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326 CHAPTER 7 TRANSPORTATION, ASSIGNMENT AND TRANSSHIPMENT PROBLEMS

Solution
We have an obvious transportation problem where the organization is looking to minimize costs. The relevant
costs are a combination of the packing costs and the transportation costs per case. For example the cost of
using the Marseille–Capetown route is E2.95 + E1.45, or E4.40 per case. We also note that supply is greater
than demand so we will need a dummy destination. Clearly, we could solve this problem in a number of ways.
We shall use the MODI method. We shall also solve the problem initially without worrying that all of Marseille’s
cases are shipped out. Setting up the transportation tableau and using the minimum cost method to find an
initial feasible solution we have:
Initial feasible solution: cases transported
Dar es Unused
From/To Capetown Salaam Nairobi (Dummy) Supply
Marseille 200 4.40 100 4.55 0 4.35 100 400
Portsmouth 200 4.20 0 5.35 400 3.70 0 600
Frankfurt 0 4.25 300 4.25 0 4.80 0 300
Capacity 400 400 400 100 1 300

The figures in the top right of each cell show the combined cost per case for each route. Because we have a
dummy destination, technically their cost coefficients are zero and so we start by allocating cases to one of the
dummy cells. So that you can check your own solution, the stages for this solution are:
l 100 cases for Marseille–Dummy. All three dummy cells have the same zero cost coefficient so we have
made an arbitrary choice. The other two Dummy cells are set to zero.
l 400 cases for Portsmouth–Nairobi as this is now the least cost route.
l Marseille–Nairobi and Frankfurt–Nairobi cells are set to zero as the maximum capacity for Nairobi is
now met.
l Next least cost route is Portsmouth–Capetown. Portsmouth is already shipping 400 cases to Nairobi so
the remaining 200 cases are allocated to the Portsmouth–Capetown route. The Portsmouth–Dar es
Salaam route is set to zero as Portsmouth has now allocated all its 600 cases.
l There are now two least cost cells: Frankfurt–Capetown and Frankfurt–Dar es Salaam. The latter has a
higher capacity so is preferred and 300 units are allocated to this route. This is all the cases from
Frankfurt so Frankfurt–Capetown is set to zero.
l This leaves Marseille–Capetown with 200 cases.
Total cost of this solution is E4930.
We now calculate the u and v values for the current solution, remembering that we must include the Dummy
column which has cost coefficients of 0. Here we have:
u 1 þ v 1 ¼ 4:40
u 1 þ v 2 ¼ 4:55
u 1 þ v 4 ¼ 0
u 2 þ v 1 ¼ 4:20
u 2 þ v 3 ¼ 3:70
u 3 þ v 2 ¼ 4:25

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