VAM and MODI Method in Solving Transportation Problems

VAM AND MODI METHODIn Solving Transportation Problems

BY: MANIEGO, Karlo & RIVERA, Ethel Bianca

TRANSPORTATION MODEL

Linear programming dealing with the issue of shipping commodities from multiple sources to multiple destinations.

Objective: To determine the shipping schedule that minimizes the total shipping cost while satisfying supply and demand constraints.

A comparative tool in business decision-making.

TRANSPORTATION MODEL Main applications:

Location decisions; Production planning; Capacity planning; and Transshipment

Major Assumptions: Items are homogeneous; Shipping cost per unit is the same; and Only one route is used from place of shipment

to the destination

TRANSPORTATION PROBLEM

The following must be known before using a method to find a low-cost solution:

Capacity Requirements of the sources (Supply) and of the destinations (Demand)

An estimation of the costs of transport per unit

Number of occupied cells/squares in the table, computed in formula of n + m – 1, where n is the number of sources (rows) and m is number of destinations (columns)

TRANSPORTATION TABLE

WAREHOUSE 1

WAREHOUSE 2

WAREHOUSE 3 SUPPLY

PLANT 1 2 4 6 300

PLANT 2 8 6 4 200

PLANT 3 2 4 6 100

DEMAND 100 200 300

DESTINATIONS

SOUR

CES

TRANSPORTATION COSTPER UNIT

CAPACITY OF SOURCES

REQUIREMENTS OF DESTINATIONS

OCCUPIED CELLS = n + m – 1 = 3 + 3 – 1 = 5 CELLS

VOGEL’S APPROXIMATION METHOD (VAM)

Solution obtained is either optimal or near to the optimal solution.

Considers the costs associated with each route alternative.

STEPS OF VAM: Find the penalties (opportunity cost), including

the dummies. Identify the row or column with the greatest

opportunity cost As many units as possible, assign it to the lowest

cost square in the row or column selected. Eliminate any row or column that has just been

completely satisfied by the assignment just made. Recompute the cost differences for the

transportation table, omitting rows or columns crossed out in the preceding step.

Return to step 2 and repeat the steps until an initial feasible solution has been obtained.

ILLUSTRATION ABC Company has 3 production

facilities S1, S2 and S3 with production capacity of 250, 300 and 400 units per week of a product, respectively. These units are to be shipped to 4 warehouses D1, D2, D3 and D4 with requirement of 200, 225, 275 and 250 units per week, respectively. The transportation costs (in peso) per unit between factories to warehouses are given in the table.

Total Supply = 250 + 300 + 400 = 950Total Demand = 200 + 225 + 275 + 250 = 950

(Total Supply = Total Demand)

Number of Occupied Cells = n + m – 1 = 3 + 4 – 1 = 6

Warehouses

D1 D2 D3 D4

Capacity

(Supply)

Production Facilities

S111 12 17 14 250

S216 18 14 10 300

S321 24 13 11

400

Requirement

(Demand)200 225 275 250 950

STEP 1: Find the penalties (opportunity cost), including the dummies, of each row and column.

STEP 2: Identify the row or column with the greatest opportunity cost

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S111 12 17 14 250 1

S2 16 18 14 10 300 4

S321 24 13 11 400 2

Requirement

(Demand)200 225 275 250 950

Penalties 5 6 1 1PENALTY/OPPORTUNITY COST is the difference between the two lowest unit shipping costs

STEP 3: Assign as many units as possible to the lowest cost square in the row or column selected.

STEP 4: Eliminate any row or column that has just been completely satisfied by the assignment just made.

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S111 17 14 250 1

S216 18 14 10 300 4

S321 24 13 11 400 2

Requirement

(Demand)200 225 275 250 950

Penalties 5 6 1 1

225

12

XX

STEP 5: Recompute the cost differences for the transportation table, omitting rows or columns crossed out in the preceding step.

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S111 17 14 250 2

S216 18 14 10 300 4

S321 24 13 11 400 2

Requirement

(Demand)200 225 275 250 950

Penalties 5 1 1

225

12

XX

STEP 6: Repeat the steps 2-5 until an initial feasible solution has been obtained.

SECOND VAM ASSIGNMENT:

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S1 17 14 250 2

S216 18 14 10 300 4

S321 24 13 11 400 2

Requirement

(Demand)200 225 275 250 950

Penalties 5 1 1

225

12

XX

25 XX11

THIRD VAM ASSIGNMENT

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S1 17 14

250 S2

18 14 10 300 4

S3 21 24 13 11

400 2 Require

ment (Demand)

200 225 275 250 950

Penalties 5 1 1

225

12

X

X

25 X X 1

1

175

16

X

FOURTH VAM ASSIGNMENT

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S1 17 14

250 S2

18 14 300 4

S3 21 24 13 11

400 2 Require

ment (Demand)

200 225 275 250 950

Penalties 1 1

225

12

X

X

25 X X 1

1

175

16

X

10125

X

FINAL VAM ASSIGNMENT

D1 D2 D3 D4

Capacity

(Supply)

Penalties

S1 17 14

250 S2

18 14 300

S3 21 24 13 11

400 2 Require

ment (Demand)

200 225 275 250 950

Penalties 1 1

225

12

X

X

25 X X 1

1

175

16

X

10125

X275

125

INITIAL FEASIBLE VAM SOLUTION

D1 D2 D3 D4

Capacity

(Supply)

S1 17 14

250 S2

18 14 300

S3 21 24 13 11

400 Require

ment (Demand)

200 225 275 250 950

225

12 25

11

175

16 1012527

5125

Total Transportation Cost:

25 units x 11 = 275225 units x 12 = 2,700175 units x 16 = 2,800125 units x 10 = 1,250275 units x 13 = 3,575125 units x 11 = 1,375

Total 11,975

MODIFIED DISTRIBUTION (MODI) METHOD

Computing improvement indices quickly for each unused square without drawing all of the closed paths.

The optimality test of the initial feasible solution

STEPS OF MODI METHOD: Obtain an initial solution using any rule or

method. Set the equations of Ri + Kj = Cij for those

squares that are currently used or occupied. After all equations have been written, set R1 = 0 Solve the system of equations for all R and K

values Compute the improvement index for each unused

square Select the largest negative index and proceed to

solve the problem as done using the stepping-stone method. If there are no negative indices computed, the Initial solution is the feasible solution.

ILLUSTRATION: Using the initial VAM solution obtained in

the previous example: Kj K1 K2 K3 K4

Ri D1 D2 D3 D4 Supply

R1 S125

11

225

12 17 14250

R2 S2175

16 18 14

125

10300

R3 S3

21 24

275

13

125

11400

Demand 200 225 275 250 950

Let:

Ri = value assigned to row i

Kj = value assigned to column j

Cij = cost in square ij (cost of shipping from source i to destination j)

STEP 1: Set the equations of Ri + Kj = Cij for those squares that are currently used or occupied.

R1 + K1 = 11 R1 + K2 = 12 R2 + K1 = 16 R2 + K4 = 10 R3 + K3 = 13 R3 + K4 = 11

Kj K1 K2 K3 K4


R1 S125

11

225

12 17 14250

R2 S2175

16 18 14

125

10300

R3 S3

21 24

275

13

125

11400

Demand 200 225 275 250 950

STEP 2: After all equations have been written, set R1 = 0STEP 3: Solve the system of equations for all R and K values

1. R1 + K1 = 110 + K1 = 11K1 =

11

2. R1 + K2 = 120 + K2 = 12K2 =

12 3. R2 + K1 = 16

R2 + 11 = 16R2 = 5

4. R2 + K4 = 10 5 + K4 = 10

K4 = 5

5. R3 + K4 = 11R3 + 5 = 11R3 = 6

6. R3 + K3 = 13

6 + K3 = 13K3 = 7

IMPROVED TABLE

Kj K1 = 11 K2 = 12 K3 = 7 K4 = 5


R1 = 0 S1 25

11

225

12 17 14250

R2 = 5 S2 175

16 18 14

125

10300

R3 = 6 S3

21 24

275

13

125

11400

Demand 200 225 275 250 950

STEP 4: Compute the improvement index for each unused square

STEP 5: Select the largest negative index and proceed to solve the problem as done using the stepping-stone method.

Improvement index formula (Iij) = Cij - Ri – Kj

S1-D3 index = 17 – R1 – K3 = 17 – 0 – 7 = 10 S1-D4 index = 14 – R1 – K4 = 14 – 0 – 5 = 9 S2-D2 index = 18 – R2 – K2 = 18 – 5 – 12 = 1 S2-D3 index = 14 – R2 – K3 = 14 – 5 – 7 = 2 S3-D1 index = 21 – R3 – K1 = 21 – 6 – 11 = 4 S3-D2 index = 24 – R3 – K2 = 24 – 6 – 12 = 6

Since there is no negative index computed, it means that using the MODI method, the VAM solution is the optimal solution of ABC Company.

ANOTHER ILLUSTRATION A certain company manufactures a

certain product in its three plants, namely Plant A, B, and C. The plants’ capacities are 700, 1000, and 1800 respectively. The company needs to meet the demand of its four distribution centers W, X, Y, and Z with 500, 800, 700, 1500 units respectively. The transportation cost per unit is provided on the table. An initial feasible solution is obtained by Matrix Minimum method as follows:

Plants

Distribution Centers

Kj K1 K2 K3 K4

Ri W X Y Z Supply

R1 A19 30 50

70012

700

R2 B300

70 60

700

40 70

1000

R3 C200

40

800

10 60

800

20

1800

Demand 500 800 700 1500 3500

Total Supply = 700 + 1000 + 1800 = 3500Total Demand = 500 + 800 + 700 + 1500 = 3500

(Total Supply = Total Demand)


SOLUTIONS: R1 + K4 =

12 R2 + K1 =

70 R2 + K3 =

40 R3 + K1 =

40 R3 + K2 =

10 R3 + K4 =

20

1. R1 + K4 = 12

0 + K4 = 12

K4 = 12

2. R3 + K4 = 20

R3 + 12 = 20

R3 = 8

3. R3 + K1 =

408 + K1 =

40K1 = 32

4. R3 + K2 = 10

8 + K2 = 10

K2 = 2

5. R2 + K1 = 70

R2 + 32 = 70

R2 = 38

6. R2 + K3 = 40

38 + K3 = 40

K3 = 2

IMPROVED TABLE

Plants

Distribution CentersKj K1 = 32 K2 = 2 K3 = 2 K4 = 12

Ri W X Y Z Supply

R1 = 0 A19 30 50

70012

700

R2 = 38 B 30070 60

70040 70

1000

R3 = 8 C200

40

80010 60

80020

1800

Demand 500 800 700 1500 3500

STEP 5: Select the largest negative index and proceed to solve the problem as done using the stepping-stone method.

The Improvement Indices (Iij) are computed as follows:

A-W index = 19 – R1 – K1 = 19 – 0 – 32 = -13A-X index = 30 – R1 – K2 = 30 – 0 – 2 = 28A-Y index = 50 – R1 – K3 = 50 – 0 – 2 = 48B-X index = 30 – R2 – K2 = 60 – 38 – 2 = 20B-Z index = 60 – R2 – K4 = 60 – 38 – 12 = 10C-Y index = 60 – R3 – K3 = 60 – 8 – 2 = 50

STEPPING STONE METHOD Beginning at the square with the best improvement index,

trace a closed path back to the original square via squares that are currently being used or occupied.

Beginning with a plus (+) sign at the unused square, place alternate minus (-) signs and plus signs on each corner square of the closed path just traced.

Select the smallest quantity found in those squares containing minus signs. Add that number to all squares on the closed path with plus signs; subtract the number from all squares assigned minus signs.

Compute new improvement indices for this new solution using the MODI method.

STEP 1: Trace a closed path back to the original square STEP 2: Place alternate plus signs and minus signs on the squares of the traced path

Plants


W X Y Z Supply

A19 30 50

70012

700

B 30070 60

70040 70

1000

C 20040

80010 60

80020

1800

Demand 500 800 700 1500 3500

-

+

+

-

STEP 3: Select the smallest quantity found in those squares containing minus signs.

Add that number to all squares on the closed path with plus signs; subtract the number from all squares assigned minus signs.

Plants


W X Y Z Supply

A19 30 50

70012

700

B 30070 60

70040 70

1000

C 20040

80010 60

80020

1800

Demand 500 800 700 1500 3500

-

+

+

-

200

200

200

200

NEW MODI SOLUTION

Plants

Distribution CentersKj K1 K2 K3 K4

Ri W X Y Z Supply

R1 A 20019 30 50

50012

700

R2 B 30070 60

70040 70

1000

R3 C40

80010 60

100020

1800

Demand 500 800 700 1500 3500

SOLUTIONS: R1 + K1 =

19 R1 + K4 =

12 R2 + K1 =

70 R2 + K3 =

40 R3 + K2 =

10 R3 + K4 =

20

1. R1 + K4 = 12

0 + K4 = 12

K4 = 12

2. R1 + K1 = 19

0 + K1 = 19

K1 = 19

3. R2 + K1 = 70

R2 + 19 = 70

R2 = 51

4. R2 + K3 = 40

51 + K3 = 40

K3 = -11

5. R3 + K4 = 20

R3 + 12 = 20

R3 = 8

6. R3 + K2 = 10

8 + K2 = 10

K2 = 2

IMPROVED NEW MODI SOLUTION

Plants

Distribution CentersKj K1 = 19 K2 = 2 K3 = -11 K4 = 12

Ri W X Y Z Supply

R1 = 0 A 20019 30 50

50012

700

R2 = 51 B 30070 60

70040 70

1000

R3 = 8 C40

80010 60

100020

1800

Demand 500 800 700 1500 3500


A-X index = 30 – R1 – K2 = 30 – 0 – 2 = 28 A-Y index = 50 – R1 – K3 = 50 – 0 – (-11) = 61 B-X index = 30 – R2 – K2 = 60 – 51 – 2 = 7 B-Z index = 70 – R2 – K4 = 70 – 51 – 12 = 7 C-W index = 40 – R3 – K1 = 40 – 8 – 19 = 13 C-Y index = 60 – R3 – K3 = 60 – 8 – (-11) = 63

Since there are no more negative indices computed, the optimal solution of the company is obtained.

TRANSPORTATION COST 200 units x 19 = 3,800 500 units x 12 = 6,000 300 units x 70 = 21,000 700 units x 40 = 28,000 800 units x 10 = 8,000 1000 units x 20 = 20,000

Total 86,800

UNBALANCED TRANSPORTATION PROBLEM

Total Supply ≠ Total Demand

To solve for this, a dummy supply or demand is added to the transportation table for the excess to make it balanced; with the transportation costs of the dummy cells valued zero (0).

SUPPLY > DEMAND = DUMMY DEMAND DEMAND > SUPPLY = DUMMY SUPPLY

ILLUSTRATION

Warehouse 1 Warehouse 2 Warehouse 3 Supply

Retailer A 10 25 10 3,000

Retailer B 15 8 8 2,000

Demand 2,000 2,000 1,500

Total Supply = 3000 + 2000 = 5000Total Demand = 2000+ 2000+ 1000 = 5500

(Total Demand > Total Supply = 500 UNITS)

BALANCED TABLEWarehouse 1 Warehouse 2 Warehouse 3 Supply

Retailer A 10 25 10 3,000

Retailer B 15 8 8 2,000

Dummy Supply 0 0 0 500

Demand 2,000 2,000 1,500 5,500


OBTAINING INITIAL VAM SOLUTION

FIRST VAM ASSIGNMENT

Warehouse 1

Warehouse 2

Warehouse 3 Supply Penalties

Retailer A10 25 10

3000 0

Retailer B15 8 8

2000 0

Dummy Supply

0 0500 0

Demand 2000 2000 1500 5500

Penalties 10 8 6

0500

X X


SECOND VAM ASSIGNMENT

Warehouse 1

Warehouse 2


Retailer A10 25 10

3000 0

Retailer B15 8 8

2000 0

Dummy Supply

0 0500

Demand 2000 2000 1500 5500

Penalties 5 17 8

0500

X X2000

X


FINAL VAM ASSIGNMENT

Warehouse 1

Warehouse 2


Retailer A10 25 10

3000 0

Retailer B15 8 8

2000 0

Dummy Supply

0 0500

Demand 2000 2000 1500 5500

Penalties 5 8

0500

X X2000

X 1500

1500 XX

INITIAL FEASIBLE VAM SOLUTION

Warehouse 1

Warehouse 2


Retailer A10 25 10

3000 0

Retailer B15 8 8

2000 0

Dummy Supply

0 0500

Demand 2000 2000 1500 5500

Penalties 5 8

0500

02000

1500

1500

SOLUTIONS USING MODI METHOD:

R1 + K1 = 10

R1 + K3 = 10

R2 + K2 = 8

R3 + K1 = 0

R3 + K2 = 0

1. R1 + K1 = 10

0 + K1 = 10

K1 = 10

2. R1 + K3 = 10

0 + K3 = 10

K3 = 10 3. R3 + K1 =

0R3 + 10 =

0 R3

= -10

4. R3 + K2 = 0

-10 + K2 = 0 K2 = 10

5. R2 + K2 = 8

R2 + 10 = 8 R2 = -2

Since there is no negative index computed, it means that using the MODI method, the VAM solution is the optimal solution of the company.


RA-W2 index = 25 – R1 – K2 = 25 – 0 – 10 = 15 RB-W1 index = 15 – R2 – K1 = 15 – (-2) – 10 = 7 RB-W3 index = 8 – R2 – K3 = 8 – (-2) – 10 = 0 RD-W3 index = 0 – R3 – K3 = 0 – (-10) – 10 = 0

TRANSPORTATION COST 1500 units x 10 = 15,000 1500 units x 10 = 15,000 2000 units x 8 = 16,000 500 units x 0 = 0

Total 46,000

Business

VAM and MODI Method in Solving Transportation Problems