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4K
8K
20K

2K

2K memory slot=job j1 (0-4) time

20k memory slot=job j2(0-10) time,j6(10-11) time.

8K memory slot=job j4(0-4) time,j5(4-8) time,j7(8-16).

4K memory slot=job j3(0-2) time.

so completion of job j7 will be 16 time unit.

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BharathiCH asked Jan 18, 2019
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Free Holes4K;8K;20K;2KProgram size2K;14K;3K;6K;10K;20K;2KTime for Execution(B.T)4 ; 10; 2; 1; 4; 1; 8Using Best Fit Allocation Policy and FCFS CPU Scheduling ...