LINEAR PROGRAMMING SENSITIVITY ANALYSIS Learning Objectives Learn sensitivity concepts Understand,...

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.

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.

Ranges of Optimality

Sensitivity Analysis Using GraphsExample 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)

Impact of price change of ReceiversIf 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 GraphsExample 1 (3 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).

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 GraphsExample 2: Galaxy Industries (1 of 5)

Recall the feasible Region

600

800

1200

400 600 800

X2

X1

We now demonstrate the search for an optimal solution Start at some arbitrary profit, say profit = $2,000...

Profit = $ 000

2,

Then increase the profit, if possible...

3,4,

...and continue until it becomes infeasible

Profit =$5040

Sensitivity Analysis Using Graphs Example 2:(2 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).

600

800

1200

400 600 800

X2

X1

The effects of changes in an objective function coefficient

on the optimal solutionMax 8x1 + 5x2

Max 4x1 + 5x2Max 3.75x1 + 5x2 Max 2x1 + 5x2


600

800

1200

400 600 800

X2

X1

The effects of changes in an objective function coefficients on the optimal solution

Max8x1 + 5x2

Max 3.75x1 + 5x2

Max8x1 + 5x2

Max 3.75 x1 + 5x2M

ax 10 x1 + 5x23.75

10

Range of optimality


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

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Sensitivity Analysis Using GraphsExample 3: Beaver Creek Pottery (1 of 4)


subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0


• 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 CoefficientSensitivity Range

objective function Z = $40x1 + $50x2 sensitivity range for:

x1: 25 c1 66.67 x2: 30 c2 80

Objective Function Coefficient Sensitivity Range for c1 and c2 Example 3 (4 of 4)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 164x1 + 3x2 24

x1, x2 0

sensitivity ranges: 4 c1 0 c2 4.5

Objective Function CoefficientSensitivity Range(for a Cost Minimization Model)

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% Rule1. 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.

The 100% Rule1. 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) ValuesIncreasing the Labor Constraint (1 of 3)Example 1: Beaver Creek


subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Changes in Constraint Quantity (RHS) ValuesSensitivity Range for Labor Constraint (2 of 3)Example 1: Beaver Creek

Sensitivity range for:30 q1 80 hr

Changes in Constraint Quantity (RHS) ValuesSensitivity Range for Clay Constraint (3 of 3)Example 1: Beaver Creek

Sensitivity range for: 60 q2 160 lb

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?

Sensitivity Analysis of Right-Hand SideValues & Shadow Prices

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.

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.

Shadow Prices (Dual Values)

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) ValuesExample 2: High Note Sound Company (2 of 6)

May change the feasible region size.

May change or move corner points.

Increase in Electricians’ Available TimeExample2: 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 TimeExample 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)

In case of electrician hours Shadow Price is $30.

For each additional hour of electrician time that firm can increase profits by $30.

Example 2: High Note Sound Company (5 of 6)

Changes in RHS of a Non-binding ConstraintExample 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 FormulationExample 3: Anderson Electronics (2 of 8)

Objective: maximize profit =

$29 V + $32 S + $72 T + $54 Dsubject to

3 V + 4 S + 4 T + 3 D 4700 (Electronic components)

2 V + 2 S + 4 T + 3 D 4500 (Non-electronic components)

1 V + 1 S + 3 T + 2 D 2500 (Assembly time in hours)

V, S, T, D 0

Where: V = number of VCRs to produce. S = number of Stereos to produce. T = number of TVs to produce. D = 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 ValuesExample 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 ValuesExample 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 FormulationExample 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

• 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

Problem Statement (1 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 1,151.43137.76 q1 89.10

Graphical Solution

Graphical Solution (2 of 2)Example Problem 5

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