GATE CSE 2023 | Question: 23

Question

Consider a $3$-stage pipelined processor having a delay of $10 \mathrm{~ns}$ (nanoseconds), $20 \mathrm{~ns}$, and $14 \mathrm{~ns},$ for the first, second, and the third stages, respectively. Assume that there is no other delay and the processor does not suffer from any pipeline hazards. Also assume that one instruction is fetched every cycle.

The total execution time for executing $100$ instructions on this processor is _____________ $\mathrm{ns}.$

Comments

Detailed Video Solution:

Pipeline 3 Stages, 100 Instructions

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https://www.youtube.com/live/WlzfxlVU6HI?feature=share&t=4714 

Answer 1

Given,

delays = $10 ns, 20ns, 14ns$

total instruction (n) = $100$

We take pipeline delay as $t_p = max(10, 20, 14) = 20$

number of stages ($k$) $ = 3$

So,

Total execution time $ = (k + (n – 1)) \times t_p$

$\implies (3 + 100- 1) \times 20 ns$

$\implies 2040ns$

Comments

@Yaman Sahu Any tips on friday’s exam ? 

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Nice

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@burriakhilreddy  @surya8370 you can not have different clock periods for different stages. Clock period is decided by system clock while keeping the pipeline in mind, i.e. all stages of pipeline should be able to able to complete their work within a clock tick. That is why $t_p = max(10, 20, 14) = 20$, i.e. if stage with $20ns$ delay can finish its work within the decided time, other stages can as well.

Answer 2

For the given pipeline system:

• Total number of stages $(k)=3$
• Total number of instructions, $(n)=100$
• Total delay ($t_p) = \max(\text{stage delay})$
  • $\implies t_p = \max (10,20,14) \ ns$
  • $\implies t_p =20 \;ns$

• $ET_{p} = [(k+(n-1))*t_p]$

  $\implies ET_p= [(3+(100-1))*20]\;ns$

  $\implies ET_p= (3+99)*20\;ns$

  $\implies ET_{pipeline}=2040\;ns$

  $\therefore$ To execute $100$ instructions in the given pipeline $2040\;ns$ time is required.

Ref: GATE CSE 2021 Set 1 | Question: 53

Answer 3

For the given pipeline system:

• Total number of stages $(k)=3$
• Total number of instructions, $(n)=100$
• Total delay ($t_p) = \max(\text{stage delay})$
  • $\implies t_p = \max (10,20,14) \ ns$
  • $\implies t_p =20 \;ns$

• $ET_{p} = [(k+(n-1))*t_p]$

  $\implies ET_p= [(3+(100-1))*20]\;ns$

  $\implies ET_p= (3+99)*20\;ns$

  $\implies ET_{pipeline}=2040\;ns$

  $\therefore$ To execute $100$ instructions in the given pipeline $2040\;ns$ time is required.

Ref: GATE CSE 2021 Set 1 | Question: 53

Answer 4

Ans = 2040 ns.

Here we have to calculate total execution time for executing 100 instructions on the processor which is 2040 ns.