a lecture on optimization

BITS PilaniHyderabad Campus

Optimization Lecture 6

BITS Pilani, Hyderabad Campus

• Formulation of Linear Programming Problem (LPP) from the real world problem

• Solving LPP using Graphical method • Solving maximization LPP using Simplex method with

less than or equal to constraints• Solving minimization LPP using Simplex method with

less than or equal to constraints• Solving maximization and minimization with greater than

or equal to constraints and equal to constraints – Big-M Method

Topics we have discussed so far

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Solving maximization and minimization with greater than or equal to constraints and equal to constraints – Two Phase Method

Today’s Topic

Second method for handling artificial variables within simplexHas two phasesPhase I : finds a basic feasible solution for the original problem, if one exists Phase II : finds an optimal solution for original problem

BITS Pilani, Hyderabad Campus

Consider a linear programming problem consisting of artificial variablesObjective: to set all artificial variables to zero at optimalityHence, solve a linear programming problem by replacing original objective function by sum of artificial variables with a minimization objectiveObserve: due to non-negativity constraint, sum of artificial variables cannot be negative, i.e., smallest possible feasible value of sum is zeroIf optimal value of modified objective function is not zero (i.e., strictly positive), conclude that problem is not feasible

Two-Phase Method

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maximize z = -3x1 + x3

subject to

x1 + x2 + x3 + x4 = 4...........(1)

-2x1 + x2 - x3 = 1...........(2)

3x2 + x3 + 3x4 = 10..............(3)

x1 , x2 , x3 , x4 0...........(4)

Observe: not in canonical form

Two-Phase Method

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maximize z = -3x1 + x3 subject to

x1 + x2 + x3 + x4 + x5 = 4...........(1’)

-2x1 + x2 - x3 + x6 = 1...........(2’)

3x2 + x3 + 3x4 + x7 = 10............(3’)

x1 , x2 , x3 , x4 , x5 , x6 , x7 0...(4’)

Two-Phase Method

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minimize r = x5 + x6+ x7

i.e., maximize w = -x5 - x6 - x7

subject to

x1 + x2 + x3 + x4 + x5 = 4...........(1’)

-2x1 + x2 - x3 + x6 = 1...........(2’)

3x2 + x3 + 3x4 + x7 = 10............(3’)

x1 , x2 , x3 , x4 , x5 , x6 , x7 0...(5’)

Two-Phase Method

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Basic Vars x1 x2 x3 x4 x5 x6 x7 RHS

w 0 0 0 0 1 1 1 0

x5 1 1 1 1 1 0 0 4

x6 -2 1 -1 0 0 1 0 1

x7 0 3 1 3 0 0 1 10

Two-Phase Method

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Basic Vars x1 x2 x3 x4 x5 x6 x7 RHS

w 1 -5 -1 -4 0 0 0 -15

x5 1 1 1 1 1 0 0 4

x6 -2 1 -1 0 0 1 0 1

x7 0 3 1 3 0 0 1 10Perform the elementary row operation R1 = R1 - (R2 + R3 + R4 ) to make the table in canonical form

Two-Phase Method

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Basic Vars x1 x2 x3 x4 x5 x6 x7

RHS Ratio

w 1 -5 -1 -4 0 0 0 -15x5 1 1 1 1 1 0 0 4 4/1=4

x6 -2 1 -1 0 0 1 0 1 1/1=1

x7 0 3 1 3 0 0 1 10 10/3=3.33

Two-Phase Method

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Basic

Varsx1 x2 x3 x4 x5 x6 x7

RHS Ratio

w -9 0 -6 -4 0 5 0 -10

x5 3 0 2 1 1 -1 0 3 3/3=1

x2 -2 1 -1 0 0 1 0 1 -----

x7 6 0 4 3 0 -3 1 7 7/6=1.3

Two-Phase Method

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Basic

Varsx1 x2 x3 x4 x5 x6 x7

RHS Ratio

w 0 0 0 -1 3 2 0 -1

x1 1 0 2/3 1/3 1/3 -1/3 0 1 1/1/3=3

x2 0 1 1/3 2/3 2/3 1/3 0 3 9/2=4.5

x7 0 0 0 1 -2 -1 1 1 1/1=1

Two-Phase Method

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Basic Vars x1 x2 x3 x4 x5 x6 x7

RHS

w 0 0 0 0 1 1 1 0x1 1 0 2/3 0 1 0 -1/3 2/3x2 0 1 1/3 0 2 1 -2/3 7/3x4 0 0 0 1 -2 -1 1 1

•Observe: all artificial variables are now non-basic, i.e., problem is feasible

•Hence x5= x6= x7 = 0

•Initial basic feasible solution: x1=2/3, x2=7/3, x4=1

•End of Phase I

Two-Phase Method

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In Phase 2 we solve the original problem Maximize z = -3x1 + x3 Subject to x1 + (⅔)x3 = ⅔ x2 + (⅓) x3 = ⅓ x4 = 1x1 , x2 , x3 ≥ 0with initial basic variables as x1 , x2 , x4

Two-Phase Method

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Basic

Varsx1 x2 x3 x4

RHS

z 3 0 -1 0 0x1 1 0 2/3 0 2/3x2 0 1 1/3 0 7/3x4 0 0 0 1 1

Drop –w rowDrop all artificial variable columns

Two-Phase Method

Not in canonical form Hence perform ERO : R1 = R1 – 3R2

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Basic

Varsx1 x2 x3 x4

RHS Ratio

z 0 0 -3 0 -2x1 1 0 2/3 0 2/3 1x2 0 1 1/3 0 7/3 7x4 0 0 0 1 1 ---

Two-Phase Method

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Basic Vars x1 x2 x3 x4

RHS

z 9/2 0 0 0 1

x3 3/2 0 1 0 1

x2-1/2 0 0 0 2

x4 0 0 0 1 1

Coefficients of all non-basic variables in the first row ≥ 0

Hence we have optimal solution

z=1, x2=2, x3=1, x4=1, x1= 0

Two-Phase Method

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Infeasible Solution

Infeasibility is detected in the simplex method when an artificial variable remains positive in the final tableau.

For Example consider the LPPMAX z = 2x1 + 6x2 s. t. 4x1 + 3x2 < 12

2x1 + x2 > 8

x1, x2 > 0

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Infeasible Solution

MAX z = 2x1 + 6x2 s. t. 4x1 + 3x2 + S1 = 12

2x1 + x2 - S2 + A1 = 8

x1, x2 , S1, S2 , A1 > 0

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Basic Vars x1 x2 S1 S2 A1

RHS

z -2 1 0 0 M 0S1 4 3 1 0 0 12A1 2 1 0 -1 1 8

Basic Vars x1 x2 S1 S2 A1

RHS Ratio

z -2-2M 1-M 0 M 0 -8MS1 4 3 1 0 0 12 3A1 2 1 0 -1 1 8 4

Infeasible Solution

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Infeasible Solution

Basic Vars x1 x2 S1 S2 A1

RHS

z 0 (5+M)/2 (1+M)/2 M 0 6-2M

x1 1 3/4 1/4 0 0 3

A1 0 -3/4 -1/4 0 0 3/2

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Infeasible Solution

Basic Vars x1 x2 S1 S2 A1

RHS

r 0 0 0 0 1 0

S1 4 3 1 0 0 12

A1 2 1 0 -1 1 8

Min r = A1

s. t. 4x1 + 3x2 + S1 = 12

2x1 + x2 - S2 + A1 = 8x1, x2 , S1, S2 , A1 > 0

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Infeasible Solution

Basic Vars x1 x2 S1 S2 A1

RHS

r -2 -1 0 1 0 8

S1 4 3 1 0 0 12

A1 2 1 0 -1 1 8

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Solve the following linear programming problem by using the simplex method:Min Z =2 X1 + 3 X2

S.t.½ X1 + ¼ X2 ≤ 4X1 + 3X2 20X1 + X2 = 10X1, X2 0

BITS Pilani, Hyderabad Campus

SolutionStep 1: standard formMin Z, s.t.Z – 2 X1 – 3 X2 - M A1 -M A2 = 0 ½ X1 + ¼ X2 + S1 = 4 X1 + 3X2 - S2 + A1 = 20 X1 + X2 + A2 = 10X1, X2 ,S1, S2, A1, A2 0 Where: M is a very large number

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M, a very large number, is used to ensure that the values of A1 and A2, …, and An will be zero in the final (optimal) tableau as follows:

1. If the objective function is Minimization, then A1, A2, …, and An must be added to the RHS of the objective function multiplied by a very large number (M).

Example: if the objective function is Min Z = X1+2X2, then the obj. function should be Min Z = X1 + X2+ MA1 + MA2+ …+ MAn

OR Z – X1 - X2- MA1 - MA2- …- MAn = 0

2. If the objective function is Maximization, then A1, A2, …, and An must be subtracted from the RHS of the objective function multiplied by a very large number (M).

Example: if the objective function is Max Z = X1+2X2, then the obj. function should be Max Z = X1 + X2- MA1 - MA2- …- MAn

OR Z - X1 - X2+ MA1 + MA2+ …+ MAn = 0

N.B.: When the Z is transformed to a zero equation, the signs are chang

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Basic variables

X1

2

X2

3

S1

0

S2

0

A1

M

A2

M

RHS

S1 ½ ¼ 1 0 0 0 4

A1 1 3 0 -1 1 0 20A2 1 1 0 0 0 1 10

Z -2 -3 0 0 -M -M 0

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To correct this violation before starting the simplex algorithm, the elementary row operations are used as follows:New (Z row) = old (z row) ± M (A1 row) ± M (A2 row)In our case, it will be positive since M is negative in the Z row, as following:Old (Z row): -2 -3 0 0 -M -M 0M (A1 row): M 3M 0 -M M o 20MM (A2 row): M M 0 0 0 M 10MNew (Z row):2M-2 4M-3 0 -M 0 0 30M

BITS Pilani, Hyderabad Campus

Basic variables

X1

2

X2

3

S1

0

S2

0

A1

M

A2

M

RHS

S1 1/2 1/4 1 0 0 0 4

A1 1 3 0 -1 1 0 20A2 1 1 0 0 0 1 10

Z 2M-2 4M-3 0 -M 0 0 30M

BITS Pilani, Hyderabad Campus

Basic variables

X1

2

X2

3

S1

0

S2

0

A1

M

A2

M

RHS

S15/12 0 1 1/12 -1/12 0 7/3

X21/3 1 0 -1/3 1/3 0 20/3

A22/3 0 0 1/3 -1/3 1 10/3

Z 2/3M-1 0 0 1/3M-1 1-4/3M 0 20+10/3M

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Basic variables

X1 X2 S1 S2 A1 A2 RHS

S10 0 1 -1/8 1/8 -5/8 1/4

X20 1 0 -1/2 1/2 -1/2 5

X11 0 0 1/2 -1/2 3/2 5

Z 0 0 0 -1/2 ½-M 3/2-M 25

BITS Pilani, Hyderabad Campus

•In the final tableau, if one or more artificial variables (A1, A2, …) still basic and has a nonzero value, then the problem has an infeasible solution•If there is a zero under one or more nonbasic variables in the last tableau (optimal solution tableau), then there is a multiple optimal solution.•When determining the leaving variable of any tableau, if there is no positive ratio (all the entries in the pivot column are negative and zeroes), then the solution is unbounded.