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Official PYQ · 2025

GATE CS 2025 Set 2 Q46 — Instruction Pipelining

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A 55-stage instruction pipeline has stage delays of 180,250,150,170180,250,150,170, and 250250, respectively, in nanoseconds. The delay of an inter-stage latch is 1010 nanoseconds. Assume that there are no pipeline stalls due to branches and other hazards. The time taken to process 10001000 instructions in microseconds is __________. (rounded off to two decimal places)
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Answer: 260.20:261.20

To find the total time taken to process 10001000 instructions in the given 55-stage pipeline, we proceed as follows:

1. Determine the Clock Cycle Time (TpT_p)

The clock cycle time of a pipelined processor is determined by the maximum delay among all stages plus the inter-stage latch delay (dd):

Tp=max⁡(t1,t2,t3,t4,t5)+dT_p = \max(t_1, t_2, t_3, t_4, t_5) + d

Given stage delays are 180,250,150,170,180, 250, 150, 170, and 250250 ns, and the latch delay is 1010 ns:

Tp=max⁡(180,250,150,170,250)+10=250+10=260 nsT_p = \max(180, 250, 150, 170, 250) + 10 = 250 + 10 = 260\text{ ns}

2. Calculate Total Execution Time

The total time TT required to execute NN instructions in a kk-stage pipeline (assuming no stalls) is given by:

T=(k+N−1)×TpT = (k + N - 1) \times T_p

Substituting the given values (k=5k = 5 stages, N=1000N = 1000 instructions):

T=(5+1000−1)×260 nsT = (5 + 1000 - 1) \times 260\text{ ns} T=1004×260 ns=261,040 nsT = 1004 \times 260\text{ ns} = 261,040\text{ ns}

3. Convert to Microseconds (μ\mus)

Since 1 μs=1000 ns1\ \mu\text{s} = 1000\text{ ns}:

T=261,0401000 μs=261.04 μsT = \frac{261,040}{1000}\ \mu\text{s} = 261.04\ \mu\text{s}

Thus, the time taken to process 10001000 instructions is 261.04261.04 microseconds.

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