CHAPTER 3
TRANSPORTATION PROBLEM
Transportation problems
are linear programming problems which involve transporting or shipping goods or products from several sources (plants) to several destinations (warehouses) at minimum costs.
- It was first presented by F. L. Hitchcock in 1941 and later expanded by T. C. Koopmans.
Balanced Transportation Problems
A
balanced transportation problem
has equal number of units of demand and supply. That is, .
Steps in solving balanced transportation problem:
- Construct a transportation model of the problem.
- The transportation model is a table showing the sources (written in rows) while the destinations (written in columns) of the items to be transported. The last row of the table shows the demand and the last column shows the supply. The cost of transportation of each source to each receiving outlet is written on the upper right corner of each cell formed.
- Obtain the initial feasible solution by using
Northwest Corner Rule
- Start allocating the units on the upper – left – hand cell (northwest corner) of the tableau. Allocate the lower of the supply and demand corresponding to that cell.
- From the northwest corner, move to the next row if the supply available in that row is not yet exhausted or move to the next column if the demand available in that column is not yet met.
- The allocation of units will end at the lower – right – hand cell (south – east corner) of the tableau.
Minimum Entry Method
- Choose the cell with the lowest available cost and allocate the shipment so as to exhaust either the production of plant or meet the requirements of warehouse or both.
- Delete the ith row or the jth column, depending on whether the allocation exhausts the supply at source or meets the requirement of warehouse . If both of these occur simultaneously (degenerate case), delete the ith row unless it is the only row remaining in which case delete the jth column.
- Adjust the supply remaining at plant or the unfilled demand of warehouse .
- If there remains two or more rows or columns not yet deleted, go to step (i); otherwise, the initial basic feasible solution is obtained.
Vogel’s Approximation Method
- Compute the difference between the costs of two cheapest routes for each origin and each destination. Each individual difference is interpreted as a penalty for not choosing the cheapest route and is marked opposite each row and column.
- Identify the row or column with the largest difference. Allocate the shipment to the lowest cost cell in that row or column so as to exhaust either the supply at a particular source or meet the demand at a warehouse.
- Drop the ith row or the jth column, depending on whether the allocation exhausts the supply at plant or meet the requirements of warehouse . If both of these occur simultaneously (degenerate case), delete the ith row unless it is the only row remaining, in which case drop the jth column.
- Adjust the supply remaining at source or the demand of the warehouse not yet met.
- Go to step (i) and repeat the procedure until an initial assignment using cells is obtained.
- Evaluate the optimal solution using the
Stepping Stone Method
.
Stepping Stone method
is a process to evaluate each vacant cell by tracing its path.
- The term
“stepping stone”
was used because the process of doing the stepping – stone is like crossing a stream by moving form one stone to another stone (occupied cells).
- Used the table done by the northwest corner rule.
- Determine the vacant cells. For each vacant cell, trace a path (horizontal and vertical directions) that will pass through the occupied cells and will end in the same vacant cell being evaluated.
- Give alternating signs (+/–) starting with a positive sign from the cost of the vacant cell being evaluated to the next costs in the path.
- Determine the improvement (a decrease in cost) for each vacant cell by adding the costs determined in step (iii). The
most negative cell
evaluation implies a most decrease in cost.
- Choose the best cell improvement. The
most negative cell
evaluation gives the best cell improvement; then shift as many units as possible to that vacant cell.
- Repeat steps (ii) to (v) until an optimal solution is obtained. An optimal solution is obtained if each cell evaluation has
positive or zero
result.
Name: _________________________________________________ Score: _____________
Course/Year: ____________________ Room: ________________ Date: _____________
Day: _______________ Time: ________________ Professor: _________________________
Exercise No. 5Balanced Transportation Problem
- Solve the following transportation problems:
- Costales manufacturing company operates plants P1, P2
and P3
in different regions. The weekly production of these plants is as follows:
Plants Weekly Production / Supply
P
1110
P
290
P
355
Total 255
The total supply of the company is absorbed by warehouse W
1, W
2and W
3. Their weekly requirement is as follows:
Warehouses Weekly Requirement / Demand
W
1100
W
270
W
385
Total 255
The transportation cost from each plant to each warehouse is given below.
Warehouses
Plants From \ To W
1W
2W
3P
1P9 P11 P7
P
28 5 10
P
314 13 9
A. Determine the minimum cost transportation schedule using an initial basic feasible solution obtained by each of the following methods:
- Northwest Corner Rule
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
- Minimum Entry Method
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
- Vogel’s Approximation Method
To From |
W1
|
W2
|
W3
|
Supply |
Row Difference |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Column Difference |
|
|
Cost = ____________________________________________________________
= _______________________
B. Find the optimal solution using stepping stone method.
Iteration 1
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Final Tableau
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
Total Cost |
- Enriquez manufacturing company must ship its products from three plants to three warehouses. The weekly production of the plants is 120, 75 and 55. The weekly requirements of the warehouses are 100, 90 and 60. The shipping cost from each plant to the warehouse is given below:
Warehouses
Plants From \ To W
1W
2W
3P
1P9 P13 P7
P
24 6 8
P
36 5 10
Find the following:
- the minimum cost transportation schedule using an initial basic feasible solution obtained by each of the following methods:
- Northwest Corner Rule
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
- Minimum Entry Method
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
- Vogel’s Approximation Method
To From |
W1
|
W2
|
W3
|
Supply |
Row Difference |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Column Difference |
|
|
Cost = ____________________________________________________________
= _______________________
- the optimal solution using stepping stone method.
Iteration 1
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 3
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Final Tableau
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
Total Cost |
Unbalanced Transportation Problems
An
unbalanced transportation problem
has unequal number of units of demand and supply. That is,
(1) Or Supply is less than Demand
(2) or Supply is greater than Demand
Steps in solving unbalanced transportation problem:
Case 1: Or Supply is less than Demand
(i) Add a dummy row to the transportation table. Give zero (0) cost to each dummy cell.
(ii) Follow the procedure in finding the optimal solution in the balanced transportation problem using the Stepping Stone method.
Case 2: or Supply is greater than Demand
(i) Add a dummy column to the transportation table. Give zero (0) cost to each dummy cell.
(ii) Follow the procedure in finding the optimal solution in the balanced transportation problem using the Stepping Stone method.
Name: _________________________________________________ Score: _____________
Course/Year: ____________________ Room: ________________ Date: _____________
Day: _______________ Time: ________________ Professor: _________________________
Exercise No. 6Unbalanced Transportation ProblemI. Solve the following transportation problems using Stepping Stone Method:
- Rendon Manufacturing Firm operates three plants that ship their goods to three regional warehouses. The production capacity of the plants and the requirements of the warehouses are given below:
Plants Capacity Warehouses Requirements
P
180 W
195
P
290 W
255
P
375 W
370
Total 245 Total 220
Warehouses
Plants From \ To W
1W
2W
3P
1P5 P4 P6
P
211 7 8
P
32 6 10
Solution:
Initial Tableau
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Iteration 1
To From |
W1
|
W2
|
W3
|
dummy |
Supply |
|
P1
|
|
|
|
|
|
|
|
P2
|
|
|
|
|
|
|
P3
|
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
W1
|
W2
|
W3
|
dummy |
Supply |
|
P1
|
|
|
|
|
|
|
|
P2
|
|
|
|
|
|
|
P3
|
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 3
To From |
W1
|
W2
|
W3
|
dummy |
Supply |
|
P1
|
|
|
|
|
|
|
|
P2
|
|
|
|
|
|
|
P3
|
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 4
To From |
W1
|
W2
|
W3
|
dummy |
Supply |
|
P1
|
|
|
|
|
|
|
|
P2
|
|
|
|
|
|
|
P3
|
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Final Tableau
To From |
W1
|
W2
|
W3
|
dummy |
Supply |
|
P1
|
|
|
|
|
|
|
|
P2
|
|
|
|
|
|
|
P3
|
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
|
Total Cost |
To From |
W1
|
W2
|
W3
|
W4
|
Supply |
|
P1
|
11 |
17 |
12 |
10 |
|
|
|
|
240
|
|
|
|
P2
|
12 |
14 |
16 |
13 |
|
|
|
|
280
|
|
|
P3
|
20 |
11 |
10 |
17 |
|
|
|
|
230
|
|
|
Demand |
250
|
200
|
340
|
110
|
900 / 750
|
|
|
Solution:
Iteration 1
To From |
W1
|
W2
|
W3
|
W4
|
Supply |
|
P1
|
11 |
17 |
12 |
10 |
|
|
|
|
|
|
|
P2
|
12 |
14 |
16 |
13 |
|
|
|
|
|
|
P3
|
20 |
11 |
10 |
17 |
|
|
|
|
|
|
dummy |
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
W1
|
W2
|
W3
|
W4
|
Supply |
|
P1
|
11 |
17 |
12 |
10 |
|
|
|
|
|
|
|
P2
|
12 |
14 |
16 |
13 |
|
|
|
|
|
|
P3
|
20 |
11 |
10 |
17 |
|
|
|
|
|
|
dummy |
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 3
To From |
W1
|
W2
|
W3
|
W4
|
Supply |
|
P1
|
11 |
17 |
12 |
10 |
|
|
|
|
|
|
|
P2
|
12 |
14 |
16 |
13 |
|
|
|
|
|
|
P3
|
20 |
11 |
10 |
17 |
|
|
|
|
|
|
dummy |
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 4
To From |
W1
|
W2
|
W3
|
W4
|
Supply |
|
P1
|
11 |
17 |
12 |
10 |
|
|
|
|
|
|
|
P2
|
12 |
14 |
16 |
13 |
|
|
|
|
|
|
P3
|
20 |
11 |
10 |
17 |
|
|
|
|
|
|
dummy |
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
|
Total Cost |
Degenerate Transportation Problems
The
degeneracy
in transportation problems occurs when
Total number of occupied cells
That is, there are lacking occupied cells to evaluate the vacant cells.
Degeneracymay occur in the following:
- In the initial feasible solution
- In the succeeding solutions.
Rules in Solving Degenerate Transportation Problem
- Degeneracy in the initial feasible solution
- Apply the North – West Corner Rule. In this step, the
stair path
pattern would be broken. Complete the stair path by using zero entry in the cell with a lower cost for a minimization problem.
- Follow the procedure in finding the optimal solution in the balanced transportation problem using the Stepping Stone method
- Degeneracy in the succeeding solutions
There will be atleast one vacant cell for which an evaluation path may not be constructed. Degeneracy occurs in the succeeding solutions using the Stepping Stone method, in this case allocate zero entry to this cell to complete the evaluation path.
Name: _________________________________________________ Score: _____________
Course/Year: ____________________ Room: ________________ Date: _____________
Day: _______________ Time: ________________ Professor: _________________________
Exercise No. 7Degenerate Transportation ProblemI. Solve the following transportation problems using Stepping Stone Method:
- Given the problem in table below, find the minimum cost.
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
12 |
10 |
13 |
|
|
150 |
P2
|
8 |
12 |
6 |
|
|
100 |
P3
|
9 |
10 |
10 |
|
|
125 |
Demand |
130 |
120 |
125 |
375 |
Solution:
Iteration 1
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Final Tableau
To From |
W1
|
W2
|
W3
|
Supply |
P1
|
|
|
P2
|
|
|
P3
|
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
Total Cost |
- Santos Printing Press delivers books from their three branches to three bookstores. Bookstore A requires 180 boxes of books each year, bookstore B requires 145, and bookstore C requires 170. Branch D can supply 120 books, branch E can supply 205 books and branch F can supply 170 books. Using the cost information given below, compute the optimal transportation cost.
From |
To Bookstore A |
To Bookstore B |
To Bookstore C |
Branch D |
P 25 / box |
18 |
22 |
Branch E |
22 |
25 |
20 |
Branch F |
15 |
24 |
20 |
Solution:
Iteration 1
To From |
A |
B |
C |
Supply |
D |
25 |
18 |
22 |
|
|
E |
22 |
25 |
20 |
|
|
F |
15 |
24 |
20 |
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
A |
B |
C |
Supply |
D |
25 |
18 |
22 |
|
|
E |
22 |
25 |
20 |
|
|
F |
15 |
24 |
20 |
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 3
To From |
A |
B |
C |
Supply |
D |
25 |
18 |
22 |
|
|
E |
22 |
25 |
20 |
|
|
F |
15 |
24 |
20 |
|
|
Demand |
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
Total Cost |
- Janet, a fruit dealer, sells fruits to customers in Manila, Pampanga and Bulacan. The monthly demand is 4000 kilos in Manila, 2500 kilos in Pampanga and 1000 kilos in Bulacan.
The fruits are shipped from three farms located in Laguna, Davao and Cebu. The monthly supply available in Laguna is 5600 kilos, in Davao 900 kilos and Cebu 3200 kilos.
The shipping cost per kilo is shown in the table below.
From |
To Manila |
To Pampanga |
To Bulacan |
Laguna |
P 5 / kilo |
P11 |
P7 |
Davao |
50 |
55 |
52 |
Cebu |
25 |
21 |
27 |
Help the dealer to decide what schedule of shipments will obtain the lowest transportation cost.
Solution:
Initial Tableau
To From |
Manila |
Pampanga |
Bulacan |
Supply |
Laguna |
|
|
Davao |
|
|
Cebu |
|
|
Demand |
|
|
Iteration 1
To From |
Manila |
Pampanga |
Bulacan |
dummy |
Supply |
|
Laguna |
|
|
|
|
|
|
|
Davao |
|
|
|
|
|
|
Cebu |
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
To From |
Manila |
Pampanga |
Bulacan |
dummy |
Supply |
|
Laguna |
|
|
|
|
|
|
|
Davao |
|
|
|
|
|
|
Cebu |
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Final Tableau
To From |
Manila |
Pampanga |
Bulacan |
dummy |
Supply |
|
Laguna |
|
|
|
|
|
|
|
Davao |
|
|
|
|
|
|
Cebu |
|
|
|
|
|
|
Demand |
|
|
|
|
|
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
|
Total Cost |
- Given the problem in table below, find the minimum cost.
C |
D |
E |
F |
G |
Supply |
A |
70 |
85 |
70 |
62 |
67 |
46 |
|
|
|
|
|
B |
75 |
98 |
64 |
58 |
72 |
88 |
|
|
|
|
|
Demand |
46 |
29 |
36 |
22 |
27 |
160 / 134 |
Solution:
Iteration 1
C |
D |
E |
F |
G |
Supply |
A |
70 |
85 |
70 |
62 |
67 |
46 |
|
|
|
|
|
B |
75 |
98 |
64 |
58 |
72 |
88 |
|
|
|
|
|
Dummy |
0 |
0 |
0 |
0 |
0 |
|
|
|
|
|
|
Demand |
46 |
29 |
36 |
22 |
27 |
160 / 160 |
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 2
C |
D |
E |
F |
G |
Supply |
A |
70 |
85 |
70 |
62 |
67 |
46 |
|
|
|
|
|
B |
75 |
98 |
64 |
58 |
72 |
88 |
|
|
|
|
|
Dummy |
0 |
0 |
0 |
0 |
0 |
|
|
|
|
|
|
Demand |
46 |
29 |
36 |
22 |
27 |
160 / 160 |
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Iteration 3
C |
D |
E |
F |
G |
Supply |
A |
70 |
85 |
70 |
62 |
67 |
46 |
|
|
|
|
|
B |
75 |
98 |
64 |
58 |
72 |
88 |
|
|
|
|
|
Dummy |
0 |
0 |
0 |
0 |
0 |
|
|
|
|
|
|
Demand |
46 |
29 |
36 |
22 |
27 |
160 / 160 |
Cost = ____________________________________________________________
= _______________________
Cell Evaluation:
Decision:
From |
To |
No. of Units for Shipment |
Cost / Unit |
Shipment Cost |
|
|
|
|
|
|
|
Total Cost |