Transportation Model(1)

8/12/2019 Transportation Model(1)

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TRANSPORTATION MODEL

DR. REKHA PRASAD


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MATHEMATICAL FORMULATION

Minimize (Total Cost) = cijxij

subject to the constraintxij=ai, i = 1,2,3,........,m(supply constraint)

xij=bj,j= 1,2,3,........,n(demand constraint)

xij0 for all i and j


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CONT...

Andheri Bandra Chinchwad

Dumdum 2 7 4

Ellora 3 3 1

Feroza 5 4 7

Guna 1 6 2


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SOLUTION 1

From the information given we can draw the following graph.

There are three different methods of finding the basic feasible solution:a) North West Corner Method

b) Least Cosy Method

c) VogelsApproximation method

A B C Supply

D 2 7 4 50

E 3 3 1 80

F 5 4 7 70

G 1 6 2 140

Demand 70 90 180 340


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CONT...

a) North West Corner Method

Total Cost = 2x50 + 3x20 + 3x60 + 4x30 + 7x40+ 2x140 = 1020

A B C Supply

D 2 7 4 50

E 3 3 1 80F 5 4 7 70

G 1 6 2 140

Demand 70 90 180 340

50

20 60

30 40

140


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CONT...

b) Least Cost Method

Total Cost = 7x20 + 4x30 + 1x80 + 4x70 +1x70 + 2x70 = 830

A B C Supply

D 2 7 4 50

E 3 3 1 80

F 5 4 7 70

G 1 6 2 140

Demand 70 90 180 340

70 30

80

70

7070


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CONT...

c) Vogels Approximation Method

A B C SUPPLY PENALTY

D 2 30 7 20 4 50 2 2 2 2 2

E 3 3 1 80 80 2 - - - -F 5 4 70 7 70 1 1 1 -

G 1 40 6 2 100 140 1 1 5 - -

DEMAND 70 90 180 340

PENALTY 1

1

3

3

1

2

3

3

2

2

-

-


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CONT...

TC = 2X30 + 7X20 + 1X80 + 4X70 + 1X70 + 2X100

= 800

v1= 2 v2= 7 v3 = 3

u1= 0

u2= -2

u3 = -3

u4= -1

A B C SUPPLY

D 2 30+ 7 20- 4 50

E 3 3 + 1 80- 80

F 5 4 70 7 70

G 1 40- 6 2 100+ 140

DEMAND 70 90 180 340


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CONT...

Occupied Cells:

v1+ u1= 2 Let u1= 0 Therefore v1= 2

v2+ u1= 7 v2= 7v3+ u2= 1 3 + u2= 1therefore u2= -2

v2+ u3= 4 7 + u3= 4 Therefore u3= -3

v1+ u4= 1 2 + u4= 1 Therefore u4= -1v3+ u4= 2 v31 = 2 therefore v3= 3


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UNOCCUPIED CELLS

2 + 2 -4 = -2

2 + 2 -3 = -3

72 - 3 = 2*2 -3 - 5 = 0

33 -7 = -7

7 -1 - 6 = 0


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CONT...

30 +

20 -

80 - 40 -

100 -

The value of which satisfies all thesituations will be 20 and the nextiteration of the table will be


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CONT...

The redistributed situation will be:

Total Cost = 760

A B C SUPPLY

D 2 50 7 4 50

E 3 3 20 1 60 80F 5 4 70 7 70

G 1 20 6 2 120 140

DEMAND 70 90 180 340


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CONT...

Occupied Cells:

u1+ v1= 2 Let v1 = 0 u1= 2

u2+ v2= 3 0 + v2= 3 v2= 3u2+ v3= 1 u2+ 1 = 1 u2= 0

u3+ v2= 4 u3+ 3 = 4 u3= 1

u4+ v1= 1 u4= 1u4+ v3= 2 1 + v3= 2 v3= 1


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UNOCCUPIED CELLS

2 + 3 - 7 = -2

2 + 1 - 4 = -1

0 + 03 = -31 + 05 = -4

1 + 1 -7 = -5

1 + 3 -6 = -2


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STEPS FOR MODIFIED DISTRIBUTION

METHOD (MODI)Step 1:Set up a feasible solution by North West corner rule, Least Cost Method or VAM.

Step 2: Assign a value of zero to any ui or vj. Calculate the remaining ui and vj values from therelationship that for all occupied cells

Cij= ui+ vj

Step 3:Calculate the opportunity costs of all unoccupied cells from the relationship: Opportunity cost= ui+ vjCij

Step 4: If the opportunity cost of all unoccupied cells is zero or negative, an optimal solution has

been reached.Step 5: In case cells have positive opportunity costs, select the cell with the maximum positive

opportunity cost. Starting from the selected cell and moving only along horizontal or verticallines, trace a closed path back to this cell, such that all the corners of the closed path areoccupied cells. Beginning with a positive sign in this cell, assign positive and negative signsalternately to the corners of the closed path.

Step 6:Determine the smallest quantity in a negative position on the closed path. Add this quantityto all corner cells with a positive sign on the closed path, and subtract it from all cells with anegative sign on the closed path. This gives us an improved solution. Go to step 2.


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THE UNBALANCED CASE

If the supply and demand or availability andrequirements are unequal, we make thesupply and demand equal by the introduction

of either a dummy destination, if the supply islarger or a dummy source if the demand islarger. The difference is allocated to thisdummy. The cost of moving units from the

dummy to any source or from sources to adummy to a dummy location is zero as nomovement actually takes place


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CONT...

Consider the following transportation

problem:

A B C SUPPLY

1 2 7 4 50

2 3 3 1 80

3 5 4 7 70

4 1 6 2 140

DEMAND 90 90 180


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CONT...

A dummy origin is added to the problem having zero

transportation cost in those routes. The difference in supply

and demand is produced by the dummy factory and then the

problem is solved in the usual way.

A B C SUPPLY

1 2 7 4 50

2 3 3 1 80

3 5 4 7 70

4 1 6 2 140

DUMMY 0 0 0 20

DEMAND 90 90 180 360


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CONT...

The solution using VAM is given by the following

table:

TC = 2x50 + 1x80 + 4x70 + 1x40 + 2x100 + 0x20

= 700A B C SUPPLY

1 2 50 7 4 50

2 3 3 1 80 80

3 5 4 70 7 70

4 1 40 6 2 100 140

DUMMY 0 0 20 0 20

DEMAND 90 90 180 360


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CHECKING THE OPTIMALITY

According to MODI this is the optimal

solution.

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Transportation Model(1)