Question
Consider the available processes given below in the ready queue for execution and with given burst
time.
Process NoArrival TimeBurst time
P102
P212
P353
P464
a) What is the time at which all the processes get executed?
b) Find the average waiting time and average turnaround time using thenon - pre-emptive FCFS
scheduling algorithm.

Answer


Image
Arrival time of P1 is 0 and it gets executed for 2 units of time. While P1 is being executed P2
has already arrived in ready queue after 1 unit of time. Completion time for P2 is 4 units. Since P3
arrives after 5 units of time, the processor remains idle for 1 unit of time. When time is 5 units P3
arrives and get executed for 3 units of time. After 6 units of time P4 is in ready queue. Thus, its
completion time is 12 units.
a) All the processes got executed at 12 units of time
b) Average waiting time $=\frac{0+1+0+2}{4}=0.75$ units
Average turnaround time $=\frac{2+3+3+6}{4}=3.5$ units

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