Question

Assume that for a program, compiler A results in a dynamic instruction count of 8.0E8 and has an execution time of 2.4 s, while compiler B results in a dynamic instruction count of 1.25E9 and an execution time of 1.8 s. a. [10] Find the average CPI for each program given that the processor has a clock cycle time of 2 ns. b. [10] Assume the compiled programs run on two different processors. If the execution times on the two processors are the same, and the clock rate of processor A is 4GHz, find the clock rate of processor B? c. [10] A new compiler is developed that uses only 6.0E8 instructions and has an average CPI of 1.25. What is the speedup of using this new compiler versus using compiler A or B on the original processor?

Answer 1

CPU clock Cycle as CPI = Instruction Count COSA = (a tomas ) (xrom) = 1.5 che = ( 1.88 V 2x1018 / 5 ) Sibskie) I 1:25*109 - 0.72 C Scanned with CamScanner

by Processor 1 is running als code. Processor 2 95 running B's code. Given that execution times are some. Execution Time = ICA * CPIA X Clock Cycle Time Execution Time = Icg * CPIE * Clock Cycle Time so, Clock Cycle Time . Ilg X CPIB Clock Cycle Time ICA XCPIA 1.25x109x0.72 8X108 x 1.5 = 0.75 is the clock of processor 1 Ys 25% faster than - the clock of processor a. Given, Clock rate of A => HGHZ CS Scanne then clock rate of B - 3GHZ CamScanner

> Let 'Ċ be the new compiler. Frec Time = IC X CPI* CPU clock Time The amount by which c's Code 95 faster as given b48 With A Performance Execution Time A Performance A Execution Time = 8x108x1.5 6x 108 x 1.25 = 1.6 Thus, c's code is about 16 times Cs Scanned with a's code. CamScannere faster than