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Inessa [10]
4 years ago
11

Consider two different implementations of the same instruction set architecture. the instructions can be divided into four class

es according to their cpi (classes a, b, c, and d). p1 with a clock rate of 2.5 ghz and cpis of 1, 2, 3, and 3, and p2 with a clock rate of 3 ghz and cpis of 2, 2, 2, and 2. given a program with a dynamic instruction count of 1.0e6 instructions divided into classes as follows: 10% class a, 20% class b, 50% class c, and 20% class d, which is faster: p1 or p2?
Mathematics
2 answers:
Lelechka [254]4 years ago
8 0
Class A: 10^5 instr. Class B: 2 x 10^5 instr. Class C: 5 x 10^5 instr. Class D: 2 x 10^5 instr.
Time = no instr x CPI/ Clock rate
Total time P1 = (10^5 + 2 x 10^5 x 2 + 5 x 10^5 x 3 + 2 x 10^5 x 3) / (2.5 x 10^9) = 10.4 x 10^-4s
Total Time P2 = (10^5 x 2 + 2 x 10^5 x 2 + 5 x 10^5 x 2 + 2 x 10^5 x 2) / (3 x 10^9) = 6.66 x 10^-4s
CPI (P1) = 10.4 x 10^-4 x 3 x 10^9/10^6 = 2.6
CPI (P2) = 6.66 x 10^-4 x 3 x 10^9/10^6 = 2.0
Therefore, P2 is faster.
Reptile [31]4 years ago
8 0
<span>p2 is the faster processor. Since we're executing 1 million instructions, let's first determine how many instructions of each class we execute. Just multiply 1 million by the percentage of each instruction class. Class a = 0.10 * 1,000,000 = 100,000 Class b = 0.20 * 1,000,000 = 200,000 Class c = 0.50 * 1,000,000 = 500,000 Class d = 0.20 * 1,000,000 = 200,000 Now let's see how many clock cycles p1 takes to execute the program. This is the sum of the product of each instruction class and the clock cycles per instruction. 100,000 * 1 + 200,000 * 2 + 500,000 * 3 + 200,000 * 3 = 100,000 + 400,000 + 1,500,000 + 600,000 = 2,600,000 And the number of clock cycles for p2: 100,000 * 2 + 200,000 * 2 + 500,000 * 2 + 200,000 * 2 = 200,000 + 400,000 + 1,000,000 + 400,000 = 2,000,000 Finally, how long does each program actually take to execute? Just divide by the clock frequency. p1: 2,600,000 / 2,500,000,000 = 0.00104 seconds p2: 2,000,000 / 3,000,000,000 = 0.000666666666666667 seconds And p2 is significantly faster than p1. In fact, p2 would still be faster than p1 even if the clock speeds were reversed with p1 being 3 GHz and p2 being 2.5 GHz. But with p2 having that higher clock speed, that's just icing on the cake.</span>
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