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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
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
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
Ri D1 D2 D3 D4 Supply
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)
Number of Occupied Cells = n + m – 1 = 3 + 4 – 1 = 6
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
Distribution Centers
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
Distribution Centers
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
The Improvement Indices (Iij) are computed as follows:
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
Number of Occupied Cells = n + m – 1 = 3 + 3 – 1 = 5
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
OBTAINING INITIAL VAM SOLUTION
SECOND 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
Demand 2000 2000 1500 5500
Penalties 5 17 8
0500
X X2000
X
OBTAINING INITIAL VAM SOLUTION
FINAL 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
Demand 2000 2000 1500 5500
Penalties 5 8
0500
X X2000
X 1500
1500 XX
INITIAL FEASIBLE VAM SOLUTION
Warehouse 1
Warehouse 2
Warehouse 3 Supply Penalties
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.
The Improvement Indices (Iij) are computed as follows:
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