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A PRODUCTION AND INVENTORY APPLICATION 327

Setting u 1 ¼ 0 and solving for the remainder gives:
u 1 ¼ 0
u 2 ¼ 0:20
u 3 ¼ 0:30
v 1 ¼ 4:40
v 2 ¼ 4:55
v 3 ¼ 3:90
v 4 ¼ 0

If we now calculate the net evaluation index for each unused route we have:

Route Net evaluation index c ij –u i –v j

Marseille-Nairobi +0.45
Portsmouth-Dar-es-Salaam +1.00
Portsmouth-Dummy +0.20
Frankfurt-Capetown +0.15
Frankfurt-Nairobi +1.2
Frankfurt-Dummy +0.30

By inspection we see that all the index values are positive, indicating that our first solution is also our
optimal, least-cost solution. If we now examine the organization’s desire to see all of Marseille’s cases
shipped out we see that currently this is not happening. There are 100 cases from Marseilles that are not
used. Because we know that the current solution is optimal, we also know that insisting on all of
Marseille’s cases being shipped must increase costs. From the net evaluation index we know what this
cost will be. If we force Marseille–Dummy to take a zero value then one of the other two cells in the
Dummy column must be used instead. By inspection we see that the lowest cost cell is Portsmouth–
Dummy at +0.20. So our new solution would be to ship only 500 units from Portsmouth. We can also
track through the changes on the rest of the solution.

Marseille–Dummy 100 cases
Portsmouth–Dummy +100 cases
Portsmouth–Capetown 100 cases (since this is the only stepping stone route for the
Portsmouth row)
Marseille–Capetown +100 cases

and a total increase in costs of E20.

Problems

Note: For some problems a variety of solution methods can be used. Where the solution
method is not specified, you may use LP or the transportation or assignment modules of
The Management Scientist or some other software package. Some problems are intended
to be solved using the special purpose algorithms of Sections 7.2 and 7.4.

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