CHAPTER 3:LINEAR PROGRAMMINGSENSITIVITY ANALYSIS

Sensitivity AnalysisWhat if there is uncertainly about one or

more values in the LP model?1. Raw material changes,2. Product demand changes, 3. Stock priceSensitivity analysis allows a manager to

ask certain hypothetical questions about the problem, such as:

How much more profit could be earned if 10 more hours of labour were available?

Which of the coefficient in model is more critical?

SENSITIVITY ANALYSIS

Sensitivity analysis allows us to determine how “sensitive” the optimal solution is to changes in data values.

This includes analyzing changes in:1. An Objective Function Coefficient

(OFC)2. A Right Hand Side (RHS) value of a

constraint

LIMIT OF RANGE OF OPTIMALITY Max Ax + BY Keeping x, Y same how the object

function behaves if A, B are changed. The optimal solution will remain

unchanged as long as An objective function coefficient lies within its range of optimality

If the OFC changes beyond that range a new corner point becomes optimal.

Generally, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.

Binding constraint Are the constraints that restrict the

feasible region

GRAPHICAL SENSITIVITY ANALYSISWe can use the graph of an LP to see

what happens when:

1. An OFC changes, or 2. A RHS changes

Recall the Flair Furniture problem

2x1 + 3x2 < 19

x1 + x2 < 8Max 5x1 + 7x2

x1 < 6Optimal solution: x1 = 5, x2 = 3

Example 1:Max 5x1 + 7x2

s.t. x1 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1 > 0 and x2 > 0

Feasibleregion

Sensitivity to Coefficients Graphical solution of Example 1

Compute the range of optimality for c1 in Example 1. The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2.

The slope of the binding third constraint, x1 + x2 = 8, is -1. The slope of the binding second constraint, 2x1 + 3x2 = 19,

is -2/3. Find the range of values for c1 (with c2 staying 7) such that

the objective-function line slope lies between that of the two binding constraints:

-1 < -c1/7 < -2/3 Multiplying by -1, 1 > c1/7 > 2/3 Multiplying by 7, 7 > c1 > 14/3

Sensitivity to Coefficients Range of optimality for c1

Example 1:Max 5x1 + 7x2

s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1 > 0 and x2 > 0

Sensitivity to Coefficients Range of optimality for c2

Example 1:Max 5x1 + 7x2

s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1 > 0 and x2 > 0

Likewise, compute the range of optimality for c2 in Example 1.

The slope of the binding third constraint is -1. The slope of the binding second

constraint is -2/3. Find the range of values for c2 (with c1 staying 5) such that

the objective-function line slope lies between that of the two binding constraints:

-1 < -5/c2 < -2/3Multiplying by -1, 1 > 5/c2 > 2/3Inverting, 1 < c2/5 < 3/2, Multiplying by 5 5 < c2 < 15/2

Example 1:Max 5x1 + 7x2

s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1 > 0 and x2 > 0

Optimal solution: S=540,D=252

Max 7S + 9D7/10S+1D<=630 (Cutting & dyeing)1/2S+5/6D <=600 Sewing1S+2/3D<=708 Finishing1/10S+1/4D<=135 Inspection

& Packaging

Sensitivity to Coefficients Graphical solution of Example 1

0 100 200 300 400 500 600 700 8000

100

200

300

400

500

600Cutting & Dyeing FinishingLine

1. objective function2. Cutting Line3.Finishing LineThis point will be an optimal solution as long as:

slope of line A <=slope of the objective function <= slope of line Ai.e. the slope of the objective function should be in between these two lines

7/10S+D=630 (C&D) D=-7/10S+630 S+2/3D=708 (Finishing) D=-3/2S+1062 -3/2<=slope of objective function <=-7/10 -3/2<=-Cs/Cd<=-7/10 if we put profit contribution of delux bag same i.e 9 -3/2<=-Cs/9<=-7/10 Cs>=3*9/2 Cs>=27/2 Cs>=13.5 Cs>=63/10 Cs>=6.3 6.3 <=Cs<=13.5 (limits for Cs with same optimal solution)

Similarly the keeping profit contribution of S bag constant. Cs=10

6.67<=Cd<=14.29 (range of optimality) If both the Cs, Cd are changed simultaneously (i.e S bags

to 13, D bags to 8) Calculate the slope again: -cs/cd =-13/8=-1.625 -3/2<=-Cs/Cs<=-7/10 now -Cs/Cd=-1.625 which is less than -3/2 which is not

acceptable according to above equation hence this means if we change the both cofficient than 540 and 252 would not be the optimal solution.

EFFECT OF CHANGE OF THE RIGH HAND SIDE OF THE CONSTRAINT.

suppose if additional 10 hrs is added to cutting and dyeing constraint

7/10S+1d<=640 ??? Feasible region extended, find extreme

point using intersection of two lines S=527.5 ,D=270.75 Max 10S+9D Profit= 7711.75 which is 7711.75-7668.00

=43.75 Increase in profit/hr =

DUAL PRICE It is the improvement in the optimal

solution per unit increase in the RHS of contraint

.if dual price is negative this means value of objective function will not improved rather it would get worse ,if value at rhs of the constrain is increased by 1 unit . for minimization problem it means cost will increase by 10.

100 % RULE FOR OBJECTIVE FUNCTION COFFICIENTS & CONSTRAINT

For all the objective functions that are changed the sum of percentages of allowable increase and allowable decrease if does not exceed by 100% then optimal solution will not change. However this does not means that if sum of the percentage is exceed by 100% then optimal solution will change , in that case the problem must be resolved . This rule is equally applicable on the constraints RHS.

EXAMPLE S is changed from 10$ to 11.50$; D is reduced from $9 to $8.25 Range of optimality from the sensitivity

analysis : allowable upper limit for S= 13.49; Value of S =$10 Allowable increase in S= 13.49-10 =3.49 for the present case D is reduced from $9

to $8.25 the increase in percentage is

1.5*100/3.49=42.86% of allowable increase

EXAMPLE For D allowable lower limit is 6.66 Value of D=9 allowable Decrease =9-6.66=2.33 for present case =0.75/2.33*100=

32.14 of allowable decrease. sum of allowable increase and

decrease is 42.86% + 32.14% <100% hence optimal solution is still valid S=540 and D-252.

FLAIR FURNITURE PROBLEMMax 7T + 5C(profit)

Subject to the constraints:3T + 4C < 2400 (carpentry hrs)2T + 1C < 1000 (painting hrs) C < 450 (max # chairs) T > 100 (min # tables)T, C > 0 (nonnegativity)

OBJECTIVE FUNCTION COEFFICIENT (OFC) CHANGES

What if the profit contribution for tables changed from $7 to $8 per table?8Max 7 T + 5 C (profit)

Clearly profit goes up, but would we want to make more tables and less chairs?(i.e. Does the optimal solution change?)

Find the range of optimality with graph.

X

0 100 200 300 400 500 T

C

500

400

300

200

100

0

Optimal Corner(T=320, C=360)

Still optimal

Feasible Region

OriginalObjective Function7T + 5 C = $4040

RevisedObjective Function8T + 5 C = $4360

C1000

600450

0 FeasibleRegion

0 100 500 800 T

What if the OFC became higher? Or lower?

11T + 5C = $5500Optimal Solution(T=500, C=0)

3T + 5C = $2850Optimal Solution(T=200, C=450)

Both have new optimal corner

points

There is a range for each OFC where the current optimal corner point remains optimal.

If the OFC changes beyond that range a new corner point becomes optimal.

Excel’s Solver will calculate the OFC range.

RIGHT HAND SIDE (RHS) CHANGESWhat if painting hours available changed

from 1000 to 1300?1300

2T + 1C < 1000 (painting hrs)

This increase in resources could allow us to increase production and profit.

X

CHARACTERISTICS OF RHS CHANGES The constraint line shifts, which could

change the feasible region

Slope of constraint line does not change

Corner point locations can change

The optimal solution can change

0 100 200 300 400 500 600 T

C500

400

300

200

100

0

Original

Feasible

Region

2T + 1 C = 1000

2T + 1 C = 1300Feasible region becomes larger

New optimalcorner point

(T=560,C=180)Profit=$4820

Old optimal corner point

(T=320,C=360)Profit=$4040

EFFECT ON OBJECTIVE FUNCTION VALUENew profit = $4,820Old profit = $4,040Profit increase = $780 from 300 additional

painting hours

$2.60 in profit per hour of painting

Each additional hour will increase profit by $2.60

Each hour lost will decrease profit by $2.60

ANDERSON ELECTRONICS EXAMPLEDecision: How many of each of 4

products to make?Objective: Maximize profitDecision Variables:

V = number of VCR’sS = number of stereosT = number of TV’sD = number of DVD players

Max 29V + 32S + 72T + 54D (in $ of profit)

Subject to the constraints:

3V + 4S + 4T + 3D < 4700 (elec. components)2V + 2S + 4T + 3D < 4500 (nonelec. components) V + S + 3T + 2D < 2500 (assembly hours)

V, S, T, D > 0 (nonnegativity)

Go to file 4-2.xls

RHS Change Questions What if the supply of nonelectrical

components changes?

What happens if the supply of electrical components increased by 400 (to 5100)? increased by 4000 (to 8700)?

What if we could buy an additional 400 elec. components for $1 more than usual? Would we want to buy them?

What if would could get an additional 250 hours of assembly time by paying $5 per hour more than usual? Would this be profitable?

DECISION VARIABLES THAT EQUAL 0We are not currently making any VCR’s

(V=0) because they are not profitable enough.

How much would profit need to increase before we would want to begin making VCR’s?

REDUCED COST OF A DECISION VARIABLE

(marginal contribution to the obj. func. value)

- (marginal value of resources used) =Reduced Cost

marginal profit of a VCR = $29 - marginal value of resources = ?

Reduced Cost of a VCR = - $1.0

Reduced Cost is: The minimum amount by which the

OFC of a variable should change to cause that variable to become non-zero.

The amount by which the objective function value would change if the variable were forced to change from 0 to 1.

OFC CHANGE QUESTIONS For what range of profit contributions

for DVD players will the current solution remain optimal?

What happens to profit if this value drops to $50 per DVD player?

ALTERNATE OPTIMAL SOLUTIONS

May be present when there are 0’s in the Allowable Increase or Allowable Decrease values for OFC values.

SIMULTANEOUS CHANGES

All changes discussed up to this point have involved only 1 change at a time.

What if several OFC’s change?Or

What if several RHS’s change?

Note: they cannot be mixed

THE 100% RULE∑ (change / allowable change) <

1

RHS Example Electrical components decrease 500

500 / 950 = 0.5263 Assembly hours increase 200

200 / 466.67 = 0.4285

0.9548The sensitivity report can still be used

PRICING NEW VARIABLESSuppose they are considering selling a

new product, Home Theater Systems (HTS)

Need to determine whether making HTS’s would be sufficiently profitable

Producing HTS’s would take limited resources away from other products

To produce one HTS requires:5 electrical components4 nonelectrical components4 hours of assembly time

Can shadow prices be used to calculate reduction in profit from other products? (check 100% rule)

5/950 + 4/560 + 4/1325 = 0.015 < 1

REQUIRED PROFIT CONTRIBUTION PER HTSelec cpnts 5 x $ 2 = $10nonelec cpnts 4 x $ 0 = $ 0assembly hrs4 x $24 = $96

$106

Making 1 HTS will reduce profit (from other products) by $106

ShadowPrices

• Need (HTS profit contribution) > $106• Cost to produce each HTS:

elec cpnts 5 x $ 7 = $35nonelec cpnts 4 x $ 5 = $20assembly hrs 4 x $10 = $40

$95(HTS profit contribution) = (selling price) - $95

So selling price must be at least $201

IS HTS SUFFICIENTLY PROFITABLE? Marketing estimates that selling price

should not exceed $175

Producing one HTS will cause profit to fall by $26 ($201 - $175)

Go to file 4-3.xls

SENSITIVITY ANALYSIS FOR A MINIMIZATION PROBLEM

Burn-Off makes a “miracle” diet drink

Decision: How much of each of 4 ingredients to use?

Objective: Minimize cost of ingredients

DATA Units of Chemical per Ounce of Ingredient

Chemical

Ingredient

RequirementA B C DX 3 4 8 10 > 280 unitsY 5 3 6 6 > 200 unitsZ 10 25 20 40 < 1050 units

$ per ounce of ingredient

$0.40 $0.20 $0.60 $0.30

Min 0.40A + 0.20B + 0.60C + 0.30D ($ of cost)

Subject to the constraintsA + B + C + D > 36 (min daily

ounces)3A + 4B + 8C + 10D > 280

(chem x min)5A + 3B + 6C + 6D > 200 (chem y

min)10A + 25B + 20C + 40D < 280 (chem z

max)A, B, C, > 0

Go to file 4-5.xls