An application executes 6.4 × 108 number of instructions in 6.3 seconds. There…
2025
An application executes 6.4 × 108 number of instructions in 6.3 seconds. There are four types of instructions, the details of which are given in the table. The duration of a clock cycle in nanoseconds is _________. (rounded off to one decimal place)
\(\begin{array}{|c|c|c|} \hline\text{Instruction type} & \text{Clock cycles required per} \\& \text{instruction (CPI)} & \text{Number of instructions executed} \\ \hline\text{Branch} & \text{2} & \text{$2.25 \times 10^{8}$} \\ \hline\text{Load} & \text{5} & \text{$1.20 \times 10^{8}$} \\ \hline\text{Store} & \text{4} & \text{$1.65 \times 10^{8}$} \\ \hline\text{Arithmetic} & \text{3} & \text{$1.30 \times 10^{8}$} \\ \hline\end{array}\)
Answer: 3 — Answer: 3.0 ns Steps: Compute cycles used by each instruction type: Branch: 2 cycles × 2.25 × 10^8 = 4.5 × 10^8 cycles Load: 5 cycles × 1.20 × 10^8 = 6.0 ×…
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Correct answer: 3
Answer: 3.0 ns
Steps:
Compute cycles used by each instruction type:
Branch: 2 cycles × 2.25 × 10^8 = 4.5 × 10^8 cycles
Load: 5 cycles × 1.20 × 10^8 = 6.0 × 10^8 cycles
Store: 4 cycles × 1.65 × 10^8 = 6.6 × 10^8 cycles
Arithmetic: 3 cycles × 1.30 × 10^8 = 3.9 × 10^8 cycles
Sum total cycles = 4.5 × 10^8 + 6.0 × 10^8 + 6.6 × 10^8 + 3.9 × 10^8 = 2.10 × 10^9 cycles
Clock period = total time / total cycles = 6.3 s / 2.10 × 10^9 = 3.0 × 10^-9 s = 3.0 ns
Rounded to one decimal place: 3.0 ns
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