(4 pts) We are comparing the performance of two different microprocessors, M1 an
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(4 pts) We are comparing the performance of two different microprocessors, M1 and M2. Assume a program P will be compiled into M2 instructions counting 20% more than M11 instructions. M1 has a 2 GHz clock rate while M2 has a 3 GHz clock rate. Instructions on these two machines can be classified into 3 classes as shown in the following table. Which machine has better MIPS (show MIPS for both machine)? Which machine has better performance (calculate better by how much percent)? Justify your answers 4. Instruction Class Cvcles needed Instruction Frequency M1 M2 25% 35% 40% 4Explanation / Answer
As the technology director for XYZ Corporation, you have been asked to evaluate two different computer systems, Machine 1 (M1) and Machine 2 (M2) which are being considered for purchase. You learn the following specifications about M1 and M2:
-----CPI-----
Instr. class M1 M2
A 2 5 M1 operates at 1.6GHz and M2 at 2.0GHz
B 3 1
C 4 2
D 2 5
a) Assuming an even workload, what are the CPI’s for these two machines?
b) You decide to compare performance by measuring the execution times of a program on both M1 and M2. The program will execute 1 billion instructions, distributed evenly over the 4 classes. Predict the execution times T1 and T2. Which machine will be faster, and by how much?
c) Later, the sales people for C2, the company that makes M2, tell you that they have achieved a speedup on M2 of 1.182 above its previous performance. If you know that they achieved this solely by improving the performance of the class A instructions, what must the new CPI of M2 be for class A instructions, in order to achieve such a speedup?
a) Since the workload is even, 25% of instructions are class A, 25% are class B, etc. CPIM1 = 0.25*2 + 0.25*3 + 0.25*4 + 0.25*2 = 2.75
CPIM2 = 0.25*5 + 0.25*1 + 0.25*2 + 0.25*5 = 3.25
b) TM1 = 109 * 2.75 * 1/(1.6 * 109) = 1.71875 seconds
TM2 = 109 * 3.25 * 1/(2.0 * 109) = 1.625 seconds
M2 is faster, it is 1.71875/1.625 times faster, or 1.0577 times faster
c) Speedup = 1.182 = Old TM2/New TM2 = Old CPIM2/New CPIM2 (since # of instructions and clock rate remain the same)
=> New CPIM2 = Old CPIM2 / 1.182 = 3.25/1.182 = 2.75
2.75 = 0.25*CPIA + 0.25*1 + 0.25*2 + 0.25*5 => 0.75 = 0.25*CPIA
New CPIA = 3 for M2, in order to achieve the given speedup.
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