COMPUTER ORGANIZATION+DESIGN >I<
COMPUTER ORGANIZATION+DESIGN >I<
5th Edition
ISBN: 9781541868397
Author: Patterson
Publisher: ZYBOOKS (CC)
Expert Solution & Answer
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Chapter 1, Problem 1.7E

a.

Explanation of Solution

Given,

The instruction count f and execution time Texec of the compiler A is 1.0E9 and 1.1s respectively.

The instruction count f and execution time Texec of the compiler B is 1.2E9 and 1.5s respectively.

The time taken by the clock cycle is 1ns.

The average Cycles Per Instruction (CPI) for each program in a compiler can be calculated using the following formula is given below:

CPI=Texec×fNumberofinstructions

Substitute, “1×109” for “f”, “1.1” for “Texec” and “1.0×109” for “Number of instructions” is given below:

CPI=1.1×1×1091

b.

Explanation of Solution

Given,

The instruction count f and execution time Texec of the compiler A is 1.0E9 and 1.1s respectively.

The instruction count f and execution time Texec of the compiler B is 1.2E9 and 1.5s respectively.

The time taken by the clock cycle is 1ns.

The average Cycles Per Instruction (CPI) for each program in a compiler can be calculated using the following formula.

CPI=Texec×fNumberofinstructions

Substitute, “1×109” for “f”, “1.1” for “Texec” and “1.0×109” for “Number of instructions” is given below:

CPI=1.1×1×1091.0×109=1.1

Thus, the average CPI for each program of the compiler A is 1.1.

Substitute, “1×109” for “f”, “1.5” for “Texec” and “1.2×109” for “Number of instructions” is given below:

CPI=1.5×1×1091

c.

Explanation of Solution

The time required by the compiler can be calculated using the formula

Time=CPI×Numberofinstructionsf

Substitute, “1.1” for “CPI”, “6×108” for “number of instructions” and “1×109” for “f” in the above formula is given below:

Tnew=1.1×6×1081×109=0.66

Thus, the time required by the compiler with 1.1 CPI and 6×108 number of instructions is 0

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