Test by Bikram | Mock GATE | Test 1 | Question: 54

Question

A pipeline system has $4$ stages and each stage takes $10ns$.

$30\%$ instructions are branch-instructions and branch-instruction introduce a delay of $3$ stall cycles. The speed-up factor compared to same non-pipelined environment if there are $100000$ instructions to be executed is ___________.

Comments

you are correct , add this range too and your marks will increase.

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@ Bikram sir, plz tell me where i m going wrong...

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in pipelined calculation, for branched you have taken 30000*3 clocks

3 clocks are stalls, so total clocks taken will be 3+1=4..

Answer 1

Speedup factor = Time to execute 100000 inst on non pipeline machine/ Time to execute 100000 inst on pipelined machine

Time to execute 100000 inst on non pipeline machine = No of inst * No of stages of pipeline * time for each stage

= 100000 * 4 * 10                       - (A)

Time to execute 100000 inst on pipelined machine:

Clock cycles per inst = 1 + %Branch inst * No of Stall cycles due to branch

= 1 + 0.3 * 3 = 1.9

-> Time to execute 100000 inst = No of instructions * Clock cycles per inst *  time for 1 clock cycle(which is same as time taken by pipeline stage having maximum delay. Here each stage takes 10 ns).

= 100000 * 1.9 * 10.                     - (B)

Now, just divide (A) & (B) to get 2.105 as the answer.

Another method:

Speedup = No of pipeline stages/ 1 + %branch * no of stall cycles due to branch (This formula is derived by using argument similar to what is given above).

= 4/(1+0.3*3)

=2.105

Comments

it is 30% only, due to latex it is not printed. ( now i will edit )

for further quires on this question, check https://gateoverflow.in/108984/pipeline

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@Shaik Masthan

If 30 was given instead of 30% ..then wil above calculation be right ?

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then it is ok