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homework 4 PATIENCE FERO 2019529628044 Version 0
👤 Author: by feropatienceoutlookcom 2021-12-28 14:28:03
My example

Consider the processes P1, P2, P3 given in the below table, arrives for execution in
the same order, with Arrival Time 0, and given Burst Time,




Total Wait Time = 0 + 24 + 27 = 51 ms
Average Waiting Time = (Total Wait Time) / (Total number of processes) = 51/3 = 17 ms
Total Turn Around Time: 24 + 27 + 30 = 81 ms
Average Turn Around time = (Total Turn Around Time) / (Total number of processes)
= 81 / 3 = 27 ms
Throughput = 3 jobs/30 sec = 0.1 jobs/sec
Another Example
Consider the processes P1, P2, P3, P4 given in the below table, arrives for execution
in the same order, with given Arrival Time and Burst Time.


Total Wait Time:= 0 + 7 + 10 + 18 = 35 ms
Average Waiting Time = (Total Wait Time) / (Total number of processes)= 35/4 = 8.75 ms
Total Turn Around Time: 8 + 11 + 19 + 23 = 61 ms
Average Turn Around time = (Total Turn Around Time) / (Total number of processes)
61/4 = 15.25 ms
Throughput: 4 jobs/26 sec = 0.15385 jobs/sec

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