Which of the following is correct?

Question

Consider all the processes are arriving at large time intervals.

Let $t$ be the time interval between two processes $p_i$ and $p_{i+1}$ for any $i$ and service time of $p_i$ is $s_i$.

If $t > s_i$ for every $i$, then find the best strategy to schedule the processes.

(a) FCFS
(b) SJN
(c) RR
(d) SRTF

Comments

since service time is given greater than time between two process ariival

so using RR ,SJF,SRTF is useless.as they are efficient when at a particular time there must exist atleast two process

so we are left with FCFS which is the answer

Answer 1

since $t > s_i$  for every $i$, we can conclude that before the next process enters the system, the first process surely completes its execution (whatever scheduling algorithm we use, it will always be true).

So, at any instant of time there will only be one process in the system and before it completes execution no other process will enter the system, as $t > s_i$.

So, we don't need any kind of preemption. Preemption will only increase the complexity. At any time, since there will only be one process, the process will preempt to itself again and again until it finishes its execution, which will surely will reduce the efficiency .

Therefore I think the answer should be FCFS.

Answer 2

I feel it should be RR
 i assumed 3 different set of values 

one set first process have highest burst time and other brst time close to each other and in the other set  the first  process has  most burst time and others have far less  than first  and in the third i took first having most burst time and others burst time as random(third process with most burst time among the remaining 3) 

in both the waiting time in RR would be least and CPU idle time is same for all in respective cases .the third case SRTF and RR performance was same.but it must hold for all cases 

 hence on the basis of waiting in  time i feel RR should be implemented.

but i am not sure ..

Answer 3

t>si for every i ,so  before the next process enters the system first process surely completes its execution ,

and for this reason all srtf sjf and fcfs will act as same as fcfs (in case of RR it`ll be more because of context switch for every time quantum, but order will be same if cs time is very less ) .