GATE Overflow Test Series | Computer Organization and Architecture | Test 1 | Question: 19

Question

Suppose the functions $A$ and $B$ can be computed in $6$ and $4$ nanoseconds by functional units $U_A$ and $U_B$, respectively. Given two instances of $U_A$ and three instances of $U_B$, it is required to implement the computation $A(B(X_i))$ for $1\leq i \leq 100$. Ignoring all other delays, the minimum time required to complete this computation (in nanoseconds) is _______

Comments

10+49*6 = 304

Answer 1

$U_A$ is a greater value than $U_B$ so the pipeline will be bottle-necked by $U_A$
    
Since, we have two instances of $U_A$ and at least two instances of $U_B$ (third instance is not required here) they can operate in parallel and $100$ computations can be done in $50$ iterations.
    
For $50$ iterations $U_A$ will take = $50 \times 6 = 300$ ns
    
But as we can see in $A(B(X_i))$ for $U_A$ to start, it has to initially wait for output from $U_B$ which comes after $4$ ns. After this execution proceeds in a pipelined fashion. Therefore, minimum execution time $= 300 + 4 = 304$ ns

Comments

Sorry, my bad!

Thanks for clearing my doubt sir.

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In the test, it was 3,6 not 4,6

now the values/test series has been changed

please check

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As we can do parallely two computations, hence 304 ns

One doubt only I have, even if the third Ub is not given, i think still we will get 304ns only