Chapter 3: Linear Programming Sensitivity Analysis
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Transcript of Chapter 3: Linear Programming Sensitivity Analysis
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