ISRO | ISRO CS 2020 | Question 6

Question

Consider a 5-segment pipeline with a clock cycle time 20 ns in each sub operation. Find out the approximate speed-up ratio between pipelined and non-pipelined system to execute 100 instructions. (If an average, every five cycles, a bubble due to data hazard has to be introduced in the pipeline.).
(A) 5
(B) 4.03
(C) 4.81
(D) 4.17

Comments

I have solved this question and got answer as 4.03 but also got a different method at a site and that method also looks right that method gives answer as 4.17 I will attach both solutions. i.e mine method and the solution from a site.
Mine method. : Mine Solution
other solution where answer is 4.17 : other Solution
site to other solution : ques - solutions adda

Answer 1

The formula for speedup of a non pipelined system with a pipelined system: $\frac{k*n}{k+(n-1)}$,
where k is the number of stages=5 and n is the number of instructions. (this formula does not take in account the hazards) for that we have to add 20% of total instructions worth cycles more in the formula because for every 5^th clock cycle we have to create a bubble.

New formula: $\frac{k*n}{k+(n-1)+\frac{n}{5}}$, now put k=5 and n=100. You get
speedup= $\frac{500}{(5+99+20)}$=$\frac{500}{124}$=4.03

But when sometimes we are not given the value of number of instructions, so we take ‘n’ to be infinity.
taking ‘n’ common in numerator and denominator: $\frac{n(5*k)}{n(\frac{5*k}{n}+6+\frac{(-5)}{n})}$ = $\frac{5*k}{6}$
(since n>>k, (k/n)-->0) = 25/6 = 4.16667 ≈ 4.17