Show your work! Two machines A and B implement the same Instruction Set Architec

Question

Show your work! Two machines A and B implement the same Instruction Set Architecture(SA), which has three classes of instructions, 1, 2 and 3. Processor A is cheaper to produce. Both processors are clocked and at the same rate, 5GHz. The CPI for each machine A and B is given in the following table for each class of instructions: class 2 4 4 processor class 1 class 3 5 2 If the average mix of instructions e class 1 to 2 parts class 2 to 3 parts class 3: a) (6 points) what is the average CPI for each machine when running this program? b) (3 points) which machine is faster? c) (3 points) by how much? (You can express this as a ratio) An engineer has found a "trick" that enables the slower pr CPI for each class of instruction is unchanged. d) (3 points) What is the average CPI for the [previouslyl slower machine now? ) (3 points) Which machine is faster now? (6 points) By how much? (Again, you can express this as a ratio)

Explanation / Answer

Solution:

The first four subparts have been answered as per Chegg guidelines, please repost others.

Total number of instructions running the package= 5 + 2 + 3= 10

For machine A:

Average CPI= ((5/10) * 1) + ((2/10) * 4) + ((3/10) * 5)= 2.8

For machine B:

Average CPI= ((5/10) * 2) + ((2/10) * 4) + ((3/10) * 1)= 2.1

b)

Machine B is faster since they are tunning on the same clock frequency and CPI of B is less.

c)

speedup= 2.8/2.1= 1.333

d)

The CPI is still same.

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