1. Transportation
Modeling
Problem
10/13/13
1

2. THE PROBLEM
The Epsilon Computers Co. sells desktop
computers to universities along University
belt, and ship them from three distribution
warehouses. The firm is able to supply the
following numbers of desktop computers to
the universities by the beginning of the
academic year:
10/13/13
2

3. Distribution
warehouse
Supply
Sta. Mesa
Taft Ave
Divisoria
150
200
50
total
400
Universities have ordered desktop computers that must be
delivered and installed by the beginning of the academic year:
University/College
Demand
CEU
FEU
UE
100
80
220
total
400
10/13/13
3

4. The shipping cost per desktop computer from each distributor to
each university are as follows:
To
From
1(Sta. Mesa)
2 (Taft Ave)
3 (Divisoria)
A
B
(CEU) (FEU)
7
10
6
5
12
3
C
(UE)
9
10
4
With cost minimization as criterion, Epsilon Company wants to determine
how many desktop computers should be shipped from each warehouse to
each university. Compare alternatives using,
a.Northwest Corner Rule (NCR)
b. Least Cost Method (LCM)
c. Stepping Stone Method
d. Vogel’s Approximation Method (VAM)
10/13/13
4

5. From
To
1
(STA. MESA)
2
(TAFT AVE)
3
(DIVISORIA)
Demand
10/13/13
A (CEU)
B (FEU)
X1a
7
X1b
5
100
9150
12
X2b
10
X3b
80
200
14
50
X2c
3
6
X3a
Supply
X1c
10
X2a
C (UE)
X3c
220
400
5

6. Objective Function:
C= cost of shipment of all
Xij= no. of computerd delivered
i= origin ; j= destination
Minimize: C= 7X1a + 5X1b + 9X1c + 10X2a + 12X2b + 10X2c + 6X3a + 3X3b + 14 X3c
Constraints:
10/13/13
X1a +X1b + X1c= 150
X2a + X2b + X2c = 200
X3a + X3b + X3c = 50
X1a + X2a + X3a= 100
X1b + X2b + X3b= 80
X1c+ X2c + x3c= 220
Xij ≥ 0
6

7. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
Supply
5
9150
12
10
50
10
2
30
(TAFT AVE)
100
80
200
170
3
6
3
(DIVISORIA)
Demand
C (UE)
50
220
14
50
400
C= 700 + 250 + 360 + 1700 + 700 = P 3710
10/13/13
7

8. From
To
A (CEU)
B (FEU)
100
7 30
5 20
2
(TAFT AVE)
10
12
3
(DIVISORIA)
6
1
(STA. MESA)
Demand
100
LEGEND:
10/13/13
12345
50
80
C (UE)
Supply
9150
200
10
14
3
220
200
50
400
C = P 3190
8

9. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
5
9150
12
10
10
30
(TAFT AVE)
100
80
200
170
3
6
3
(DIVISORIA)
10/13/13
Supply
50
2
Demand
C (UE)
50
220
Table derived through Northwest Corner
Rule
14
50
400
9

10. Compute for Improvement Indices:
Improvement Index- the increase/decrease in total cost that
would result from reallocating one unit to an unused square.
Unused
Square
Closed Path
Improvement
Indices
X1c
+X1c-X1b+X2b-X2c
+9-5+12-10= +6
X2a
+X2a-X1a+X1b-X2b
+10-7+5-12= -4
X3a
+X3a-X1a+X1b-X2b+X2c-X3c
+6-7+5-12+10-14= -12
X3b
+X3b-X2b+X2c-X3c
+3-12+10-14= -13
10/13/13
10

11. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
5
9150
12
10
10
30
(TAFT AVE)
100
80
200
170
3
6
3
(DIVISORIA)
10/13/13
Supply
50
2
Demand
C (UE)
50
220
14
50
400
11

12. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
C (UE)
5
9150
12
10
200
14
50
50
10
2
(TAFT AVE)
6
3
(DIVISORIA)
Demand
Supply
100
30
80
3
200
20
220
400
Improved Transportation Tableau @ C= P 3,320
10/13/13
12

13. NOTE: Test the solution for improvement by computing again the
improvement indices of the improved transportation tableau. If all indices
are positive then an optimal solution has been obtained, otherwise, repeat
the same steps for SSM.
Unused
Square
Closed Path
Improvement
Indices
X1c
+X1c-X1b+X3b-X3c
+9-5+3-14= -7
X2a
+X2a-X1a+X1b-X3b+X3c-X2b
+10-7+5-3+14-10= +9
X2b
+X2b-X2c+X3c-X3b
+12-10+14-3= +13
X3a
+X3a-X1a+X1b-X3b
+6-7+5-3= +1
10/13/13
13

14. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
C (UE)
5
9150
12
10
200
14
50
50
10
2
(TAFT AVE)
6
3
(DIVISORIA)
Demand
Supply
100
30
80
3
200
20
220
400
Previously improved transportation tableau
10/13/13
14

15. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
100
C (UE)
5
30
12
10
200
14
50
200
(TAFT AVE)
6
3
(DIVISORIA)
Demand
9150
20
10
2
Supply
100
50
80
3
220
400
New improved Transportation Tableau @ C= P 3180
10/13/13
15

16. Again, computing for improvement indices:
Unused
Square
X2a
+X2a-X1a+X1c-X2c
Improvement
Indices
+10-7+9-10= +2
X2b
+X2b-X1b+X1c-X2c
+12-5+9-10= +6
X3a
+X3a-X1a+X1b-X3b
+6-7+5-3= +1
X3c
+X3c-X3b+X1b-X1c
+14-3+5-9= +7
Closed Path
Since all improvement indices are all positive, optimal solution
is obtained @ C= P 3,180.
10/13/13
16

17. This method is an algorithm that finds an initial feasible
solution to transportation problem by considering the
“penalty cost” of not choosing cheapest available route.
Opportunity Cost- the cost of opportunities that are
sacrificed in order to take a specific action.
10/13/13
17

18. 1. For each row with an available supply and each column with an unfilled
demand, calculate an opportunity/penalty cost by subtracting the smallest
cost per unit entry from the second smallest entry for a minimization
problem. While, for maximization problem, opportunity cost is calculated
by getting the difference between the highest and the second highest
entry.
2. Identify the row or column with largest opportunity(difference) cost.
3. Allocate maximum amount possible to the available route with the lowest
cost for minimization or highest revenue for maximization in the
row/column selected from step 2.
4.Reduce appropriate supply and demand by the amount allocated in step 3.
5. Remove any rows with zero available supply and columns with unfilled
demand for further consideration.
6.Return to step 1.
10/13/13
18

19. From
To
A (CEU)
7
1
(STA. MESA)
X1a
2
(TAFT AVE)
3
(DIVISORIA)
Demand
B (FEU)
C (UE)
5
X1b
12
X2b
100
10
X3b
80
200
14
50
X2c
3
6
X3a
9 150
X1c
10
X2a
Supply
X3c
220
400
Original Transportation Tableau
10/13/13
19

20. OPPORTUNITY COST FOR THE FIRST ALLOCATION
Row/ Column
Lowest Cost
Opportunity
Cost
1
7
5
2
2
10
10
0
3
6
3
3
A
7
6
1
B
5
3
2
C
10/13/13
Second
Lowest Cost
10
9
1
20

21. From
To
A (CEU)
B (FEU)
C (UE)
Supply
7
5
9150
2
10
12
10
200
(TAFT AVE)
3
(DIVISORIA)
6
14
50
1
(STA. MESA)
Demand
100
50 3
80
220
400
First allocation
10/13/13
21

22. OPPORTUNITY COST FOR THE SECOND ALLOCATION
Row/ Column
Lowest Cost
Opportunity
Cost
1
7
5
2
2
10
10
0
A
10
7
1
B
12
5
7
C
10/13/13
Second
Lowest Cost
10
9
1
22

23. From
To
A (CEU)
B (FEU)
7
1
(STA. MESA)
Supply
5
9150
12
10
200
14
50
30
2
10
(TAFT AVE)
3
(DIVISORIA)
6
Demand
C (UE)
100
50 3
80
220
400
Second allocation
10/13/13
23

24. OPPORTUNITY COST FOR THE THIRD ALLOCATION
Row/ Column
Lowest Cost
Opportunity
Cost
1
9
7
2
2
10
10
0
A
10
7
1
C
10/13/13
Second
Lowest Cost
10
9
1
24

25. From
To
1
(STA. MESA)
A (CEU)
B (FEU)
7
100
30
(TAFT AVE)
3
(DIVISORIA)
6
12
80
220
10
200
14
200
50 3
Third allocation
10/13/13
9150
20
10
100
Supply
5
2
Demand
C (UE)
50
400
C= P 3180
25

26. • DEGENERATE PROBLEMS
>> (column + row) – 1 , SSM not applicable
>> put zero (0) to any unused square and
proceed with SSM steps
• UNBALANCED PROBLEMS
>> demand ≠ supply
>> add dummy column/row
10/13/13
26

27. Otsukaresamadeshita!!!
Thank You for listening!!! 
10/13/13
27