LINEAR PROGRAMMING
SENSITIVITY ANALYSIS
Learning Objectives

Learn sensitivity concepts

Understand, using graphs, impact of changes in
objective function coefficients, right-hand-side values,
and constraint coefficients on optimal solution of a
linear programming problem.

Generate answer and sensitivity reports using Excel's
Solver.

Interpret all parameters of reports for maximization
and minimization problems.

Analyze impact of simultaneous changes in input data
values using 100% rule.

Analyze the impact of addition of new variable using
pricing-out strategy.
Introduction (1 of 2)
 Optimal solutions to LP problems have been
examined under deterministic assumptions.
 Conditions in most real world situations are
dynamic and changing.
 After an optimal solution to problem is found, input
data values are varied to assess optimal solution
sensitivity.
 This process is also referred to as sensitivity
analysis or post-optimality analysis.
Introduction (2 of 2)
 Sensitivity analysis determines the effect on optimal
solutions of changes in parameter values of the
objective function and constraint equations
 Changes may be reactions to anticipated uncertainties
in the parameters or the new or changed information
concerning the model
The Role of Sensitivity Analysis of the
Optimal Solution
 Is the optimal solution sensitive to changes in input
parameters?
 Possible reasons for asking this question:
Parameter values used were only best estimates.
Dynamic environment may cause changes.
“What-if” analysis may provide economical and
operational information.
1. Sensitivity Analysis of
Objective Function Coefficients.
Ranges of Optimality
The optimal solution will remain unchanged as long as:
an objective function coefficient lies within its range of
optimality
there are no changes in any other input parameters.
The value of the objective function will change if the
coefficient multiplies a variable whose value is nonzero.
Sensitivity Analysis Using Graphs
Example 1: High Note Sound Company(HNSC) (1 of 4)
 HNSC Manufactures quality CD players and stereo receivers.
 Each product requires skilled craftsmanship.
 LP problem formulation:
Objective: maximize profit = $50C + $120R
subject to
2C + 4R  80
(Hours of electricians' time available)
3C + R  60
(Hours of audio technicians' time available)
C, R  0
(Non-negativity constraints)
Where:
C = number of CD players to make.
R = number of receivers to make.
Sensitivity Analysis Using Graphs
Example 1(2 of 4)
Sensitivity Analysis Using Graphs
Example 1 (3 of 4)
Impact of price change of Receivers
If unit profit per stereo receiver (R) increased
from $120 to $150, is corner point a still the
optimal solution? YES !
But Profit is $3,000 = 0 ($50) + 20 ($150)
Sensitivity Analysis Using Graphs
Example 1 (4 of 4)
Impact of price change of Receivers
If receiver’s profit coefficient changed from
$120 to $80, slope of isoprofit line changes
causing corner point (b) to become optimal.
But Profit is $1,760 = 16 ($50) + 12 ($80).
Sensitivity Analysis Using Graphs
Example 2: Galaxy Industries (1 of 5)
Max 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = 1200 (Plastic)
3X1 + 4X2 < = 2400 (Production Time)
X1 + X2 < = 800
(Total production)
X1 - X2 < = 450
(Mix)
Xj> = 0, j = 1,2
(Nonnegativity)
Sensitivity Analysis Using Graphs Example 2:(2 of 5)
We now demonstrate the search for an optimal solution
Start at X2
some arbitrary profit, say profit = $2,000...
Then increase the profit, if possible...
...and continue until it becomes infeasible
1200
800
Profit
4,
Profit
= $=$5040
2,
3,
000
600
X1
400
600
800
Sensitivity Analysis Using Graphs
Example 2 (3 of 5)
MODEL SOLUTION
Space Rays = 480 dozens
Zappers
= 240 dozens
Profit
= $5040
– This solution utilizes all the plastic and all the production hours.
– Total production is only 720 (not 800).
– Space Rays production exceeds Zapper by only 240 dozens (not
450).
Sensitivity Analysis Using Graphs
Example 2 (4 of 5)
1200
800
X2
The effects of changes in an objective
function coefficient
on the optimal solution
600
X1
400
600
800
Sensitivity Analysis Using Graphs
Example 2 (5 of 5)
1200
X2
Range of optimality
800
600
400
600
800
The effects of changes in an objective function coefficients
on the optimal solution
X1
The optimality range for an objective
coefficient is the range of values over which
the current optimal solution point will remain
optimal
For two variable LP problems the optimality
ranges of objective function coefficients can be
found by setting the slope of the objective
function equal to the slopes of each of the
binding constraints
Sensitivity Analysis Using Graphs
Example 3: Beaver Creek Pottery (1 of 4)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Sensitivity Analysis Using Graphs
Example 3 (2 of 4)
Maximize Z = $100x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Sensitivity Analysis Using Graphs
Example 3 (3 of 4)
Maximize Z = $40x1 + $100x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Objective Function Coefficient
Sensitivity Range
• The sensitivity range for an objective function coefficient is
the range of values over which the current optimal solution
point will remain optimal.
• The sensitivity range for the xi coefficient is designated as ci.
Objective Function Coefficient Sensitivity Range
for c1 and c2
Example 3 (4 of 4)
objective function Z = $40x1 + $50x2
sensitivity range for:
x1: 25  c1  66.67
x2: 30  c2  80
Objective Function Coefficient
Sensitivity Range
(for a Cost Minimization Model)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
sensitivity ranges:
4  c1  
0  c2  4.5
Multiple changes
– The range of optimality is valid only when a single
objective function coefficients changes.
– When more than one variable changes we turn to
the 100% rule.
The 100% Rule
1. For each increase (decrease) in an objective function
coefficient, calculate (and express as a percentage) the
ratio of the change in the coefficient to the maximum
possible increase (decrease) as determined by the
limits of the range of optimality.
2. Sum all these percent changes. If the total is less
than 100 percent, the optimal solution will not
change. If this total is greater than or equal to 100%,
the optimal solution may change.
Reduced Costs
The reduced cost for a variable at its lower bound
(usually zero) yields:
The amount the profit coefficient must change
before the variable can take on a value above its
lower bound.
The amount the optimal profit will change per
unit increase in the variable from its lower
bound.
2. Sensitivity Analysis of Right Hand
Side Values
Changes in Right-Hand-Side Values of
Constraints
The sensitivity range for a RHS value is the range
of values over which the quantity (RHS) values
can change without changing the solution variable
mix, including slack variables.
Sensitivity Analysis of
Right-Hand SideValues
Any change in the right hand side of a
binding constraint will change the optimal
solution.
Any change in the right-hand side of a
nonbinding constraint that is less than its
slack or surplus, will cause no change in the
optimal solution.
Changes in Constraint Quantity (RHS) Values
Increasing the Labor Constraint (1 of 3)
Example 1: Beaver Creek
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Changes in Constraint Quantity (RHS) Values
Sensitivity Range for Labor Constraint (2 of 3)
Example 1: Beaver Creek
Sensitivity range for:
30  q1  80 hr
Changes in Constraint Quantity (RHS) Values
Sensitivity Range for Clay Constraint (3 of 3)
Example 1: Beaver Creek
Sensitivity range for:
60  q2  160 lb
Sensitivity Analysis of
Right-Hand SideValues & Shadow Prices
In sensitivity analysis of right-hand sides of
constraints we are interested in the following
questions:
Keeping all other factors the same, how much
would the optimal value of the objective function
(for example, the profit) change if the right-hand
side of a constraint is changed by one unit?
For how many additional or fewer units this per
unit this change will be valid?
Changes in Right-hand-side (RHS) &
Shadow Prices
RHS of Binding Constraint • If RHS of non-redundant constraint changes, size of feasible
region changes.
– If size of region increases, optimal objective function
improves.
– If size of region decreases, optimal objective function
worsens.
• Relationship expressed as Shadow Price.
• Shadow Price is change in optimal objective function value
for one unit increase in RHS.
Shadow Prices (Dual Values)
Defined as the marginal value of one additional
unit of resource.
The sensitivity range for a constraint quantity
value is also the range over which the shadow
price is valid.
Range of Feasibility
The set of right - hand side values for which same set of
constraints determines the optimal point.
Within the range of feasibility, shadow prices remain
constant; however, the optimal solution will change.
Example 2: High Note Sound Company Problem
(1 of 6)
 HNSC Manufactures quality CD players and stereo receivers.
 Each product requires skilled craftsmanship.
 LP problem formulation:
Objective: maximize profit = $50C + $120R
subject to
2C + 4R  80
(Hours of electricians' time available)
3C + R  60
(Hours of audio technicians' time available)
C, R  0
(Non-negativity constraints)
Where:
C = number of CD players to make.
R = number of receivers to make.
Sensitivity Analysis of Right-hand-side (RHS) Values
Example 2: High Note Sound Company (2 of 6)
May change the
feasible region size.
May change or move
corner points.
Increase in Electricians’ Available Time
Example2: High Note Sound Company (3 of 6)
• As available electricians’ time increases,
corner points a and b will move closer to one
other.
• Further increases in available electricians’ time
may make this constraint redundant.
Decrease in Electricians’ Available Time
Example 2: High Note Sound Company (4 of 6)
• As available electricians’ time decreases, corner
points b and c move closer to one another from their
current locations.
• Corner points b and c will no longer be feasible, and
intersection of electricians’ time constraint with
horizontal (C) axis will become a new feasible corner
point.
 Primary information is provided by Shadow Price
 Resources available:
 80 hours of electricians’ time.
 60 hours of audio technicians’ time.
 Final Values in table reveal optimal solution requires:
 all 80 hours of electricians’ time.
 Only 20 hours of audio technicians’ time.
 Electricians’ time constraint is binding.
 Audio technicians’ time constraint is non-binding.
 40 unused hours of audio technicians’ time are referred to as
slack.
Changes in Right-Hand-Side (RHS)
Example 2: High Note Sound Company (5 of 6)

In case of electrician hours Shadow Price is $30.
 For each additional hour of electrician time that
firm can increase profits by $30.
Changes in RHS of a Non-binding Constraint
Example 2: High Note Sound Company (6 of 6)

Audio technicians’ time has 40 unused hours.
 No interest in acquiring additional hours of resource.
 Shadow price for audio technicians’ time is zero.
 Once 40 hours is lost (current unused portion, or slack) of
audio technicians’ time, resource also becomes binding.
 Any additional loss of time will clearly have adverse effect
on profit.
 Reduced Cost value - shows amount one will ‘lose’ if solution
is forced to make an additional unit.
Sensitivity Analysis For A Larger
Maximization Example
Example 3: Anderson Electronics (1 of 8)
Considering producing four potential products: VCRs, stereos,
televisions (TVs), and DVD players:
Profit per unit:
VCR
Stereo
TV
DVD
$41
$32
$72
$54
LP Formulation
Example 3: Anderson Electronics (2 of 8)
Objective: maximize profit =
$29 V + $32 S + $72 T + $54 D
subject to
3 V + 4 S + 4 T + 3 D  4700
2 V + 2 S + 4 T + 3 D  4500
1 V + 1 S + 3 T + 2 D  2500
V, S, T, D  0
Where: V
S
T
D
=
=
=
=
(Electronic components)
(Non-electronic components)
(Assembly time in hours)
number of VCRs to produce.
number of Stereos to produce.
number of TVs to produce.
number of DVD players to produce.
Excel Solver Sensitivity Report
Example 3: Anderson Electronics (3 of 8)
Non Zero value decision variables, Stereos and
DVDs:
Produce 380 Stereos with unit profit of $32.
• Decision should not change as profit is between
$31.33 and $72:
Produce 1060 DVDs with unit profit of $54.
• Decision should not change as profit is between $49
and $64:
Example 3: Anderson Electronics (4 of 8)
Zero value decision variables, VCRs and TVs:
Produce 0 VCRs (Reduced cost of $1.00)
• Decision to make 0 should not change as profit is below $29
– but should change over and $30:
Produce 0 TVs with unit cost of $8.00 (Reduced Cost).
• Decision to make 0 should not change as profit is below $72
– but should change over and $80:
Simultaneous Changes In Parameter Values
Example 3: Anderson Electronics (5 of 8)
Possible to analyze impact of simultaneous changes on optimal
solution only under specific condition:
 (Change / Allowable change)  1
• If decrease RHS from 4,700 to 4,200, allowable decrease is
950.
The ratio is: 500 / 950 = 0.5263
• If increase 200 hours (from 2,500 to 2,700) in assembly time,
allowable increase is 466.67.
The ratio is: 200 / 466.67 = 0.4285
• The sum of these ratios is:
Sum of ratios = 0.5263 + 0.4285 = 0.9548 < 1
Since sum does not exceed 1, information provided in
sensitivity report is valid to analyze impact of changes.
Simultaneous Changes In Parameter Values
Example 3:Anderson Electronics (6 of 8)
• Decrease of 500 units in electronic component
availability reduces size of feasible region and causes
profit to decrease.
– Magnitude of decrease is $1,000 (500 units x $2 per unit).
• Increase of 200 hours of assembly time results in larger
feasible region and net increase in profit.
– Magnitude of increase is $4,800 (200 hours x $24 per
hour).
• Net impact of both changes simultaneously is an
increase in profit by $3,800 ( $4,800 - $1,000).
Simultaneous Changes In Parameter Values
Example 3: Anderson Electronics (7 of 8)
Simultaneous Changes in OFC Values
• What is impact if selling price of DVDs drops by $3 per unit
and at same time selling price of stereos increases by $8 per
unit?
• For current solution to remain optimal, allowable decrease in
DVD players is $5, while allowable increase in OFC for
stereos is $40.
– Sum of ratios is:
Sum of ratios = $3 / $5 + $8 / $40 = 0.80 < 1
– $3 decrease in profit per DVD player causes total profit to decrease by
$3,180 (i.e., $3 x 1,060).
– $8 increase in unit profit of each stereo results in total profit of $3,040
(i.e., $8 x 380).
• Net impact is a decrease in profit of only $140 to a new value
of $69,260.
Checking Validity of the 100% Rule
Example 3: Anderson Electronics Example (8 of 8)
• Calculate ratio of reduction in each resource’s availability
to allowable decrease for that resource.
Sum of ratios = 5/950 + 4/560 + 4/1325 = 0.015 < 1
• Required Profit on Each HTS:
5 x shadow price of electronic components +
4 x shadow price of non-electronic components +
4 x shadow price of assembly time
or 5 x $2 + 4 x $0 + 4 x $24 = $106
• Profit contribution of each HTS has to at least make up
shortfall in profit.
• OFC for HTS must be at least $106 in order for optimal
solution to have non-zero value.
Sensitivity Analysis - Minimization Example
Example 4: Burn-Off Diet Drink Example (1 of 2)
• Plans to introduce miracle drink that will magically burn
fat away.
LP Formulation
Example 4: Burn-Off Diet Drink Example (2 of 2)
Objective: minimize daily dose cost in cents.
4A + 7B + 6C + 3D
Subject to
A + B + C + D  36
(Daily dose requirement)
3A + 4B + 8C + 10D  280
(Chemical X requirement)
5A + 3B + 6C + 6D  200
(Chemical Y requirement)
10A + 25B + 20C + 40D  1050 (Chemical Z max limit)
A, B, C, D  0
Problem Statement (1 of 2)
Example Problem 5
• Two airplane parts: no.1 and no. 2.
• Three manufacturing stages: stamping, drilling, milling.
• Decision variables: x1 (number of part no.1 to produce)
x2 (number of part no.2 to produce)
• Model: Maximize Z = $650x1 + 910x2
subject to:
4x1 + 7.5x2  105 (stamping,hr)
6.2x1 + 4.9x2  90 (drilling, hr)
9.1x1 + 4.1x2  110 (finishing, hr)
x1, x2  0
Graphical Solution (2 of 2)
Example Problem 5
Maximize Z = $650x1 + $910x2
subject to:
4x1 + 7.5x2  105
6.2x1 + 4.9x2  90
9.1x1 + 4.1x2  110
x1, x2  0
s1 = 0, s2 = 0, s3 = 11.35 hr
485.33  c1  1,151.43
137.76  q1  89.10
Graphical Solution