or note farhan idzni

8/2/2019 Or Note Farhan Idzni

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Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering

change to

standard model

The Simplex MethodStandard model :

a) All variables must be non-negativei) If xi is any valueSubstitute xi with xi+ - xi++ii) If xi


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Since the entering variable is x1.

Leaving variable = min { 24/6, 6/1, -, - }

= min { 4, 6, -, - }

Thus, the leaving variable is s1.

Step 1 : determine the pivot equation

New Pivot Equation = ( Current Pivot Equation ) / (Pivot Element)

New x1-row = ( 6, 4, 1, 0, 0, 0, 24 ) / ( 6 )

= ( 1, 2/3, 1/6, 0, 0, 0, 4 )

Step 2 : other equation

New Equation = Current Equation ( Pivot Column Coefficient )( New Pivot Equation )

New z-row = ( -5, -4, 0, 0, 0, 0, 0 )( -5 )( 1, 2/3, 1/6, 0, 0, 0, 4 )= ( 0, -2/3, 5/6, 0, 0, 0, 20 )

New s2-row = ( 1, 2, 0, 1, 0, 0, 6 )

( 1 )( 1, 2/3, 1/6, 0, 0, 0, 4 )= ( 0, 4/3, -1/6, 1, 0, 0, 2 )New s3-row = ( -1, 1, 0, 0, 1, 0, 1 )( -1 )( 1, 2/3, 1/6, 0, 0, 0, 4 )

= ( 0, 5/3, 1/6, 0, 1, 0, 5 )

New s4-row = ( 0, 1, 0, 0, 0, 1, 2 )( 0 )( 1, 2/3, 1/6, 0, 0, 0, 4 )= ( 0, 1, 0, 0, 0, 1, 2 )

Step 3 : arrange the data in the new table

Basic x1 x2 s1 s2 s3 s4 solution

z 0 -2/3 5/6 0 0 0 20 z-row

x1 1 2/3 1/6 0 0 0 4 s1-row

s2 0 4/3 -1/6 1 0 0 2 s2-rows3 0 5/3 1/6 0 1 0 5 s3-row

s4 0 1 0 0 0 1 2 s4-row

Since the entering variable is x2.Leaving variable = min { 4/(2/3), 2/(4/3), 5/(5/3), 2/1 }

= min { 6, 1.5, 3, 2 }


Repating step 1 and 2, arranging the data in the table.

For MAXIMIZATION

problem, since -2/3 is

more negative than 5/6,

the entering variable is x2

For MAXIMIZATION

problem, the all

coefficient in the z-row is

NON-NEGATIVE


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change to

standard model

Basic x1 x2 s1 s2 s3 s4 solution

z 0 0 3/4 1/2 0 0 21 z-row

x1 1 0 1/4 -1/2 0 0 3 s1-row

s2 0 1 -1/8 3/4 0 0 3/2 s2-row

s3 0 0 3/8 -5/4 1 0 5/2 s3-row

s4 0 0 1/8 -3/4 0 1 1/2 s4-row

The M Method

THE M METHODStandard model :

a) When all constraints or = thenArtificial variables, R that play the role of slacks is added.

Artificial variable objective coefficient

i) M in maximization problem.ii) M in minimization problem.

Example of M Method :

Minimize z = 4x1 + x2 Minimize z = 4x1 + x2 + MR1 + MR2Subject to Subject to

3x1 + x2 = 3 3x1 + x2 + R1 = 3

4x1 + 3x2 => 6 4x1 + 3x2 + R2s2 = 6x1 + 2x2 0 x1,x2,R1,R2,s2,s3 => 0

arrange the equation in the table

Basic x1 x2 s2 s3 R1 R2 solution

z -4 -1 0 0 -M -M 0 z-rowR1 3 1 0 0 1 0 3 s1-row

R2 4 3 -1 0 0 1 6 s2-row

s3 1 2 0 1 0 0 4 s3-row

For MAXIMIZATION problem, the all coefficient in the z-row is NON-NEGATIVE. Since all the coefficient in

z-row is NON-NEGATIVE, thus the solution is optimum

Change z-row equation

into new z-row equation

by eliminating the M


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New z-row = Old z-row + M( R1-row ) + M( R2-row )

= ( -4, -1, 0, -M, 0, -M, 0 ) + M( 4, 3, -1, 0, 0, 1, 6 ) + M( 3, 1, 0, 0, 1, 0, 3 )= [( -4 + 7M ), ( -1 + 4M ), -M, 0, 0, 0, 9M ]

Arrange new table


z (7M-4) (4M-1) -M 0 0 0 9M z-rowR1 3 1 0 0 1 0 3 s1-row

R2 4 3 -1 0 0 1 6 s2-row

s3 1 2 0 1 0 0 4 s3-row


Leaving variable = min { 3/3, 6/4, 4/1 }= min { 1, 3/2, 4 }

Thus, the leaving variable is R1.

By using step 1 and step 2, we get the new table


z 0 (1+5M)/3 -M 0 (4-7M)/3 0 (4+2M) z-row

x1 1 1/3 0 0 1/3 0 1 s1-row

R2 0 5/3 -1 0 -4/3 1 2 s2-rows3 0 5/3 0 1 -1/3 0 3 s3-row


Leaving variable = min { 1/(1/3), 2/(5/3), 3/(5/3) }

= min { 3, 1.2, 1.8 }Thus, the leaving variable is R2.

For MINIMIZATION

problem, the all


NON-POSITIVE

For MINIMIZATION problem, since

7M-4 is more positive than 4M-1

and -M, the entering variable is x1

For MINIMIZATION

problem, the all


NON-POSITIVE


1+5M/3 is more positive than 4-7M/3

and -M, the entering variable is x2


5/18




z 0 0 1/5 0 (8/5) - M (-1/5) -M 18/5 z-row

x1 1 0 1/5 0 3/5 -1/5 3/5 s1-row

x2 0 1 -3/5 0 -4/5 3/5 6/5 s2-row

s3 0 0 1 1 1 -1 1 s3-row

Since the entering variable is s2.

Leaving variable = min { (3/5)/(1/5), -, 1/1 }

= min { 3, -, 1 }Thus, the leaving variable is s3.



z 0 0 0 -1/5 (7/5) - M -M 17/5 z-row

x1 1 0 0 -1/5 2/5 0 2/5 s1-row

x2 0 1 0 3/5 -1/5 0 9/5 s2-row

s2 0 0 1 1 1 -1 1 s3-row

THE Two Phase METHODPhase I

a) Objective : MINIMIZE all the artificial variables in the modelb) If at optimum tablo, the optimum objective function is zero, then can proceed to Phase II

Phase II

a) Use original objective function.b) All artificial variable in Phase I is omittedc) Feasible solution from Phase I is used as starting feasible solution. Then proceed as usual

For MINIMIZATION

problem, the all


NON-POSITIVE

For MINIMIZATION problem, since 1/5 is

more positive than (8/5) - M and (-1/5) - M,

the entering variable is s2

For MINIMIZATION problem, the all coefficient in the z-row is NON-POSITIVE. Thus, the table is now optimum


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change to

standard model

Example of Two Phase Method :

PHASE I

Minimize z = 4x1 + x2 Minimize R = R1 + R2

Subject to Subject to3x1 + x2 = 3 3x1 + x2 + R1 = 3

4x1 + 3x2 => 6 4x1 + 3x2 + R2s2 = 6

x1 + 2x2 0 x1,x2,R1,R2,s2,s3 => 0



R 0 0 0 0 -1 -1 0 z-row

R1 3 1 0 0 1 0 3 s1-row

R2 4 3 -1 0 0 1 6 s2-row

s3 1 2 0 1 0 0 4 s3-row

New R-row = R-row + R1-row + R2-row

= ( 0, 0, 0, 0, -1, -1, 0 ) + ( 4, 3, -1, 0, 0, 1, 6 ) + ( 3, 1, 0, 0, 1, 0, 3 )= ( 7, 4, -1, 0, 0, 0, 9 )

Arrange new table

Basic x1

x2

s2

s3

R1

R2

solution

z 7 4 -1 0 0 0 9 z-row

R1 3 1 0 0 1 0 3 s1-row

R2 4 3 -1 0 0 1 6 s2-row

s3 1 2 0 1 0 0 4 s3-row


Leaving variable = min { 3/3, 6/4, 4/1 }= min { 1, 3/2, 4 }


Change R-row equation into new

z-row equation by eliminating

the R1 and R2 column

For PHASE I problem, all the coefficient

in the z-row is NON-POSITIVE


7is more positive than 4 and -1,

the entering variable is x1


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z 0 5/3 -1 0 -7/3 0 2 z-row

x1 1 1/3 0 0 1/3 0 1 s1-row

R2 0 5/3 -1 0 -4/3 1 2 s2-row

s3 0 5/3 0 1 -1/3 0 3 s3-row


Leaving variable = min { 1/(1/3), 2/(5/3), 3/(5/3) }

= min { 3, 1.2, 1.8 }




z 0 0 0 0 -1 -1 0 z-row

x1 1 0 1/5 0 3/5 -1/5 3/5 s1-row

x2 0 1 -3/5 0 -4/5 3/5 6/5 s2-row

s3 0 0 1 1 1 -1 1 s3-row

PHASE II

Minimize z = 4x1 + x2Subject to

x1 + (1/5)s2 = 3/5

x2(3/5)s2 = 6/5

s2 + s3 = 1

x1,x2,s2,s3 => 0


Basic x1

x2

s2

s3

solution

z -4 -1 0 0 0 z-row

x1 1 0 1/5 0 3/5 s1-row

x2 0 1 -3/5 0 6/5 s2-row

s3 0 0 1 1 1 s3-row

For MINIMIZATION problem, since 5/3

is more positive than -7/3 and -1, the

entering variable is x2

For PHASE I problem, the all coefficient in the z-row is NON-POSITIVE. The table is now optimum.

Change z-row equation

into new z-row equation

by eliminating the value in

x1 and x2 column

For PHASE I problem, all

the coefficient in the z-

row is NON-POSITIVE


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New z-row = Old z-row + 4( x1-row ) + ( x2-row )

= ( -4, -1, 0, 0, 0 ) + 4( 1, 0, 1/5, 0, 3/5 ) + ( 0, 1, -3/5, 0, 6/5 )= ( 0, 0, 1/5, 0, 18/5 )

Arrange new table

Basic x1 x2 s2 s3 solutionz 0 0 1/5 0 18/5 z-row

x1 1 0 1/5 0 3/5 s1-row

x2 0 1 -3/5 0 6/5 s2-row

s3 0 0 1 1 1 s3-row

Since the entering variable is s2.

Leaving variable = min { (3/5)/(1/5), -, 1/1 }= min { 3, -, 1 }



Basic x1 x2 s2 s3 solution

z 0 0 0 -1/5 17/5 z-row

x1 1 0 0 -1/5 2/5 s1-row

x2 0 1 0 3/5 9/5 s2-rows2 0 0 1 1 1 s3-row

THE DUAL SIMPLEX METHOD

Different between DUAL SIMPLEX METHOD and M-METHOD or TWO PHASE METHODDUAL SIMPLEX METHOD M-METHOD or TWO PHASE METHOD

No slack artificial variable is used, only slack

variable is used.

Both artificial variable and slack variable can

be included

Initial solution is optimum but not feasible Initial solution is feasible but not optimum

For MINIMIZATION

problem, the all


NON-POSITIVE

For MINIMIZATION problem, the

entering variable is s2

For MINIMIZATION problem, the all coefficient in the z-row is NON-POSITIVE. Thus, the table is now optimum


9/18


change to

standard model

Example of Dual Simplex Method :

Minimize z = 3x1 + 2x2 Minimize z = 3x1 + 2x2

Subject to Subject to

3x1 + x2 => 3 -3x1 - x2 + s1 = -3

4x1 + 3x2 => 6 -4x1 - 3x2 + s2 = -6x1 + x2 0 x1,x2,s1,s2,s3 => 0


Basic x1 x2 s1 s2 s3 solution

z -3 -2 0 0 0 0 z-row

s1 -3 -1 1 0 0 -3 s1-row

s2 -4 -3 0 1 0 -6 s2-row

s3 1 1 0 0 1 3 s3-row

The entering variable is,( z-row ) / ( s2-row ) = ( -3, -2, 0, 0, 0 ) / ( -4, -3, 0, 1, 0 )

= ( 3/4, 2/3, -, -, - )

Since the SMALLEST RATIO is 2/3, thus the entering variable is x 2.


Basic x1 x2 s1 s2 s3 solution

z -1/3 0 0 -2/3 0 4 z-row

s1 -5/3 0 1 -1/3 0 -1 s1-row

x2 4/3 1 0 -1/3 0 2 s2-row

s3 -1/3 0 0 1/3 1 1 s3-row


all coefficient in the z-row is

NON-POSITIVE. The table seems

to look optimum but it is not

feasible since the solution

column have negative value (not

feasible)

Since -6 is more negative than -3 and 3,

s2-row it is chosen to be leaving

variable. The entering variable is the

non-basic variable with SMALLEST

RATIO (for MINIMIZATION) and

SMALLEST ABSOLUTE RATIO (for

MAXIMIZATION)


all coefficient in the z-row is

NON-POSITIVE. The table seems

to look optimum but it is not

feasible since the solution

column have negative value (not

feasible)

Since -1 is more negative than 2 and 1, s1-

row it is chosen to be leaving variable.

The entering variable is the non-basic

variable with SMALLEST RATIO (for

MINIMIZATION) and SMALLEST

ABSOLUTE RATIO (for MAXIMIZATION)


10/18


The entering variable is,

( z-row ) / ( s1-row ) = ( -1/3, 0, 0, -2/3, 0 ) / ( -5/3, 0, 1, -1/3, 0 )

= ( 1/5, -, -, 2, - )

Since the SMALLEST RATIO is 2/3, thus the entering variable is x 1.

By using step 1 and step 2, we get the new tableBasic x1 x2 s1 s2 s3 solution

z 0 0 -1/5 -3/5 0 21/5 z-row

s1 1 0 -3/5 -1/5 0 3/5 s1-row

x2 0 1 4/5 -3/5 0 6/5 s2-row

s3 0 0 -1/5 2/5 1 6/5 s3-row

The table is now optimum and feasible


11/18


A)DEGENERACY- More than 1 basic variable can be the non-basic variable at the next iteration


z -12 -4 0 0 0 z-row

s1 4 1 1 0 8 s1-row

s2 2 2 0 1 4 s2-row

B)ALTERNATIVE OPTIMA- More than 1 solution point which can assume the same optimal value.


z 0 0 2 0 10 z-rows1 1/2 1 0 5/2 s1-row

s2 0 -1/2 1 3/2 s2-row


z 0 0 2 0 10 z-row

s1 0 1 1 -1 1 s1-row

s2 1 0 -1 2 3 s2-row

Alternative optima occurs when the objective function is parallel with one of the constraints

C) UNBOUNDED SOLUTION- At least 1 basic variable with a value of infinity.


z -2 -4 0 0 0 z-row

s1 1 -1 1 0 10 s1-row

s2 1 0 0 1 40 s2-row

Both s1-row and s2-row can be choose at the next iteration

The table is optimum but x1 can enter the basis without changing the value of z

For MAXIMIZATIONPROBLEM, -4 is chosen

to leave because -4 is

more negative than -2.

But, the problem occurs here. Since 10/-1

and 40/0 cant give any value, so there are

no variable can leave the basis


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D)NON-EXISTING ( OR FEASIBLE ) SOLUTION- The region for the solution does not exist

Basic x1 x2 s1 s2 R2 solution

z 1+5M 0 M 2+4M 0 4-4M z-row

x2 2 1 0 1 0 2 s1-row

R2 -5 0 -1 -4 1 4 s2-row

SENSITIVITY ANALYSISDetermine whether change in parameters of the model within certain limits would cause the

optimum solution to change

Changes due to :a) Parameters value which have changedb) Conditions which have changedReason for changes :a) New priceb) New technologyc) New optiond) Inaccurate parameters estimatesAdvantages :

a) Determine the accuracy of the data inputb) Determine the control range for the parameters values so that obtained solutions are

always optimumc) Used in process planning

For MAXIMIZATION PROBLEM, the table is now optimum. BUT value of

M cant be exists at solution. THUS THE SOLUTION IS NOT FEASIBLE


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*objective of the model is to DETERMINE THE UNKNOWN xij that will MINIMIZE the total

transportation cost.

THEEE steps of NORTHWEST CORNER METHOD

i) Allocate as much as possible to the selected cell and adjust the associated amount ofsupply and demand by substracting the allocated amount.

ii) Cross out the row or column with zero supply or demand to indicate that no furtherassignment can be made to that row or column. If both a row and a column net to 0

simultaneously, cross out one only, and leave a zero supply (or dema

iii) nd) in the un-crossed row(or column)iv) If exactly one row or column is left uncrossed, stop. Otherwise, move to the cell to

the right if a column has just been crossed out or below if a row just was crossed out.

Go to step 1.

EXAMPLE TRANSPORTATIONAL PROBLEM

An electric company has 4 electric power plants which uses coal. The coals are supplied by 3coalmines. The cost of transport ting a unit of coal from the coalmines to the power plants are :

Power plants Supply

Coalmines

1 2 3 4

1 2 3 4 5 10

2 5 4 3 1 15

3 1 3 3 2 21

Demand 6 11 17 12

Using step 1 :Power plants Supply

Coalmines

1 2 3 4

1 6 2 4 3 4 5 10

2 5 7 4 8 3 1 15

3 1 3 9 3 12 2 21

Demand 6 11 17 12

Cost = (6x2) + (4x3) + (7x4) + (8x3) + (9x3) + (12x2) = 127 units

A. Determination of entering variable.x11 : u1 + v1 = c11 = 2x12 : u1 + v2 = c12 = 3

x22 : u2 + v2 = c22 = 4x23 : u2 + v3 = c23 = 3

x33 : u3 + v3 = c33 = 3

x34 : u3 + v4 = c34 = 2

The 1st

step is choose 6 and put in 1st

column 1st

row since 6 is the minimum value in the demand and supply.


14/18


let u1 = 0 then

v1 =2, v2 =3, v3 =2, v4 =1, u2 =1, u3 =1

B. Determination of optimality.Since the problem is MINIMIZATION, thus all the entering variable must be NON-NEGATIVE.

x13 : u1 + v3 - c11 = 0 + 2

4 = -2x14 : u1 + v4 - c14 = 0 + 1 5 = -4

x21 : u2 + v1c21 = 1 + 2 5 = -2

x24 : u2 + v4 - c24 = 1 + 1 1 = 1x31 : u3 + v1c31 = 1 + 2 1 = 2 since x31 have the larger positive number, the loop start at x31

x32 : u3 + v2c32 = 1 + 3 3 = 1

Power plants Supply

Coalmines

1 2 3 4

1 6 - 2 4 + 3 4 5 10

2 5 7 - 4 8 + 3 1 15

3 + 1 3 9 - 3 12 2 21Demand 6 11 17 12

C. Determination of leaving variable.a) The leaving variable is the basic variable with the SMALLEST negative sign. In this case

6 is SMALLER than 7 and 9.

b) Basic variable with +ve sign, add the value of leaving variable.c) Basic variable with -ve sign, minus the value of leaving variable.

Construct new table.

Power plants Supply

Coalmines

1 2 3 41 2 10 3 4 5 10

2 5 1 4 14 3 1 15

3 6 1 3 3 3 12 2 21

Demand 6 11 17 12

Cost = (6x1) + (10x3) + (1x4) + (14x3) + (3x3) + (12x2) = 115 unit

The step A, B and C is repeated.

A. Determination of entering variable.B. Determintaion of optimality

Power plants Supply

Coalmines

1 2 3 4

1 -2 2 10 3 -2 4 -4 5 10

2 -4 5 1 - 4 14 + 3 1 1 15

3 6 1 1 + 3 3 - 3 12 2 21

Demand 6 11 17 12

The entering variable is either x32 or x24. In this case we choose x32.


15/18


A. Determination of leaving variable.The leaving variable is x22 since it is SMALLER than x33.


Power plants Supply

Coalmines

1 2 3 41 2 10 3 4 5 10

2 5 4 15 3 1 15

3 6 1 1 3 2 3 12 2 21

Demand 6 11 17 12

Cost = (6x1) + (10x3) + (1x3) + (15x3) + (2x3) + (12x2) = 114 unit


C. Determination of entering variable.D. Determination of optimality

Power plants Supply

Coalmines

1 2 3 41 -1 2 10 3 -1 4 -3 5 10

2 -4 5 -1 4 15 - 3 1 + 1 15

3 6 1 1 3 2 + 3 12 - 2 21

Demand 6 11 17 12

The entering variable is x24

B. Determination of leaving variable.The leaving variable is x34 since it is SMALLER than x23.


Power plants Supply

Coalmines

1 2 3 41 2 10 3 4 5 10

2 5 4 3 3 12 1 15

3 6 1 1 3 14 3 2 21

Demand 6 11 17 12

Cost = (6x1) + (10x3) + (1x3) + (3x3) + (14x3) + (12x1) = 102unit


A. Determination of entering variable.B. Determination of optimality

Power plants Supply

Coalmines

1 2 3 4

1 -1 2 10 3 -1 4 -4 5 10

2 -4 5 -1 4 3 3 12 1 15

3 6 1 1 3 14 3 -1 2 21

Demand 6 11 17 12

The table is optimum.


16/18


Assign n workers to n jobs cij is the cost of assigning worker i to job j. MINIMIZING the cost.

The HUNGARIAN METHOD :

i) From original cost matrix, identify each row minimum, and substract it from allentries of the row.

ii) For the matrix resulting from (i), identify each column minimum and substract it fromall the entries of the column.

iii) Identify the optimal solution as the feasible assignment associated with the zeroelement of the matrix obtained in (ii)

iv) If no feasible solution (with 0 entries) :-a) Draw the minimum number of horizontal and vertical line in the last reducedmatrix that will cover all the 0 entries.b) Select the smallest uncovered element and then add it to every element at the

intersection of 2 lines.

c) Repeat this step until feasible solution is obtained.EXAMPLE OF ASSIGNMENT MODEL

1 2 3 Worker

1 5 7 9 1

2 14 10 12 1

3 15 13 16 1

Machine 1 1 1Find the optimal assignment.

1 2 3 Worker MIN

1 5 7 9 1 5

2 14 10 12 1 10

3 15 13 16 1 13

Machine 1 1 1

Subtract each row with minimum number of the row.

1 2 3 Worker

1 0 2 4 12 4 0 2 1

3 2 0 3 1

Machine 1 1 1

MIN 0 0 2

Substract each column with the minimum number of the column.


17/18


Optimum table

1 2 3 Worker

1 0 2 2 1

2 4 0 0 1

3 2 0 1 1

Machine 1 1 1Thus,Worker 1 can be assign to machine 1 with cost of 5 units.

Worker 2 can be assign to machine 3 with cost of 12 units.


The MINIMUM COST = 30 units

EXAMPLE OF ASSIGNMENT MODEL

1 2 3 4 Worker

1 1 4 6 3 1

2 9 7 10 9 13 4 5 11 7 1

4 8 7 8 5 1

Machine 1 1 1 1

Find the optimal assignment.

1 2 3 4 Worker MIN

1 1 4 6 3 1 1

2 9 7 10 9 1 7

3 4 5 11 7 1 4

4 8 7 8 5 1 5

Machine 1 1 1 1Subtract each row with minimum number of the row.

1 2 3 4 Worker

1 0 3 5 2 1

2 2 0 3 2 1

3 0 1 7 3 1

4 3 2 3 0 1

Machine 1 1 1 1

MIN 0 0 3 0

Substract each column with the minimum number of the column.


18/18


1 2 3 4 Worker

1 0 3 2 2 1

2 2 0 0 2 1

3 0 1 4 3 1

4 3 2 0 0 1

Machine 1 1 1 1

The table is not optimum yet. DRAW THE HORIZONTAL AND VERTICAL LINE THAT

COVER ALL THE 0 ENTRIES.

The SMALLEST UNCOVERED element is 1. Then substract each uncovered element with 1

and add the intersection of line with 1.

1 2 3 4 Worker

1 0 2 1 1 1

2 3 0 0 2 1

3 0 0 3 2 14 4 2 0 0 1

Machine 1 1 1 1

The table is now optimum.Thus,



Worker 3 can be assign to machine 2 with cost of 10 units.Worker 4 can be assign to machine 4 with cost of 5 units.

The MINIMUM COST = 21 units

or note farhan idzni

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