Andrew S. Tanenbaum (OS) Edition 4 Exercise 2 Question 38 (Page No. 177)

Question

The $CDC\: 6600$ computers could handle up to $10\: I/O$ processes simultaneously using an interesting form of round-robin scheduling called processor sharing. A process switch occurred after each instruction, so instruction $1$ came from process $1$, instruction $2$ came from process $2$, etc. The process switching was done by special hardware, and the overhead was zero. If a process needed $T$ sec to complete in the absence of competition, how much time would it need if processor sharing was used with n processes?

Answer 1

The time taken to complete will be $\mathrm {nT} \;\text {seconds}$

Assume, time quantum$=\mathrm q \;\text{seconds}$

Assume that the process runs for $\mathrm q$ seconds

$\therefore 1$ cycle completion time $=\mathrm {nq} \;\text{seconds}$

Let total time $= \mathrm T$

Now, time for completion $ = \mathrm {nq + nq + nq + \dots} = \mathrm{n\underbrace{(q+q+q+\dots)}_\text{=T}} = \mathrm {nT}$