Part II: Solutions Guide
55
2.1
For program 1, M2 is 2.0 (10/5) times as fast as M1. For program 2, M1 is 1.33 (4/3)
times as fast as M2.
2.2
Since we know the number of instructions executed and the time it took to execute
the instructions, we can easily calculate the number of instructions per second while
running program 1 as (200
´
10
6
)/10 = 20
´
10
6
for M1 and (160
´
10
6
)/5 = 32
´
10
6
for
M2.
2.3
We know that Cycles per instruction = Cycles per second / Instructions per sec-
ond. For M1 we thus have a CPI of 200
´
10
6
cycles per second / 20
´
10
6
instructions
per second = 10 cycles per instruction. For M2 we have 300/32 = 9.4 cycles per instruc-
tion.
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
cy-
cles/second = 600
´
10
6
cycles per program / 10 cycles per instruction = 60
´
10
6
in-
structions per program. For M2, we have 4 seconds
´
300
´
10
6
cycles/second = 1200
´
10
6
cycles per program / 9.4 cycles per instruction = 127.7
´
10
6
instructions per pro-
gram.
2.5
M2 is twice as fast as M1, but it does not cost twice as much. M2 is clearly the ma-
chine to purchase.
2.6
If we multiply the cost by the execution time, we are multiplying two quantities,
for each of which smaller numbers are preferred. For this reason, cost times execution
time is a good metric, and we would choose the machine with a smaller value. In the
example, we get $10,000
´
10 seconds = 100,000 for M1 vs. $15,000
´
5 seconds = 75,000
for M2, and thus M2 is the better choice. If we used cost divided by execution time and
assume we choose the machine with the larger value, then a machine with a ridiculous-
ly high cost would be chosen. This makes no sense. If we choose the machine with the
smaller value, then a machine with a ridiculously high execution time would be cho-
sen. This too makes no sense.
2.7
We would define cost-effectiveness as performance divided by cost. This is essen-
tially (1/Execution time)
´
(1/Cost), and in both cases larger numbers are more cost-
effective when we multiply.
2.8
We can use the method in Exercise 2.7, but the execution time is the sum of the two
execution times.
So M1 is slightly more cost-effective, specifically 1.04 times more.
2
Solutions
Executions
per
second
per
dollar
for
M1
1
13 10,000
´
----------------------------
1
130,000
-------------------
==
Executions
per
second
per
dollar
for
M2
1
9 15,000
´
-------------------------
1
135,000
-------------------
==
