Assignment no 4
a)
Let us consider that CPU remains idle for first 2 unit of time. Draw the Gantt chart and calculate the average turnaround time for above processes in this situation by using non preemptive Shortest Job First (SJF) scheduling strategy?
Solution:
| CPU Idle | Process A | Process C | Process B |
0 2 11 13 18
| Process | Arrival Time | End Time | Burst Time | Turned around Time |
| A | 1.0 | 11 | 9 | 10 |
| B | 1.4 | 18 | 5 | 16.6 |
| C | 2.0 | 13 | 2 | 11 |
Now,
Average turnaround time = 10 + 16.6 + 11 / 3 = 12.5
b) Let us consider that CPU remains idle for first 1.5 unit of time. Draw the Gantt chart and calculate the average waiting time for above processes in this situation by using non preemptive Shortest Job First (SJF) scheduling strategy?
Solution:
| CPU Idle | Process A | Process C | Process B |
0 1.5 10.5 12.5 17.5
| Process | Arrival Time | F Time | Burst Time | Waiting Time |
| A | 1.0 | 10.5 | 9 | 0.5 |
| B | 1.4 | 17.5 | 5 | 11.1 |
| C | 2.0 | 12.5 | 2 | 8.5 |
Now,
Average waiting time = 0.5 + 11.1 + 8.5 / 3 = 6.7 Time units
c).Let us consider that CPU remains idle for first 1 unit of time and let us consider that a fourth process ‘D’ enters the system at time 3 and needs 1 unit of CPU time. Draw the Gantt chart and Calculate the average waiting time for four processes A, B, C and D in this situation by using preemptive SJF i.e. Shortest Remaining Time scheduling strategy.
Solution:
| Process A | Process B | Process C | Process D | Process C | Process B | Process A |
1 1.4 2 3 4 5 9.4 17
| Process | Arrival Time | F Time | Burst Time | Waiting Time |
| A | 1.0 | 17 | 9 | 7 |
| B | 1.4 | 9.4 | 5 | 3 |
| C | 2.0 | 5 | 2 | 1 |
| D | 3.0 | 4 | 1 | 0 |
Now,
Average waiting time = 7 + 3 + 1 + 0 / 4 = 2.75 unit Time
d.)Let us consider that CPU remains idle for first 3 unit of time. In this situation, draw the Gantt chart and calculate the average waiting time for a round robin algorithm with q (time quantum) set at 2. Ignore the effect of context switching time.
Solution:
Round Robin ( quantum Time = 2 )
| PA | PB | PC | PA | PB | PA | PB | PA | PA |
3 5 7 9 11 13 15 16 18 19
| Process | Arrival Time | End Time | Burst Time | T.T | Waiting Time |
| A | 1.0 | 19 | 9 | 18 | 9 |
| B | 1.4 | 16 | 5 | 14.6 | 9.6 |
| C | 2.0 | 9 | 2 | 5 | 3 |
Now,
Average waiting time = 9 + 9.6 + 3 / 3 = 7.2 unit time