . . . sk . . . sm

basic1 a11 . . . a1n 11 . . . 1k . . . 1m b1...

.

.

....

.

.

....

.

.

....

basick ak1 . . . akn k1 . . . kk . . . km bk

... ... ... ... ... ... ...

basicm am1 . . . amn m1 . . . mk . . . mm bmZ c1 . . . cn cn+1 . . . cn+k . . . cn+m z0

As this is a canonical form, aij and ij in the final tableau will be structured so that one basic variable isisolated in each constraint.


General Discussion
http://find/


29/35

Variations in the Righthand-Side values. II

We can change the coefficient bk in the kth righthand-side by bkwith all the other data held fixed, simply bysubstituting sk bys

k = sk+ bk in the initial tableau.

x1

. . . xn s1

. . . s

k . . . sm

s1 a11 . . . a1n 1 . . . 0 . . . 0 b1...

.

.

....

.

.

....

.

.

....

sk ak1 . . . akn 0 . . . 1 . . . 0 bk+ bk...

.

.

....

.

.

....

.

.

....

sm am1 . . . amn 0 . . . 0 . . . 1 bm

Z c1 . . . cn 0 . . . 0 . . . 0 0

In the final tableau: (sk = sk+ bk)x1 . . . xn s1 . . . s

k . . . sm

basik1 a11 . . . a1n 11 . . . 1k . . . 1m b1 + 1kbk...

.

.

....

.

.

....

.

.

....

basick ak1 . . . akn k1 . . . kk . . . km bk+kkbk...

.

.

....

.

.

....

.

.

....

basicm am1 . . . amn m1 . . . mk . . . mm bm + mkbkZ c1 . . . cn cn+1 . . . cn+k . . . cn+m z0 + cn+kbk

The current basis is optimal whenever bi+ikbk 0 for (i = 1, 2, . . . , m).

Equivalently Maxibiik

| ik >0 bk Mini

biik

| ik < 0


General Discussion
http://find/


30/35

Variations in the Righthand-Side values. III

When bkreaches either its upper or lower bound any further increase (or decrease) in its value

makes one of the updated righthand sides negative.Consider that bk is such that br+rkbk


31/35

Alternative optimal solutions

As in the case of the objective function and righthand-side ranges,the final tableau of the linear program tells us somethingconservative about the possibility of alternative optimal solutions.

If all reduced costs of the nonbasic variables are strictly negative

(positive) in a maximization (minimization) problem, then there isno alternative optimal solution, because introducing any variableinto the basis at a positive level would reduce (increase) the value ofthe objective function.

If one or more of the reduced costs are zero, there may exist

alternative optimal solutions.

http://find/


32/35

Alternative shadow pricesIndependent of the question of whether or not alternative optimalsolutions exist in the sense that different values of the decisionvariables yield the same optimal value of the objective function,there may exist alternative optimal shadow prices.

If all righthand-side values in the final tableau are positive, thenthere do not exist alternative optimal shadow prices.

If one or more of these values are zero, then there may existalternative optimal shadow prices.

http://find/


33/35

Alternative Shadow PricesE II


34/35

Example II

To isolate the entering variable, we must perform a pivot operation, and a multiple, say t, of row

3 in the final tableau will be subtracted from the objective-function row.

xVP xA xF s1 s2 s3

xVP 1 0 5

81

32 0 3

16 40

s2 0 0 78

332

1 716

60

xA 0 1 58 3160 0 516 0

Z 0 0 48 34

716

0 9 38 2480

t t 58

+t 3160

t 516

t 0

Coefficient forXA: t 0 t 0Coefficient forXF: 48

34 t

58 0 t

2885

Coefficient for s1: 716 +t 3160 0 t 703Coefficient for s3: 9

38 t

516 0 t 30

0 t 703


Alternative Shadow PricesE III
http://find/


35/35

Example III

Since the coefficient ofs1 is most constraining on t, s1 will enter the basis and the new tableauwill be the following1:

xVP xA xF s1 s2 s3xVP 1

53

32

0 0 1724

40

s2 0 5 4 0 1 9

8 60

s1 0 160

3 100

3 1 0 50

3 0

Z 0 703 190

3 0 0 50

3 2480

In this example we have for the same optimal solution alternative shadow prices.The optimal solution is (xvp, xA, xF, s1, s2, s3) = (40, 0, 0, 0, 60, 0) and the shadow prices are:

(yO1, yW1, yD1) = (48 34 , 0, 9 38 ) and (yO2, yW2, yD2) = (0, 0, 503 )

1In order to obtain the alternative shadow prices there would be no need to built the whole tableau, the line of

the objective function would be sufficient as you may see.(FEUP | DEGI) Operations Research March 19, 2014 35 / 35
http://find/http://goback/

Documents

03.1 Chapter3_SensitivityAnalysis.pdf