1) Consider the data of this table of CPUs running a bench mark program and comp

Question

1) Consider the data of this table of CPUs running a bench mark program and compute and fill in the missing entries in the table. MIPS = millions of instructions per second. Benchmarks compile differently depending on the processors's ISA. Fill out the empty cells with answers to questions below.

(a) find the execution time of each processor

(b) Find the mips rating of each processor

(c) find the perfromace ratio of P2 to that of P1 (remember: Performance = 1/ Execution time)

2) a program of 2*10^9 instructions has 40% arithmetic instructions. A new arithmetic unit used in the new CPU architecture is 15% faster. Using Amdahl's law, what is the speed up of the new machine.

speed up = performance after modification /performance before modification

Processor Instructions clockrate CPI (a) Execution time (sec) (b) MIPS rating (c) R = p2/p1 #1 10*10^9 4GHz 1.5 #2 8*10^9 3GHz 1.2

Explanation / Answer

1)

(a)
Execution time = (No of instructions) * (clock cycle) * (CPI)

processor 1:

execution time(Sec) = (10^10) * (1/(4*10^9)) * (1.5) = 3.75 sec

processor 2:
execution time(Sec) = (8*10^9) * (1/(3*10^9)) * (1.2) = 3.2 sec

(b)

MIPS rating = (Instruction count )/(execution time * 10^6)

processor 1:
MIPS rating = (10^10)/(3.75*(10^6)) =2666.67

processor 2:
MIPS rating = (8*10^9)/(3.2*(10^6)) =2500

(C)

performance = 1/(Execution time)

p1: performance = 1/3.75

p2:performance = 1/3.2

ratio of performance of p2 to that of p1 = p2/p1 = 3.75/3.2 = 1.171

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.