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8/2/2019 Or Note Farhan Idzni
1/18
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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8/2/2019 Or Note Farhan Idzni
2/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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 }
Thus, the leaving variable is s2.
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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8/2/2019 Or Note Farhan Idzni
3/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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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8/2/2019 Or Note Farhan Idzni
4/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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
Basic x1 x2 s2 s3 R1 R2 solution
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
Since the entering variable is x1.
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
Basic x1 x2 s2 s3 R1 R2 solution
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
Since the entering variable is x2.
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
coefficient in the z-row is
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
coefficient in the z-row is
NON-POSITIVE
For MINIMIZATION problem, since
1+5M/3 is more positive than 4-7M/3
and -M, the entering variable is x2
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8/2/2019 Or Note Farhan Idzni
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Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
By using step 1 and step 2, we get the new table
Basic x1 x2 s2 s3 R1 R2 solution
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.
By using step 1 and step 2, we get the new table
Basic x1 x2 s2 s3 R1 R2 solution
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
coefficient in the z-row is
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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8/2/2019 Or Note Farhan Idzni
6/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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
arrange the equation in the table
Basic x1 x2 s2 s3 R1 R2 solution
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
Since the entering variable is x1.
Leaving variable = min { 3/3, 6/4, 4/1 }= min { 1, 3/2, 4 }
Thus, the leaving variable is R1.
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
For MINIMIZATION problem, since
7is more positive than 4 and -1,
the entering variable is x1
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8/2/2019 Or Note Farhan Idzni
7/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
By using step 1 and step 2, we get the new table
Basic x1 x2 s2 s3 R1 R2 solution
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
Since the entering variable is x2.
Leaving variable = min { 1/(1/3), 2/(5/3), 3/(5/3) }
= min { 3, 1.2, 1.8 }
Thus, the leaving variable is R2.
By using step 1 and step 2, we get the new table
Basic x1 x2 s2 s3 R1 R2 solution
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
arrange the equation in the table
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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8/2/2019 Or Note Farhan Idzni
8/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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 }
Thus, the leaving variable is s3.
By using step 1 and step 2, we get the new table
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
coefficient in the z-row is
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
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8/2/2019 Or Note Farhan Idzni
9/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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
arrange the equation in the table
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.
By using step 1 and step 2, we get the new table
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
For MINIMIZATION problem, the
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)
For MINIMIZATION problem, the
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)
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8/2/2019 Or Note Farhan Idzni
10/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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
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8/2/2019 Or Note Farhan Idzni
11/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
A)DEGENERACY- More than 1 basic variable can be the non-basic variable at the next iteration
Basic x1 x2 s1 s2 solution
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.
Basic x1 x2 s1 s2 solution
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
Basic x1 x2 s1 s2 solution
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.
Basic x1 x2 s1 s2 solution
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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8/2/2019 Or Note Farhan Idzni
12/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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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8/2/2019 Or Note Farhan Idzni
13/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
*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.
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8/2/2019 Or Note Farhan Idzni
14/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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.
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8/2/2019 Or Note Farhan Idzni
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Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
A. Determination of leaving variable.The leaving variable is x22 since it is SMALLER than x33.
Construct new table.
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
The step A, B and C is repeated.
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.
Construct new table.
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
The step A, B and C is repeated.
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.
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8/2/2019 Or Note Farhan Idzni
16/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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.
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8/2/2019 Or Note Farhan Idzni
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Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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.
Worker 3 can be assign to machine 2 with cost of 13 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.
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8/2/2019 Or Note Farhan Idzni
18/18
Farhan Idzni bin Ahmad Shuhami 100853 Mechanical Engineering
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 1 can be assign to machine 1 with cost of 1 units.
Worker 2 can be assign to machine 3 with cost of 5 units.
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
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