Recently, Yadaiah and Haragopal published in the American Journal of Operations Research a new approach to solving the unbalanced assignment problem. They also provide a numerical example which they solve with their approach and get a cost of 1550 which they claim is optimum. This approach might be of interest; however, their approach does not guarantee the optimal solution. In this short paper, we will show that solving this same example from the Yadaiah and Haragopal paper by using a simple textbook formulation to balance the problem and then solve it with the classic Hungarian method of Kuhn yields the true optimal solution with a cost of 1520.
The assignment problem is a standard topic discussed in operations research textbooks (See for example, Hillier and Lieberman [
Problems in which there are more jobs than machines and more than one job can be assigned to a machine can easily be handled as a balanced assignment problem with a little modeling effort. The idea is to “make copies” or “clone” the machines. This approach is discussed in Hillier and Lieberman [
In Yadaiah and Haragopal [
Their numerical example is given in
The first sub-problem solved byYadaiah and Haragopal [
Their Lexi-search solution to this sub-problem is the following assignment:
J3 assigned to M1, J4 assigned to M2, J5 assigned to M3, J6 assigned to M5, and J7 assigned to M4 with a cost of 870. The second sub-problem is given in
Their Lexi-search solution to this sub-problem is the following assignment:
J1 assigned to M4, J2 assigned to M5, and J8 assigned to M2 with a cost of 680. The final assignment cost is
Jobs | Machines | ||||
---|---|---|---|---|---|
M1 | M2 | M3 | M4 | M5 | |
J1 | 300 | 290 | 280 | 290 | 210 |
J2 | 250 | 310 | 290 | 300 | 200 |
J3 | 180 | 190 | 300 | 190 | 180 |
J4 | 320 | 180 | 190 | 240 | 170 |
J5 | 270 | 210 | 190 | 250 | 160 |
J6 | 190 | 200 | 220 | 190 | 140 |
J7 | 220 | 300 | 230 | 180 | 160 |
J8 | 260 | 190 | 260 | 210 | 180 |
Jobs | Machines | ||||
---|---|---|---|---|---|
M1 | M2 | M3 | M4 | M5 | |
J3 | 180 | 190 | 300 | 190 | 180 |
J4 | 320 | 180 | 190 | 240 | 170 |
J5 | 270 | 210 | 190 | 250 | 160 |
J6 | 190 | 200 | 220 | 190 | 140 |
J7 | 220 | 300 | 230 | 180 | 160 |
870 + 680 = 1550. It can easily be checked using the Hungarian method, that these sub-problems were solved optimally. In their paper,
Yadaiah and Haragopal have a minor typo on page 88, they have 870 + 670 = 1550. It should read 870 + 680 = 1550.
We will now solve the original problem of assigning these eight jobs to five machines such that each machine is used at least once, but not more than twice.
Using the approach suggested in both Hillier and Lieberman [
Solving the balanced assignment problem given in
Machines | |||
---|---|---|---|
Jobs | M2 | M4 | M5 |
J1 | 290 | 290 | 210 |
J2 | 310 | 300 | 200 |
J8 | 190 | 210 | 180 |
M1 | M2 | M3 | M4 | M5 | M1 | M2 | M3 | M4 | M5 | |
---|---|---|---|---|---|---|---|---|---|---|
J1 | 300 | 290 | 280 | 290 | 210 | 300 | 290 | 280 | 290 | 210 |
J2 | 250 | 310 | 290 | 300 | 200 | 250 | 310 | 290 | 300 | 200 |
J3 | 180 | 190 | 300 | 190 | 180 | 180 | 190 | 300 | 190 | 180 |
J4 | 320 | 180 | 190 | 240 | 170 | 320 | 180 | 190 | 240 | 170 |
J5 | 270 | 210 | 190 | 250 | 160 | 270 | 210 | 190 | 250 | 160 |
J6 | 190 | 200 | 220 | 190 | 140 | 190 | 200 | 220 | 190 | 140 |
J7 | 220 | 300 | 230 | 180 | 160 | 220 | 300 | 230 | 180 | 160 |
J8 | 260 | 190 | 260 | 210 | 180 | 260 | 190 | 260 | 210 | 180 |
DUM1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
DUM2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
M1 | M2 | M3 | M4 | M5 | M1 | M2 | M3 | M4 | M5 | |
---|---|---|---|---|---|---|---|---|---|---|
J1 | 40 | 30 | 20 | 30 | 0 | 40 | 30 | 20 | 30 | 0 |
J2 | 0 | 60 | 40 | 50 | 0 | 0 | 60 | 40 | 50 | 0 |
J3 | 0 | 10 | 120 | 10 | 50 | 0 | 10 | 120 | 10 | 50 |
J4 | 140 | 0 | 10 | 60 | 40 | 140 | 0 | 10 | 60 | 40 |
J5 | 80 | 20 | 0 | 60 | 20 | 80 | 20 | 0 | 60 | 20 |
J6 | 0 | 10 | 30 | 0 | 0 | 0 | 10 | 30 | 0 | 0 |
J7 | 40 | 120 | 50 | 0 | 30 | 40 | 120 | 50 | 0 | 30 |
J8 | 70 | 0 | 70 | 20 | 40 | 70 | 0 | 70 | 20 | 40 |
DUM1 | 0 | 0 | 0 | 0 | 50 | 0 | 0 | 0 | 0 | 50 |
DUM2 | 0 | 0 | 0 | 0 | 50 | 0 | 0 | 0 | 0 | 50 |
From
J1 assigned to M5 at a cost of 210,
J2 assigned to M1 at a cost of 250,
J3 assigned to M1 at a cost of 180,
J4 assigned to M2 at a cost of 210,
J5 assigned to M3 at a cost of 190,
J6 assigned to M5 at a cost of 140,
J7 assigned to M4 at a cost of 180,
J8 assigned to M2 at a cost of 190,
For a total minimum cost of 1520 (not 1550).
Both of these assignments use each machine at least once and no more than twice, but the standard textbook formulation solved with the Hungarian method gets the guaranteed optimal solution of 1520―maybe simpler is better. The flaw with their approach is that their assignment is not optimal because their decomposition into sub- problems does not guarantee an optimal solution to the original problem as illustrated by this example.
Nathan Betts,Francis J. Vasko, (2016) Solving the Unbalanced Assignment Problem: Simpler Is Better. American Journal of Operations Research,06,296-299. doi: 10.4236/ajor.2016.64028