11
Transportation and assignment problems
In this section we study two classical and very useful linear programming
models: the “transportation problem” and a special case, the “assignment
problem.” Incidentally, we shall see an important class of problems that ex
hibit massive degeneracy. We begin with a simple example of a transporta
tion problem, modified from one in the seminal book on linear programming
[3].
We consider a company with plants in Seattle and San Diego, with ca
pacities 350 and 625 cases per week respectively. The company must satisfy
demand in New York, Chicago and Topeka of 325, 300, and 275 cases respec
tively. Transportation costs (in dollars per case) are as follows.
New York
Chicago
Topeka
Seattle
225
153
162
San Diego
225
162
126
We must decide how many cases to send on each route in order to satisfy the
demand at minimum total transportation cost.
In the general transportation problem, we have
m
“origins,” labeled
i
=
1
,
2
, . . . , m
: origin
i
has a supply of
s
i
units. We also have
n
“destinations,”
labeled
j
= 1
,
2
, . . . , n
: destination
j
has demand
d
j
units. To transport one
unit from origin
i
to destination
j
costs
c
ij
, and our aim is to satisfy all the
demands using the available supplies, at minimum total transportation cost.
For the general transportation problem to be feasible, the total supply
must be at least as large as the total demand.
In fact, we do not really
restrict the model if we make the following assumption:
(11.1)
total supply
m
i
=1
s
i
=
total demand
n
j
=1
d
j
.
If, in our problem, the total supply strictly exceeded the total demand, we
could simply introduce an extra “dummy” destination. The demand at this
destination is the difference between supply and demand, and the transporta
tion cost from any origin to this destination is zero. Clearly, this new prob
lem is equivalent to the original problem, and it satisfies assumption (11.1).
Henceforth, we therefore assume that condition (11.1) holds. We introduce
variables
x
ij
that measure the number of units we transport from origin
i
to
56
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destination
j
. Our transportation problem then becomes the following linear
program.
minimize
m
i
=1
n
j
=1
c
ij
x
ij
subject to
n
j
=1
x
ij
=
s
i
(
i
= 1
,
2
, . . . , m
)
m
i
=1
x
ij
=
d
j
(
j
= 1
,
2
, . . . , n
)
x
≥
0
.
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 Fall '08
 TODD
 Linear Programming, Optimization, Seattle

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