Lecture Notes for Computer Architecture II - St. Cloud State University

Question 9
Yet another user has the following requirements for the machines discussed in Question
2.1: program 1 must be executed 200 times each hour. Any remaining time can be used
for running program 2. If the machine has enough performance to execute program 1 the
required number of times per hour, performance is measure by the throughput for
program 2. Which machine is faster for this workload? Which machine is more costeffective?
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Solution
3600 seconds = 1 hour
M1
P1 10*200 = 2000sec
P2 = 3600-2000
= 1600
M2
5*200 = 1000sec
Remaining time P2 = 3600-1000
= 2600
Number of times P2 on M1 = 1600/3
= 533
Number of times P2 on M2 = 2600/4
= 65
Cost effective M1 = 533/10000
= 0.0533
Cost effective M2 = 650/15000
= 0.0433
M1 is slightly better