Consider four processes P, Q, R, and S scheduled on a CPU as per round robin…

Final answer: P = 3, Q = 7, R = 7, S = 3 is not possible.

Explanation by simulating round-robin with time quantum = 4 (all processes arrive at t = 0 in order P, Q, R, S):

• P runs from t = 0 to t = 3 and terminates (burst 3).

• Q runs from t = 3 to t = 7 (uses a full quantum of 4), remaining = 3. CPU then switches from Q to R (this is one Q→R switch).

• R runs from t = 7 to t = 11 (uses 4), remaining = 3. CPU then switches from R to S (an R→S switch).

• S runs from t = 11 to t = 14 and terminates (burst 3). After S terminates, the next ready process is Q, so this is one S→Q switch (counts as the required single S→Q).

• Q resumes from t = 14 to t = 17 (remaining 3) and then terminates. After Q terminates, the CPU switches to R, producing a second Q→R switch.

• R runs from t = 17 to t = 20 (remaining 3) and terminates. After R terminates there is no ready process to switch to.

Observed context-switch counts from this simulation:

• S→Q = 1 (matches requirement).

• Q→R = 2 (matches requirement).

• R→Q = 0 (does NOT match the required 1).

Because R never transfers control to Q in any valid round-robin progression for these burst times, the required single R→Q switch cannot be obtained. Therefore the burst combination P = 3, Q = 7, R = 7, S = 3 is impossible.

By contrast, the other given burst combinations allow Q and R to alternate with enough remaining quanta to produce one S→Q, one R→Q and two Q→R switches, so they are achievable.