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.

(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.