Question

We have two jobs A and B. The execution pattern for A is CPU first IO next, and of B IO first CPU next. Each one needs CPU time of 40mn. The IO time of A is 50% of total execution time of A. Likewise for B. Ignore context switch time or initial program load time (assume that it is zero). Total execution time is defined to be CPU time plus IO time. The unit mn stands for minute.

(a) Serial execution. What is the elapsed time (i.e. turnaround time) of A? What is it of B? Assume that A arrives ’ahead’ of B at the same time with B. What is the CPU utilization when the jobs are done? (Express it as a percentage.)

(b) Multiprogramming. Assuming that both arrive at the same time, what is the elapsed time of A? What is it of B? What is the CPU utilization (percentage) ?

(c) Multitasking (timesharing) with slice of 20mn. Multiprogramming is not available. Assuming that both arrive at the same time but A has a slightly faster arrival. What is the elapsed time of A? Of B? CPU utilization (percentage)?

(d) Multitasking (timesharing) with slice of 1mn. Multiprogramming is not available. Assuming that both arrive at the same time but A has a slightly faster arrival. What is the elapsed time of A? Of B? CPU utilization?

Answer 1

Gievn both have the IO time of A is 50% of total execution time and cpu time is 40 mn which means io time is also 40 mn.

a) Serial Execution: First the A job will finish first and then B job.

Turn around time for A = 80mn and for B = 160mn.

CPU utilization : In first 80 mn cpu is used for 40 min and from 80 to 160mn it is again used for 40 mn. Which gives us 50% cpu utilization.

b) Multiprogramming : Initiall A uses cpu while B uses io for first 40 mn and then for next 40 mn A uses io and B uses cpu. which gives us time used for both A and B as 80 mn and cpu utilization of 100%

c) Multitasking( no multiprogramming, time slice of 20mn):

time	cpu	io
0-20	A	-
20-40	-	B
40-60	A	-
60-80	-	B
80-100	-	A
100-120	B	-
120-140	-	A
140-160	B	-

Cpu utilization = 50 %

turn around A=140,B=160

d) This part is similar to c)

cpu utilization = 50%

turn around time for A=159mn

B=160mn