Process Management

Question

Q.Consider the processes P1, P2, P3, P4 whose arrival times are 0, 1, 2, 3 and Burst times are 5, 2, 13, 7. If the Context Switching time is 1 unit (ms), what is the average waiting time if Shortest Job Next scheduling algorithm is used? 

(a) 7.5 
(b) 6.25 
(c) 6.5 
(d) None

Comments

Avg. waiting time : = (0+6+17+9)/4

Ans; (d)

Raj Kumar 7  wrong bro

 

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@Magma

The first context switch needn't be taken.

Even if you take it assuming that there was some process P0 which ended at t=0 then :

Suppose P0 ended at t=0 and the short term scheduler finds P1 waiting in the ready queue. It will start the process of saving state of P0 to PCB of P0 and loading the state of P1 from PCB of P1 and brings P1 to CPU at t=1. At this moment P2 arrives. But P1 is already taken by the CPU and since this is a non pre-emptive scheduling so P1 won't be put back to the queue.

If you ask whether this will be done in case of pre-emptive then I am not sure :v :P

But anyway the context switch at the beginning is not considered.

I did like this :

P1 (0-5)

CS (5-6)

P2(6-8)

CS(8-9)

P4(9-16)

CS(16-17)

P3(17-30)

Waiting time of :

P1 : 0-0 =0

P2: 6-1=5

P3: 17-2=15

P4: 9-3=6

Avg : (0+5+15+6)/4 = 6.5

 

 

 

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hmmm got it !!

Answer 1

my ans is (0+4+12+4)/4=5 . 

P1

P2

P4

P3

0                                5                                 7                               14                                27

 5 +3*1ms(Context-Switch)=8ms.  Is this correct??