−1
= k
k If k≥0, then return to Step 1.
Step 8. Set C C= +1, j=0,πm =φ,k = 0, am= −T (m− ×1) C and return to Step 1.
Step 9: The makespan is Cmax. Assign the jobs in S to i* M , i i m= K , and the remaining , , 2 jobs to M . Set the machines processing the composite job as the high-level machines. 1 The job order on each machine is arbitrary.
We illustrate Algorithm B with an example. Consider a four-machine problem where two high-level machines are included. The data of nine jobs are given in Table 2. We can calculate
h 22
T = , 23Tl = and Cmax =12. Then the valid range of the assigned workload on the high-level is between 10 and 12. Repeating Steps 1-3, jobs J and 1 J are assigned and the workload is 11 4 falling into the valid range. Then (J J ) and the other high-level jobs (1, 4 J J ) are treated as two 2, 3 composite jobs J and C1 J respectively. After performing the m-IMO’ algorithm, we obtain C2 three sets of jobs (J ), (C1 J ), and (C2 J J ) and the makespan 5, 7 Cmax′ =12=Cmax. Then job sets (J J ) and (1, 4 J J ) are assigned to the two high-level machines and (2, 3 J J ) and (5, 7 J J J ) are 6, ,8 9 assigned to the two low-level machines. The optimal makespan is 12.
3. Computational results
Computational experiments are conducted to evaluate both the effectiveness of the LG-LPT heuristic algorithm proposed by Hwang et al (2004) and the efficiency of the proposed algorithm, which are coded in Visual Basic and run on a Pentium 3.0G PC. In this experiment, the job processing times are randomly generated from a discrete uniform distribution U(1, )b with
25, 50,
b= 100. Three values of total machine number (m=3,4,5) are test and the numbers of high- and low-level machine are set as ml >mh ≥ . Different problem sizes of jobs (n=10, 15, 20, 1 30, 50, 100, 500, 1000) each of which include the number of high-level jobs
n =h ⎢⎣mh/(mh+ml)×n⎥⎦ are tested. For example, for problems with m=4 and mh = , there are 8 1 high-level jobs for processing in the problem with n=30. The combinations of the three factors give a total of 120 set of problems. For each problem set, 100 replications are made. Hence, we report the results of the total 12,000 problems solved.
Table 3 gives the set CPU time of 100 problems by the proposed Algorithm. Examining Table 3, we see that the algorithm appears to perform rather efficiently in deriving the optimal solution, although it has an exponential time complexity. In summary, the set CPU time increases as each of number of jobs; number of machines; and the range of job processing time increases. Table 4 gives the mean percent deviations (MPD) from optimum for LG-LPT. In summary, the MPD tend to decrease as the number of jobs and the number of machines increase. For the same number of machines; problems with few high-level machines have smaller percent deviations.
Table 2
Processing times for the Algorithm B example
Job J 1 J 2 J3 J4 J5 J6 J 7 J 8 J9
Level h h h h l l l l l
Processing time 8 6 5 3 7 6 5 3 2
*h =high, l=low
36
ble 3 s for the proposed algorithm 1hm= 2hm= 3m= 4m= 5m= 4m= 5m= n (1,25)(1,50)(1, 100) (1,25)(1,50)(1, 100)(1,25)(1,50)(1, 100) (1,25)(1,50)(1, 100)(1,25)(1,50)(1, 100) 10 0.03 0.02 0.05 0.08 0.17 0.56 0.13 0.20 0.25 0.08 0.13 0.48 0.23 0.20 0.50 15 0.02 0.03 0.03 0.00 0.38 0.17 0.20 1.39 6.38 0.14 0.53 0.84 0.53 0.64 6.92 20 0.05 0.05 0.05 0.06 0.02 0.050.02 0.06 0.14 0.06 0.19 0.81 0.09 0.73 3.00 30 0.06 0.00 0.02 0.02 0.03 0.05 0.05 0.00 0.06 0.02 0.03 0.09 0.05 0.02 0.06 50 0.06 0.05 0.06 0.05 0.06 0.08 0.08 0.08 0.09 0.06 0.03 0.03 0.05 0.11 0.13 100 0.06 0.03 0.08 0.13 0.13 0.06 0.08 0.08 0.11 0.05 0.08 0.09 0.06 0.13 0.11 500 0.73 0.75 0.81 0.81 0.73 0.81 0.77 0.81 0.81 0.83 0.84 0.81 0.84 0.83 0.81 2.53 2.56 2.55 2.44 2.48 2.44 2.72 2.70 2.86 2.58 2.58 2.44
37
Table 4 Mean percentage deviations of LG-LPT 1hm= 2hm= 3m= 4m= 5m= 4m= 5m= n (1,25)(1,50)(1, 100) (1,25)(1,50)(1, 100)(1,25)(1,50)(1, 100) (1,25)(1,50)(1, 100)(1,25)(1,50)(1, 100) 10 1.42%1.90%1.37% 1.58%1.73%1.37%0.09%0.49%0.12% 4.06%3.12%3.94%1.36%1.68%1.52% 15 0.90%0.86%1.06% 2.44%3.16%2.37%2.04%1.96%1.93% 2.65%3.40%3.73%3.22%3.19%3.46% 20 0.66%0.75%0.85% 1.00%1.03%1.39%1.89%1.86%1.76% 2.18%2.15%2.43%3.05%3.60%3.06% 30 0.19%0.30%0.25% 0.45%0.54%0.81%0.70%0.89%0.79% 1.05%1.17%1.26%1.31%1.26%1.76% 50 0.05%0.09%0.11% 0.11%0.17%0.20%0.13%0.28%0.30% 0.26%0.36%0.37%0.34%0.53%0.61% 100 0.01%0.02%0.03% 0.02%0.03%0.05%0.02%0.06%0.06% 0.03%0.08%0.08%0.06%0.10%0.13% 500 0.00%0.00%0.00% 0.00%0.00%0.00%0.00%0.00%0.00% 0.00%0.00%0.00%0.00%0.00%0.00% 1000 0.00%0.00%0.00% 0.00%0.00%0.00%0.00%0.00%0.00% 0.00%0.00%0.00%0.00%0.00%0.00%
4. Conclusions
In the third part of this research, we consider a makespan minimization parallel machine problem with nested machine eligibility restraints. In our problem, the machines and jobs are labeled two levels, low-level and high-level. The high-level jobs can be only processed by the high-level machines and the low-level jobs are available for processing on any machine. We propose an optimal algorithm with the developed lower bounds for the addressed problem.
Due to the machine eligibility restriction, the high-level workload processed only on the high-level machines can be treated as the composite jobs. Then the addressed problem can be regarded a parallel machine problem for processing the low-level jobs and the composite jobs. We proposed an optimal algorithm, called m-IMO’, modified from m-IMO (Lin and Liao, 2004) to solve this parallel machine problem. Although the proposed algorithm has an exponential time complexity, the results show that it can find the optimal solution for various-size problems in a short time.
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