CHAPTER 3:
LINEAR PROGRAMMING
SENSITIVITY ANALYSIS
Sensitivity Analysis
What if there is uncertainly about one or more
values in the LP model?
1.
2.
3.
Raw material changes,
Product demand changes,
Stock price
Sensitivity 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 ANALYSIS
We can use the graph of an LP to see what
happens when:
1.
2.
An OFC changes, or
A RHS changes
Recall the Flair Furniture problem
Sensitivity to Coefficients
Graphical solution of Example 1
x1 + x2 < 8
Max 5x1 + 7x2
Example 1:
Max
5x1 + 7x2
s.t.
x1
6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
x1 < 6
Optimal solution:
x1 = 5, x2 = 3
2x1 + 3x2 < 19
Feasible
region
Sensitivity to Coefficients





Range of optimality for c1
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:
Example 1:
-1 < -c1/7 < -2/3
Max
5x1 + 7x2
Multiplying by -1, 1 > c1/7 > 2/3 s.t.
x1
< 6
Multiplying by 7,
7 >
c1
> 14/3
2x1 + 3x2 < 19
x 1 + x2 < 8
x1 > 0 and x2 > 0
Sensitivity to Coefficients
Range of optimality for c2
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:
Example 1:
-1 < -5/c2 < -2/3
Example 1:
Max
5x1 + 7x2
Max
5x + 7x
Multiplying by -1, 1 > 5/c2 > 2/3
s.t.
x1
< 6
s.t.
x
< 6
Inverting,
1 < c2/5 < 3/2
2x
+ 3x < 19
2x1 + 3x 2 < 19
xx1 ++ xx2 << 8 8
,
x1x >>00 and
and xx2 >>00
Multiplying by 5
5 < c2 < 15/2

1
2
1
1
1
2
1
2
2
Sensitivity to Coefficients
Graphical solution of Example 1
1200
Cutting & Dyeing
0,1062
FinishingLine
Max 7S + 9D
7/10S+1D<=630 (Cutting
& dyeing)
1/2S+5/6D <=600 Sewing
1S+2/3D<=708 Finishing
1/10S+1/4D<=135 Inspection
& Packaging
1000
800
0, 630
600
400
200
708,0
0
0
200
400
600
800
900,0
1000
Optimal solution:
S=540,D=252
1. objective function
2. Cutting Line
3.Finishing Line
This point will be an optimal solution as long as:
slope of line A <=slope of the objective function <= slope of line A
i.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 PROBLEM
Max 7T + 5C
(profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000(painting hrs)
C < 450 (max # chairs)
T
> 100
T, C > 0
(min # tables)
(nonnegativity)
OBJECTIVE FUNCTION
COEFFICIENT (OFC) CHANGES
What if the profit contribution for tables changed
from $7 to $8 per table?
8
Max 7 T + 5 C
(profit)
X
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.
C
Original
Objective Function
7T + 5 C = $4040
Revised
Objective Function
8T + 5 C = $4360
Optimal Corner
(T=320, C=360)
Still optimal
500
400
300
200
Feasible
Region
100
0
0
100
200
300
400
500 T
C
1000
What if the OFC
became higher?
Or lower?
11T + 5C = $5500
Optimal Solution
(T=500, C=0)
Both have new
optimal corner
points
600
450
3T + 5C = $2850
Optimal Solution
(T=200, C=450)
Feasible
Region
0
0 100
500
800 T

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) CHANGES
What if painting hours available changed from
1000 to 1300?
1300
X
2T + 1C < 1000
(painting hrs)
This increase in resources could allow us to
increase production and profit.
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
Old optimal
corner point
(T=320,C=360)
Profit=$4040
C
500
Feasible region
becomes larger
400
New optimal
corner point
(T=560,C=180)
Profit=$4820
300
200
Original
100
Feasible
Region
0
0
100
200
300
400
500
600 T
EFFECT ON OBJECTIVE FUNCTION VALUE
New profit
Old profit
Profit increase
= $4,820
= $4,040
= $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 EXAMPLE
Decision: How many of each of 4 products
to make?
Objective: Maximize profit
Decision Variables:
V = number of VCR’s
S = number of stereos
T = number of TV’s
D = number of DVD players
Max 29V + 32S + 72T + 54D
(in $ of profit)
Subject to the constraints:
3V + 4S + 4T + 3D < 4700
2V + 2S + 4T + 3D < 4500
V + S + 3T + 2D < 2500
V, S, T, D > 0
(elec. components)
(nonelec. components)
(assembly hours)
(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 0
We 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
0.9548
The sensitivity report can still be used
PRICING NEW VARIABLES
Suppose 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 components
4 nonelectrical components
4 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 HTS
elec cpnts
5 x $ 2 = $10
nonelec cpnts 4 x $ 0 = $ 0
assembly hrs 4 x $24 = $96
$106
Shadow
Prices
Making 1 HTS will reduce profit (from other
products) by $106
• Need (HTS profit contribution) > $106
• Cost to produce each HTS:
elec cpnts
5 x $ 7 = $35
nonelec cpnts 4 x $ 5 = $20
assembly 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
Ingredient
X
A
3
B
4
C
8
D
10
> 280 units
Y
5
3
6
6
> 200 units
Z
10
25
20
40
< 1050 units
Chemical
$ per ounce of ingredient
$0.40
$0.20 $0.60
$0.30
Requirement
Min 0.40A + 0.20B + 0.60C + 0.30D
($ of cost)
Subject to the constraints
A+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