Consider a processor designed with 14nm technology. It is designed to be instruc

Question

Consider a processor designed with 14nm technology. It is designed to be instruction-level-parallel (ILP) with the goal of 4 instructions completing every cycle: Ideal CPI= ¼. However, on a given program with 100,000 instruction count, the actual CPI is 0.5. The clock frequency of this processor is 2 GHz (0.5ns clock period). Now consider 7nm technology node. There are two design options- (1) increase the ILP to 8 – complete 8 instructions per cycle with ideal CPI of 1/8; Assume that for this program actual CPI is 0.25; the clock period however behaves CP (ns) = (1/448)*(technology feature size in nm)* (ILP)2. Hence for 14 nm technology, CP = (1/448)*(14)* (4)2 = 0.5ns. Compute the clock period for 7nm processor with ILP=8. Does the program speed-up?; (2) Take the extra transistors to carve out identical 4 cores with with ILP=4 each; clock period half as much that of 14 nm processor core; Does the program speed-up in this processor? Assume that the same program can be parallelized on 4 cores for a speed-up of 2. Which processor is faster for this program (1) or (2)?

Explanation / Answer

ANSWER:

1. CP7 = (1/448*)*(size of chip)*ILP2 = 1/448 * 7 * 8*8 = 1ns with ILP 8

time for 1 instruction in the 7nm chip = 1/8 ns

CP14 = 0.5 ns with ILP 4 .

time for 1 instruction in the 14 nm chip = 0.5 /4 = 1/8 ns

So,there is no speed up.

2. Clock period in this case = 0.5 / 2 ns = 0.25 ns. ILP = 4 , so time per instruction = 0.25/4 ns = 1/16 ns. For a program with 100,000 instruction count , total time (including the parallel speedup of 2) =

100000/(16*2) ns= 100000/32 = 3125 ns compared to the case (1) where total time for 100,000 instruction count = 100000/8 ns = 12500 ns.

So, Processor (2) is faster for this program

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