A transportation branch and bound algorithm for solving the generalized assignment problem

Abstract

This paper presents a transportation branch and bound algorithm for solving the generalized assignment problem. This is a branch and bound technique in which the sub-problems are solved by the available efficient transportation techniques rather than the usual simplex based approaches. A technique for selecting branching variables that minimize the number of sub-problems is also presented.

Appendix: Details of GAP numerical illustration

In this appendix, detailed working of the numerical illustration for the transportation branch and bound algorithm for solving the GAP model is presented. For completeness and ease of reference, some essential elements of the numerical illustration are reproduced again in this appendix. The problem is given by Eq. (21):

$$ \begin{gathered} Z_{GAP} = {\text{Min}}28x_{11} + 76x_{12} + 52x_{14} + 28x_{15} + 98x_{21} + 40x_{23} + 92x_{24} + 98x_{25} + 90x_{32} + 32x_{33} + 20x_{34}\end{gathered} $$

Subject to:

$$ \left. \begin{gathered} 24x_{11} + 38x_{12} + 22x_{14} + 36x_{15} \le 56 \hfill \\ 12x_{21} + 22x_{23} + 30x_{24} + 36x_{25} \le 56 \hfill \\ 20x_{32} + 28x_{33} + 44x_{34} \le 56 \hfill \\ \end{gathered} \right\} $$

$$ \left. \begin{gathered} x_{11} + x_{21} = 1 \hfill \\ x_{12} + x_{32} = 1 \hfill \\ x_{23} + x_{33} = 1 \hfill \\ x_{14} + x_{24} + x_{34} = 1 \hfill \\ x_{15} + x_{25} = 1 \hfill \\ \end{gathered} \right\} $$

$$ x_{ij} = 0\,{\text{or 1}}\,\forall ij $$

(21)

Arranging the resource coefficients of constraints in (first three constraints of Eq. 21) in ascending order, we have Eq. (22).

$$ \left. \begin{gathered} \{ 22,24,36,38\} \hfill \\ \{ 12,22,30,36\} \hfill \\ \{ 20,28,44\} \hfill \\ \end{gathered} \right\} $$

(22)

The γ _i values are easily calculated from Eq. (21) and Eq. (22).

$$ \left. \begin{gathered} 22 + 24 = 46 \le 56:\,\gamma_{1} = 2 \hfill \\ 12 + 22 = 34 \le 56:\,\gamma_{2} = 2 \hfill \\ 20 + 28 = 48 \le 56:\,\gamma_{3} = 2 \hfill \\ \end{gathered} \right\} $$

(23)

The transportation model becomes as shown in Table 4.

Table 4 Transportation model for numerical illustration

Table 5 Balancing the transportation model (by adding a dummy column)

Any efficient transportation technique can be used to solve the model and an optimal solution to the relaxed model is obtained as presented in Table 6. The solution in Table 6 is a second order degenerate solution. The optimality solution can be easily verified by using cells (1,6) and (3,3) as basic with zero allocation.

Table 6 Optimal solution to the relaxed model (lower bound)

Using the resource constraints, one can easily verify that row one is infeasible, since 24 +36 = 60 > 56

i.e., x ₁₁ + x ₁₅ ≤ 1

This implies: either x ₁₁ = 0 or x ₁₅ = 0

Similarly the third row is also infeasible, because 20 + 44 = 64 > 56

i.e. x ₃₂ + x ₃₄ ≤ 1

Which implies: either x ₃₂ = 0 or x ₃₄ = 0.

We select the branches from row 3 as it is more restricted compared to row 1. This gives either x ₃₂  = 0 or x ₃₄ = 0 This results in the following Fig. 1.

Fig. 1

Initial branching with respect to cells (3,2) and (3,4)

Now let us consider the case when x ₃₂ = 0, the transportation Table 6 is modified by replacing the assignment cost 90 in the cell (3,2) by L. This is given in Table 7.

Table 7 Node corresponding to the restriction of no assignment in cell (3,2)

Now once again, for the above solution, the resource constraint 3 is not satisfied because 28 + 44 = 72 > 56. Hence, either x ₃₃ = 0 or x ₃₄ = 0. This will lead to nodes 4 and 5 respectively.

Similarly, at node 3 we deal with the restriction that allocation in the cell (3,4) is restricted to zero, in other words we modify the Table 6 and replace the cost element in the cell (3,4) by L. This is shown in Tables 8, 9, 10, 11, 12 and 13.

Table 8 Node corresponding to the restriction that allocation in cell (3, 4) is zero

Table 9 Node 4

Table 10 Node 5

Table 11 Node 6

Table 12 Node 7

Table 13 Node 9

Form the solution obtained in Table 8, it is noted that the resource constraint 1 is not satisfied, since for the allocation in cells (1,1) and (1,5) resource requirement is 24 + 36 = 60 > 56. Hence branching from node 3 will be either x ₁₁ = 0 or x ₁₅ = 0. These restrictions lead to nodes 10 and 11 in Fig. 2.

Fig. 2

Full search tree for the given numerical illustration

In the following, we have given the transportation cost tables under various restrictions as shown in the tree diagram shown in Fig. 2.

These results have been summarized in the tree diagram given in Fig. 2, where the following interpretations have been used.

Note:

• A terminal node is said to be feasible if the optimal solution to the transportation sub-problem is feasible to the original GAP problem.

• A terminal node is said to be infeasible if the optimal solution to the transportation is infeasible to the GAP model.

• DNE means the transportation sub-problem does not have a feasible optimal solution

• The numbers in the circles denote the order of solution

From the search tree given in Fig. 2 the optimal solution to the GAP problem is given as shown in (4A) and (5A).

$$ Z_{GAP} = \hbox{min} [Z_{1}^{T} ,Z_{2}^{T} ,Z_{3}^{T} ,Z_{4}^{T} ,Z_{5}^{T} ,Z_{6}^{T} ] = 300 $$

(24)

$$ \begin{gathered} x_{11} = x_{14} = x_{25} = x_{32} = x_{33} = 1\,{\text{and}} \hfill \\ x_{12} = x_{15} = x_{21} = x_{23} = x_{24} = x_{32} = x_{33} = x_{34} = 0 \hfill \\ \end{gathered} $$

(25)

The transportation optimal solutions for the nodes 2–7 and node 9 are given below in the appendix. Node 1 is optimal solution to the relaxed model and is given in Table 6. Thus a total of 10 nodes (starting from node 2) are required to verify the optimal solution value.