Assume that to spell check a large file, 820,000,000 instructions are needed. The instructions in the program are broken down into 4 different classes, and each class requires N clock cycles to execute. Specific information is given in the table below. (Here, N is the same as in the MIPS multi-cycle datapath discussed in class.)

Instruction class Clock cycles per instruction Number of instruction
Branch 3 150,000,000
Store 4 185,000,000
Load 5 260,000,000
ALU / R-type 4 225,000,000

If the total execution time for this program is found to be 1.57 seconds, what is the clock cycle time of the computer on which it was run?

(A) 1.6 GHz
(B) 1.82 GHz
(C) 2.16 GHz
(D) 3.32 GHz


Answer: (C)

Explanation: We can apply the CPU time formula: CPU time = IC x CPIavg x CC Time, and solve for CC Time.
820,000,000 * {3*(150,000,000 / 820,000,000) + 4*(185,000,000 / 820,000,000) + 5*(260,000,000 / 820,000,000) + 4*(225,000,000 / 820,000,000)} × N = 1.57s

Thus, the clock cycle time is 4.63 x 10-10s.

Which is equal to,

1 / (4.63 x 10-10) 
= 0.215982721 x 1010
= 2.15982721 x 109 cycle per second
= 2.16 GHz 

This is a 2.16 GHz machine.

So, option (C) is correct.


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  • Last Updated : 26 Dec, 2018

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