Although the Hungarian method is an efficient methodfor solving an assignment problem, the branch-and-boundmethod can also be used to solve an assignment problem.Suppose a company has five factories and five warehouses.Each factory’s requirements must be met by a singlewarehouse, and each warehouse can be assigned to only onefactory. The costs of assigning a warehouse to meet afactory’s demand (in thousands) are shown in Table 77.Let xij 1 if warehouse i is assigned to factory j and 0otherwise. Begin by branching on the warehouse assigned tofactory 1. This creates the following five branches: x11 1,x21 1, x31 1, x41 1, and x51 1. How can we obtaina lower bound on the total cost associated with a branch?Examine the branch x21 1. If x21 1, no furtherassignments can come from row 2 or column 1 of the costmatrix. In determining the factory to which each of theunassigned warehouses (1, 3, 4, and 5) is assigned, we cannotdo better than assign each to the smallest cost in thewarehouse’s row (excluding the factory 1 column). Thus, theminimum-cost assignment having x21 1 must have a totalcost of at least 10 10 9 5 5 39.Similarly, in determining the warehouse to which eachof the unassigned factories (2, 3, 4, and 5) is assigned, wecannot do better than to assign each to the smallest cost inthe factory’s column (excluding the warehouse 2 row). Thus,the minimum-cost assignment having x21 1 must have atotal cost of at least 10 9 5 5 7 36. Thus, thetotal cost of any assignment having x21 1 must be at leastmax(36, 39) 39. So, if branching ever leads to a candidatesolution having a total cost of 39 or less, the x21 1 branchmay be eliminated from consideration. Use this idea to solvethe problem by branch-and-bound.
Although the Hungarian method is an efficient method
for solving an assignment problem, the branch-and-bound
method can also be used to solve an assignment problem.
Suppose a company has five factories and five warehouses.
Each factory’s requirements must be met by a single
warehouse, and each warehouse can be assigned to only one
factory. The costs of assigning a warehouse to meet a
factory’s demand (in thousands) are shown in Table 77.
Let xij 1 if warehouse i is assigned to factory j and 0
otherwise. Begin by branching on the warehouse assigned to
factory 1. This creates the following five branches: x11 1,
x21 1, x31 1, x41 1, and x51 1. How can we obtain
a lower bound on the total cost associated with a branch?
Examine the branch x21 1. If x21 1, no further
assignments can come from row 2 or column 1 of the cost
matrix. In determining the factory to which each of the
unassigned warehouses (1, 3, 4, and 5) is assigned, we cannot
do better than assign each to the smallest cost in the
warehouse’s row (excluding the factory 1 column). Thus, the
minimum-cost assignment having x21 1 must have a total
cost of at least 10 10 9 5 5 39.
Similarly, in determining the warehouse to which each
of the unassigned factories (2, 3, 4, and 5) is assigned, we
cannot do better than to assign each to the smallest cost in
the factory’s column (excluding the warehouse 2 row). Thus,
the minimum-cost assignment having x21 1 must have a
total cost of at least 10 9 5 5 7 36. Thus, the
total cost of any assignment having x21 1 must be at least
max(36, 39) 39. So, if branching ever leads to a candidate
solution having a total cost of 39 or less, the x21 1 branch
may be eliminated from consideration. Use this idea to solve
the problem by branch-and-bound.
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