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.