Consider two different machines, with two different instruction sets, both of which have a clock rate of 200 mhz. the following measurements are recorded on the two machines running a given set of benchmark programs: instruction type instruction count (millions) cycles per instruction machine a arithmetic and logic 8 1load and store 4 3branch 2 4others 4 3 machine barithmetic and logic 10 1load and store 8 2branch 2 4others 4 3(a) determine the effective cpi, mips rate, and execution time for each machine. (b) comment on the results.

(A) CPIₐ = 2.22, MIPSₐ = 90, CPUₐ = 0.2 s

CPIₙ = 1.92, MIPSₙ, CPUₙ = 0.23 s

(B) Even though machine B has a higher MIPS than machine A, it needs a longer CPU time to execute the similar set of benchmark programs instructions.

Explanation:

To start with, we solve for CPI ∨ A,

Where ∨ = superscript

CPIₙ = Machine B, that is (ₙ = B),

Therefore,

a) CPIₐ  = Σ CPI ∨i × I ∨i ÷ I ∨c

= (8 × 1 + 4 × 3 + 2 ×4 + 4 × 3 ) × 10 ⁶ ÷ ( 8 +4 +2+4) × 10 ⁶

≈ 2.22

MIPSₐ = f / CPIₐ × 10 ⁶ = 200 × 10 ⁶ ÷ 2.22 × 10 ⁶

≈ 90

CPUₐ = I ∨c × CPIₐ ÷ f  

= 18 × 10 ⁶ × 2.2 ÷ 200 × 10 ⁶

= 0.2 s

CPIₙ = Σ CPI ∨i × I ∨i ÷ I ∨c

= (10 × 1 + 8  × 2 + 2 × 4 + 4 × 3) × 10 ⁶ ÷ (10 + 8+ 2 + 4) × 10 ⁶

≈ 1.92

MIPSₙ = f / CPIₙ × 10 ⁶ = 200 × 10 ⁶ / 1.92 × 10 ⁶

= 104

CPUₙ = I ∨c × CPIₙ ÷ f  

=24 × 10 ⁶ × 1.92 ÷ 200 × 10 ⁶

≈ 0.23 S

b) Even though machine B has a higher MIPS than machine A, it needs a longer CPU time to execute the similar set of benchmark programs instructions.

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explanation: