Help with performance and speed-up question related to number of cores and their areas

Question

There is a question in my exam that said:

Consider the following three processors (x, y, and z) that are all of varying areas. Assume that the single-thread performance of a core increases with the square root of its area.

1. Processor X, core area = A

2. Processor Y, core area = 4A

3. Processor Z, core area = 16A

You are given a workload where S fraction of its work is serial and 1 - S of its work is infinitely parallelizable.

a. If executed on a die composed of 16 Processor X's, what value of S would give a speedup of 4 over the performance of the workload on just Processor X?

b. Given a homogenous die of area 16A, which of the three processors would you use on your die to achieve maximal speedup? What is the speedup over just a single Processor X? Assume the same work load as in part b.

My progress with this part so far:

In an area of 16A, 4 cores of Y would fit. The square root of its area is $2\sqrt{A}$, which each of the four cores' performance would increase by a factor of two? Using Amdahl's law,

Overall speed-up for this choice $=\dfrac{1}{\dfrac{1 - S}{4(2)?} + S}$

And likewise, the overall speed-up for choosing 1 core of Z $=\dfrac{1}{\dfrac{1 - S}{1(4)?} + S}$

I am assuming that 1 core of area 4A is like 2 cores of area A, is this a valid assumption? For S = 0.2, 4 cores of Y would yield a higher speedup than one core of Z. Is this reasoning correct?

Answer 1

The problem in an exam is that you need the answer that is expected from you - something I’d have a real problem with. So keep this solution to yourself.

Speed up is always hard to calculate - calculating execution time avoids lots of fractions and is equivalent. Assume your task would take one second on one processor X.

Since S is serial, you need at least S seconds. In that time, your 15 remaining cores can do 15S worth of work. If 15S >= 1-S or S >= 1/16 then all work is done after S. if S <= 1/16 then all cores can work on the task simultaneously, so the work takes 1/16 time. So time = max(S, 1/16) and speed up = min (1/S, 16). Speed up = 4 when S = 1/4.

You can have four processors of type Y at twice the speed or 1 of type Z at four times the speed. The work takes half of max(S, 1/4), that is max (S/2, 1/8) on 4 processors of type Y and it takes 1/4 on one processor of type Z.

X is fastest when S <= 1/8. Y is fastest for 1/8 <= S <= 1/2, and Z is fastest for S >= 1/2.