Problem 1:

a) P1 : Clock rate 2 5 . GH2 P2 : Clock rate 3GH2 .

classes : CPI : classes : CPI :

A 1 A L

B 2 B 2

C 3 C L

D 3 D 2

CPIpy = 1x 0 1 . + 2x 0 2 . + 3x0 5 .
+ 3x0 2 . = 2 6
. .

CPIp = 2x0 .
1 + 2x0 2 .
+ 2x0 5 . + 2x0 2. = .
2

cycle (1)
b) instructions 10"x2 .
Clock
Cycle (P) of instructions
number CPI 10 . Clock xCPI
2 6
cycles number
of
= x =
= x .
=

execution time (P) Clock 106 2 6 04x10s Execution time (B) Clock
=
cycles =
x .

= 1 . =
1 Oh
. ms. =
cycles = 6 66
.
x 10-s = 0 66
. ms.

Clock ratio 2 .
5x109 Clock rate

PSP ( . 04 > 0 .

66) => Po is therefore faster

Problem 2 :

a) initial total time 250s

Y
= .

reduced total time = 70x8 8 . = 56s .

Atime-time reduced =
70-56 = 245 .

b) reduced total time = 250x0 8 . : 200s .

and
considering only INT operations were modified ,
then INT operations time was reduced
by 250-200 = 50s .

in order to reduce total time by 20 %, we have Hence , Since 40s < 50s then 0 it's
C) time for branch instructions is hos .
however ,
to reduce
by 50s .

,
.
impossible

Problem 3 :

A/p execution time for processor

·
p = C: execution time for processer = + h =

100 + h = 5h ·
p = 16 : :

100 + h = 10. 2

specclup of single processor-1 85 .

speed up of single processor- = 9 75.

relation 0 92. relation between actual

a ideal
185 :
between actual
a ideal =
1 05 = .
= .
0
.
=

execution time for processor execution time for processor

p=4
: :
·
·
: + h =1 p = 32 : + h =
7 . 14 o

speed up of single processor-103h speed up of single processor- = 14 . o

relation relation between actual

3 hh
actual ideal a
ideal th
between = =B.

execution time for processor execution time for processor

p-8 :+ 16 5 p = 64 :
10
· ·
: h = . : + h = 5 56.

speed up of single processor-1 = 6 0 . speed up of single processor

: = 17 97.

relation
Gob
between actual
a ideal relation between actual
ideal-
=
= 0 .
7. a = 0
.

p 120 execution time for processor

: 4 7
·
- : = .

speed up of single processor- = 20. 2

relation between actual

Cog
x ideal 0
.
=
=