sensitivity analysis - koç...

INDR 262 Introduction to Optimization Methods

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 1

SENSITIVITY ANALYSIS

Sensitivity analysis is carried out after the optimum solution of an LP problem is

obtained. In most problems:

-data are not known exactly

-constraints may not be rigid

-a new constraint or activity may be added

The objective of sensitivity analysis is to find new optimal solution for an LP when some

of the problem data changes without solving the problem from scratch.

Change in the objective function coefficients, c

I) xk is a nonbasic variable

• ck is replaced by ck′ in the LP model

• zk-ck is replaced by zk-ck′ in the simplex tableau

• current zk-ck in the simplex tableau is negative

• if zk-ck′=(zk-ck)+(ck-ck′) is negative, xk must enter into the basis (for maximization).

• if zk-ck′=(zk-ck)+(ck-ck′) is positive, the optimal solution does not change.

II) xk is a Basic variable (xk=xBt)

• cBt is replaced by cBt′ in the LP model

• Row Ø is affected by this change

• Calculate new zj-cj as zj′-cj

zj′-cj = cB′B-1aj–cj = (cBB-1aj–cj) + (0,…, 0, cBt′ – cBt, 0, …0) yj

= (zj–cj) + (cBt′–cBt)ytj for all j.

• Update row Ø by multiplying the row xBt by the net change in the cost coefficient of xBt

and add the current row Ǿ.

Changes in the Right-Hand-Side, b

• Right-hand-side vector b is replaced by b′.

• B-1b will be replaced by B-1b′

• B-1b′ = B-1b + B-1(b′–b)

• If B-1b′ ≥ 0, then z* = cBB-1b′


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 2

• If B-1b′ < 0, then the problem becomes infeasible, use the dual simplex method to find the

new optimal solution.

Changes in the Constraint Matrix, A

I. Nonbasic Variables

• Technological coefficient vector aj is replaced by aj′

• The new column under xj in the simplex tableau is B-1aj′

• The new reduced cost for xj is zj′–cj = cBB-1aj′–cj

• If zj′–cj ≤ 0, the optimal solution is the same

• If zj′–cj > 0, then the nonbasic variable xj must enter the basic

II. Basic Variables

• Technological coefficient vector aj is replaced by aj′

• Basis matrix, B, and consequently its inverse, B-1

• Update B:

ü Add a new column xj′ whose entries will be aj′

zj′–cj = cBB-1aj′–cj and yj′ = B-1aj′

ü eliminate the old column xj from the basis

• continue with the simplex method

Adding a New Activity

• A new activity xn+1 with a cost of cn+1 and technology coefficient of an+1 is introduced

• If zn+1–cn+1≤ 0, let xn+1 enter the basis and continue with the simplex method (for

maximization)

• If zn+1–cn+1> 0, the optimal solution does not change

Adding a New Constraint

• A new row am+1 x ≤ bm+1 is added

• The new row is rewritten as aBm+1xB + aNm+1xN + xn+1= bm+1

• The new system of equation is:

z + (cBB-1N–cN) xN = cBB-1b

xB + B-1NxN = B-1b


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 3

(aNm+1–aBm+1B-1N) xN + xn+1 = bm+1–aBm+1B-1b

• If aNm+1–aBm+1B-1b ≥ 0, the optimal solution does not change

• If aNm+1–aBm+1B-1b < 0, the current solution becomes infeasible, solve the dual simplex

method to find a feasible solution.

REVIEW OF SENSIVITY ANALYSIS:

max z = c j x jj =1

n

∑

s.t. aij x jj =1

n

∑ ≤ bi ∀i=1,...,m

Sensivity Analysis is identification of sensitive parameters. * determine the range of values of the parameters over which the optimal solution will

remain unchanged. *determine the range of values over which the optimal BF solution will remain feasible. Example: Original Model Revised Model max z = 3x1 + 5x2 max z = 4x1 + 5x2

s.t. x1 ≤ 4 s.t. x1 ≤ 4 2x2 ≤ 12 2x2 ≤ 24 3x1 + 2x2 ≤ 18 2x1 +2x2 ≤ 18 x1, x2 ≥ 0 x1, x2 ≥ 0 The simplex tablaeu of the revised problem is given as follows:

Iter B.V. Eqn. z x1 x2 x3 x4 x5 RHS z 0 1 -4 -5 0 0 0 0 x3 1 0 1 0 1 0 0 4 0 x4 2 0 0 2 0 1 0 24 x5 3 0 2 2 0 0 1 18

• we have a solution to the original problem.

Iter B.V. Eqn. z x1 x2 x3 x4 x5 RHS z 0 1 0 0 0 3/2 1 36 x3 1 0 0 0 1 1/3 -1/3 2 2 x2 2 0 0 1 0 1/2 0 6 x1 3 0 1 0 0 -1/3 1/3 2


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 4

How to find the optimal solution to the revised problem. • Revised Simplex Method:

Eqn. z Original variables Slack Variables RHS 1 1 z*- = y* a - y* z*=y*b

1,...,m 0 A* = S* S* b*=S* where , , are for the revised problem.

A = = =

y* = S*=

z*- = y* - = [ ]0 3/ 2 1 - = [ ]2 0−

z* = y* = = 54

A*= S* =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

3/13/1002/103/13/11 1 0

0 22 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

= 1/ 3 00 12 / 3 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

b*= S* = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

3/13/1002/103/13/11 4

2418

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

= 6122

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

c ca b

a c b

1 00 22 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

c [ ]4 5 b42418

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

[ ]0 3/ 2 1⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

3/13/1002/103/13/11

c A c1 00 22 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

[ ]4 5

b [ ]0 3/ 2 142418

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A

b


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 5

Revised Final Tableu:

B.V. Eqn. z x1 x2 x3 x4 x5 RHS z 0 1 -2 0 0 3/2 1 54 x3 1 0 1/3 0 1 1/3 -1/3 6 x2 2 0 0 1 0 1/2 0 12 x1 3 0 2/3 0 0 -1/3 1/3 -2

Ratio - - - 1/2 - B.V. Eqn. z x1 x2 x3 x4 x5 RHS

z 0 1 1 0 0 0 5/2 45 x3 1 0 1 0 1 0 0 4 x2 2 0 1 1 0 0 1/2 9 x4 3 0 -2 0 0 1 -1 6


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 6

APPLYING SENSIVITY ANALYSIS Case 1 - changes in bi Case 2 - a) changes in the coefficents of a nonbasic variable b) introduction of a new variable Case 3 - changes in the coefficients of a basic variable Case 4 - introduction of a new constraints Case 1- Changes in bi The only revision in the model is the changes in RHS. RHS of final row 0: z*= y* RHS at final rows 1,...,m = S* Example:

b= 41218

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

, = 42418

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

z*= y* = = 54

B* = S* =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

3/13/1002/103/13/11

= 612−2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Equivalently incremental analysis:

∆z*=y*∆b = y*Δb1Δb2Δb3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= y*0120

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∆b*=S*∆b = S*1

2

3

bbb

Δ⎡ ⎤⎢ ⎥Δ⎢ ⎥⎢ ⎥Δ⎣ ⎦

= S*0120

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∆z* = 3/2 (12) = 18 z* = z + ∆z* = 36 +18 = 54 ∆b1*= 1/3 (12) = 12 b1*= 2 + 4 = 6

bb

b

b [ ]0 3/ 2 142418

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

b42418

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 7

∆b2*= 1/2 (12) = 6 b2*= 6 + 6 = 12 ∆b3*= -1/3 (12) = -4 b3*= 2 – 4 = -2 Revised Final Tableu

B.V. Eqn. z x1 x2 x3 x4 x5 RHS

z 0 1 0 0 0 3/2 1 54 x3 1 0 0 0 1 1/3 -1/3 6 x2 2 0 0 1 0 1/2 0 12

x1 3 0 1 0 0 -1/3 1/3 -2 entering b.v.

Ratio - - - 9/2 B.V. Eqn. Z x1 x2 x3 x4 x5 RHS

z 0 1 9/2 0 0 0 5/2 45 x3 1 0 1 0 1 0 0 4 x2 2 0 3/2 1 0 0 1/2 9 x4 3 0 -3 0 0 1 -1 6

The allowable range to stay feasible: ∆b2 = 12 is too large for original BF solution to stay feasible. Second column at S* ↑ b1* = 2 + 1/3∆b2 b2* = 6 + 1/2 ∆b2 b3* = 2 – 1/3 ∆b2 these quantities must be non-negative for the revised solution to be feasible 2 + 1/3∆b2 ≥ 0 ∆b2 ≥ -6 6 + 1/2∆b2 ≥ 0 ∆b2 ≥ -12 2 – 1/3∆b2 ≥ 0 ∆b2 ≤ 6 min ∆b2 is -6 max ∆b2 is 6 so -6≤ ∆b2≤6 6≤b2≤18 → allowable range of values for b2 to stay feasible.


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 8

Case 2a- change in the coefficient of Nonbasic Variables:

• one or more of the following changes are made cj → aj → new coefficients of xj in final row 0: *

jjz c− = y* -

new coefficients of xj in final rows 1...m: A* = S* Example: c1=3 → = 4

a1= ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

301

→ =

y*= S*=1 0 00 0 1 20 1 1

/⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

z1* - = y* – = - 4 = 1

a1* = S* = 1 0 00 0 1 20 1 1

/⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

= 112

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

• allowable range to say optimal: the range of values over which the current optimal solution remains optimal. Criteria: z1-c1 ≥ 0 Since zj* = y* aj → cjy*aj

c1≤y*a1 = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

301

= 15/2

• c1>15/2, x1 will be a basic variable of the optimal solution.

jc

ja

ja jc

ja

1c

1a⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

201

[ ]0 0 5/ 2

1c 1a 1c [ ]0 0 5/ 2⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

201

1a⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

201

[ ]0 0 5/ 2


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 9

Case 2b: Introduction of a New Variable:

• Introduce a new variable for the model • Check whether the current optimal solution stays optimal

Example: max z= 3x1 + 5x2 + 4x6

s.t. x1 + 2x6 ≤4 + 2x2 + 3x6 ≤12 3x1 + 2x2 + x6 ≤18 x1, x2, x6 ≥0 Current Optimal Solution:

x=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

00262

[y1 y2 y3 z1-c1 z2-c2] = [0 3/2 1 0 0]

write the dual constraint for the new variable x6 2y1 + 3y2 + y3 ≥ 4 • if the dual is feasible 2(0) + 3(3/2) + 1(1) ≥ 4 Yes, the dual constaint is feasible. So the primal remains optimal. Case3: Changes in the coefficients of a basic variable xj is a basic variable. cj → aj → Example: c2 = 5 → 2c = 3

jc

ja


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 10

a2= ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

220→ 2a =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

430

z2– 2c = y* – 2c = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

430

- 3= 7 → non-zero

a2* = S* = 1 0 00 0 1/ 20 1 1

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

430

= 021

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

→ not a basic variable

make it a basic variable

B.V. Eqn. z x1 x2 x3 x4 x5 RHS z 0 1 9/2 7 0 0 5/2 45 x3 1 0 1 0 1 0 0 4 x2 2 0 3/2 2 0 0 1/2 9 x4 3 0 -3 -1 0 1 -1 6 z 0 1 -3/4 0 0 0 3/4 27/2 x3 1 0 1 0 1 0 0 4 x2 2 0 3/4 1 0 0 1/4 9/2 x4 3 0 -9/4 0 0 1 -3/4 21/2 z 0 1 0 0 3/4 0 3/4 33/2 x1 1 0 1 0 1 0 0 4 x2 2 0 0 1 -3/4 0 1/4 3/2 x4 3 0 0 0 9/4 1 -3/4 39/2

Allowable Range To Stay Optimal: zj*–cj must be positive for variable xj to stay optimal. zj*– = 0

= cj + ∆cj zj*-cj - ∆cj = 0 Example: Consider the optimal simplex tableu Row 0 = [1: 0 0 ¾ 0 ¾ : 33/2] * change c2= 3 to 2c = c2 + ∆c2 New Row 0= [1: 0 -∆c2 ¾ 0 ¾ : 33/2] *for x2 to stay a basic z2*- 2c must be equal to 0.

2a [ ]0 0 5/ 2

2a

jc

jc


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 11

Gaussian elimination: Row 0: [1: 0 ∆c2 ¾ 0 ¾ : 33/2] + ∆c2xRow 2: [0: 0 -∆c2 -¾∆c2 0 1/4∆c2 : 3/2∆c2] New Row0: [1: 0 0 ¾-¾∆c2 0 ¾+1/4∆c2 : 33/2+3/2∆c2] 3/4-3/4∆c2≥0, 3/4≥3/4∆c2, ∆c2≤1 3/4+1/4∆c2≥0, 1/4∆c2≥3/4, ∆c2≥-3 -3≤∆c2≤1 So 0≤c2≤4 is the allowable range for c2 to stay optimal. Case 4. Introduction of a New Constraint: • when a new constraint is added • check if the optimal solution satisfies the constraint • if the constraint is satisfied, the optimal solution is still the optimal • if the constraint makes the optimal solution infeasible:

- introduce the constraint into the final simplex tableu of the original model ( the slack

variable for the new constraint becomes the basic variable for this constraint) - convert this constraint to proper form by making the coefficients of the basic variables 0. - Proceed with regualr simplex ( or dual simplex if necessary) until you confirm

optimality. Example: A new constraint is added. 2x1 + 3x2 ≤ 24 The optimal solution is given in the following simplex tableu. Find if the introduction of this new constraint changes the optimal solution or not.

B.V. Eqn. z x1 x2 x3 x4 x5 RHS z 0 1 9/2 0 0 0 5/2 45 x3 1 0 1 0 1 0 0 4 x2 2 0 3/2 1 0 0 1/2 9 x4 3 0 -3 0 0 1 -1 6

- check for feasibility: 2x1 + 3x2≤ 24 2(0) + 3(9) =27 infeasible - the optimal solution is infeasible


⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Metin Turkay 12

B.V. Eqn. z x1 x2 x3 x4 x5 x6 RHS z 0 1 9/2 0 0 0 5/2 0 45 x3 1 0 1 0 1 0 0 0 4 x2 2 0 3/2 1 0 0 1/2 0 9 x4 3 0 -3 0 0 1 -1 0 6 x6 4 0 2 3 0 0 0 1 24 z 0 1 9/2 0 0 0 5/2 0 45 x3 1 0 1 0 1 0 0 0 4 x2 2 0 3/2 1 0 0 1/2 0 9 x4 3 0 -3 0 0 1 -1 0 6 x6 4 0 -5/2 0 0 0 -3/2 1 -3

RATIO 9/5 - - - 5/3 - z 0 1 0 0 0 0 0 5/3 40 x3 1 0 1 0 1 0 0 0 4 x2 2 0 2/3 1 0 0 0 1/3 8 x4 3 0 -4/3 0 0 1 0 -2/3 8 x5 4 0 5/3 0 0 0 1 -2/3 2

sensitivity analysis - koç...

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