GO Classes DPP | GATE CS | Operating System | CPU Scheduling

Answer 1

Using FCFS scheduling:

$|^{0} P1 |^{8} P2 |^{12} P3 |^{14} P4 |^{15}$

Average waiting time = (0+8+12+14)/4 = 8.5

Using non-preemptive SJF scheduling :

$|^{0} P4 |^{1} P3 |^{3} P2 |^{7} P1 |^{15}$

Average waiting time = (0+1+3+7)/4 = 2.75

So, difference between average waiting time of both : 8.5 - 2.75 = 5.75

Answer 2

Under FCFS, the order is:

$P_1 \rightarrow P_2 \rightarrow P_3 \rightarrow P_4$

Waiting times are:

$P_1 = 0$

$P_2 = 8$

$P_3 = 8+4 = 12$

$P_4 = 8+4+2 = 14$

Average waiting time:

$\frac{0+8+12+14}{4} = \frac{34}{4} = 8.50$

Under non-preemptive SJF, the order is:

$P_4 \rightarrow P_3 \rightarrow P_2 \rightarrow P_1$

Waiting times are:

$P_4 = 0$

$P_3 = 1$

$P_2 = 1+2 = 3$

$P_1 = 1+2+4 = 7$

Average waiting time:

$\frac{0+1+3+7}{4} = \frac{11}{4} = 2.75$

Therefore, the required difference is:

$8.50 - 2.75 = 5.75$

Answer : $\boxed{5.75}$