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Linear Programming – Sensitivity Analysis PowerPoint PPT Presentation


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Linear Programming – Sensitivity Analysis. How much can the objective function coefficients change before the values of the variables change? How much can the right hand side of the constraints change you obtain a different basic solution?

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Linear Programming – Sensitivity Analysis

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Linear programming sensitivity analysis l.jpg

Linear Programming – Sensitivity Analysis

How much can the objective function coefficients change

before the values of the variables change?

How much can the right hand side of the constraints change

you obtain a different basic solution?

How much value is added/reduced to the objective function if I have

a larger/smaller quantity of a scarce resource?


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Linear Programming – Sensitivity

Leo Coco Problem

Max 20x1 + 10 x2

s.t. x1 - x2 <= 1

3x1 + x2 <= 7

x1, x2 >= 0

Solution:

x1 = 0, x2 = 7

Z = 70

Issue: How much can you change a cost coefficient without changing the solution?


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Sensitivity – Change in cost coefficient

What if investment 2 only

pays $5000 per share?

Max 20x1 + 5 x2

s.t. x1 - x2 <= 1

3x1 + x2 <= 7

x1, x2 >= 0

New objective function isobar

Optimal solution is now the point (2,1).

Issue: At what value of C2 does solution change?


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Sensitivity – Change in cost coefficient

Issue: At what value of C2 does solution change?

Ans.: When objective isobar is parallel to the binding constraint.

Max 20x1 + C2* x2

s.t. 3x1 + x2 <= 7

3/1 = 20/C2 or C2 = 20/3 or 6.67


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Sensitivity – Change in cost coefficient

Lindo Sensitivity Analysis Output – Leo Coco Problem

LP OPTIMUM FOUND AT STEP 1

OBJECTIVE FUNCTION VALUE

1) 70.00000

VARIABLE VALUE REDUCED COST

X1 0.000000 10.000000

X2 7.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES

1) 8.000000 0.000000

2) 0.000000 10.000000

NO. ITERATIONS= 1

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

X1 20.000000 10.000000 INFINITY

X2 10.000000 INFINITY 3.333333

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

1 1.000000 INFINITY 8.000000

2 7.000000 INFINITY 7.000000


Sensitivity change in right hand side l.jpg

Sensitivity – Change in right hand side

What if only 5 hours

available in time constraint?

Max 20x1 + 5 x2

s.t. x1 - x2 <= 1

3x1 + x2 <= 5

x1, x2 >= 0

New time constraint.

Optimal solution is now the point (5,0). But, basis does not change.

Issue: At what value of the r.h.s. does the basis change?


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Sensitivity – Change in right hand side

Issue: At what value of the r.h.s.

does the basis change?

Max 20x1 + 5 x2

s.t. x1 - x2 <= 1

3x1 + x2 <= ?

x1, x2 >= 0

Basis changes at this constraint.

x2 becomes non-basic at the origin. Or, when the constraint is:

3x1 + x2 < 0


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Sensitivity – Change in right hand side

Lindo Sensitivity Analysis Output – Leo Coco Problem

LP OPTIMUM FOUND AT STEP 1

OBJECTIVE FUNCTION VALUE

1) 70.00000

VARIABLE VALUE REDUCED COST

X1 0.000000 10.000000

X2 7.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES

1) 8.000000 0.000000

2) 0.000000 10.000000

NO. ITERATIONS= 1

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

X1 20.000000 10.000000 INFINITY

X2 10.000000 INFINITY 3.333333

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

1 1.000000 INFINITY 8.000000

2 7.000000 INFINITY 7.000000


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Sensitivity – Shadow or Dual Prices

Issue, how much are you willing to pay for one additional unit of a limited resource?

Max 20x1 + 10 x2

s.t. x1 - x2 <= 1 (budget constraint

3x1 + x2 <= 7 (time constraint

x1, x2 >= 0

Knowing optimal solution is (0,7) and time constraint is binding:

Not willing to increase budget constraint (shadow price is $0).

If time constraint increase by one unit (to 8), solution will change to (0,8) and Z=80. Therefore should be willing to pay up to $10(000s) for each additional unit of time constraint.


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Sensitivity – Change in right hand side

Lindo Sensitivity Analysis Output – Leo Coco Problem

LP OPTIMUM FOUND AT STEP 1

OBJECTIVE FUNCTION VALUE

1) 70.00000

VARIABLE VALUE REDUCED COST

X1 0.000000 10.000000

X2 7.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES

1) 8.000000 0.000000

2) 0.000000 10.000000

NO. ITERATIONS= 1

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASE

X1 20.000000 10.000000 INFINITY

X2 10.000000 INFINITY 3.333333

RIGHTHAND SIDE RANGES

ROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASE

1 1.000000 INFINITY 8.000000

2 7.000000 INFINITY 7.000000


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Linear Programming – Sensitivity Analysis

What if more than one coefficient is changed?:

100% Rule (for objective function coefficients):

if <= 1, the optimal solution will not change,

where is the actual increase (decrease) in the coefficient

and is the maximum allowable increase (decrease) from

the sensitivity analysis.


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Linear Programming – Sensitivity Analysis

Example obj. function coefficient changes


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Linear Programming – Sensitivity Analysis

Simultaneous variations in multiple coefficients:

100% Rule (for RHS constants):

if <= 1, the optimal basis and product mix will not change,

where is the actual increase (decrease) in the coefficient

and is the maximum allowable increase (decrease) from

the sensitivity analysis.


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Linear Programming – Sensitivity Analysis

Example RHS constant changes


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Linear Programming – Duality Theory

Every LP has an associated dual problem. The Dual is

essentially the inverse of the Primal (original problem).

The optimal dual solution produces the dual price or shadow

price reported in the Lindo output reports.


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Integer Programming (IP)

Similar to Linear Programming with the exception that

decision variables must take on integer values.

Drawbacks: solution effort considerably more difficult than

Linear programming.

Should we have used IP to solve the microwave oven

homework problem? (solution was 147.8 model 1 and 100.0

model II ovens).

Lindo solves both binary and integer programming problems.


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