SOLUTION OF SAMPLE EXERCISES IN CHAPTER 2
2.4
We are given the number of cycles per second and the number of seconds, so we
can calculate the number of required cycles for each machine. If we divide this by the
CPI we’ll get the number of instructions.
For M1, we have
3 seconds
×
200
×
10
6
cycles/second = 600
×
10 cycles per program / 10 cycles per
instruction
= 60
×
10
6
instructions per program.
For M2, we have
4 second
×
300
×
10
6
cycles/second = 1200
×
10
6
cycles per program / 9.4 cycles per
instruction = 127.7
×
10
6
instructions per program.
2.9
We do this problem by finding the amount of time that program 2 can be run in an
hour and using that for executions per second, the throughput measure.
seconds
seconds
3600
200
hour
Execution of P1
Executions of P2 per hour=
seconds
Execution of P2
−
seconds
3600
200
100
1600
hour
Executions of P2 per hour on M1=
53
33
−×
=
=
seconds
3600
200
5
2600
hour
Executions of P2 per hour on M2=
650
44
==
With performance measured by throughput for program 2, machine M2 is
650
533
=1.2 times
faster than M1. The costeffectiveness of the machines is to be measured in units
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 Spring '09
 Jame
 Ratio

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