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Compuetrr Architecture

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Subject
Computer Science
School
Harvard University
Type
Homework
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RUNNING HEAD 1
Title
Name
Institution
Course number: Course name
Instructor name
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RUNNING HEAD 2
Question 1:
For the non-pipelined CPU;
The average CPI = Total program execution time/ Instruction Count
Instruction count = (5*30 + 4*15 + 3*10 + 2*5 + 4*40)
= 410
Therefore, the average CPI= 410/ 100
=4.1
The average instruction execution time = Average CPI * clock cycle
Clock cycle = 1/f
=1/ 2* 10
8
= 5* 10
-10
So, IET = 4.1
* 5* 10
-10
= 2.05 * 10
-10
ns
For the pipelined CPU;
The average instruction execution time = clock cycle + overhead
=5* 10
-10
+ 0.1
= 0.1 ns
Speed up = average IET (non-pipelined) /average IET (pipelined)

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RUNNING HEAD 1 Title Name Institution Course number: Course name Instructor name Assignment due date RUNNING HEAD 2 Question 1: For the non-pipelined CPU; The average CPI = Total program execution time/ Instruction Count Instruction count = (5*30 + 4*15 + 3*10 + 2*5 + 4*40) = 410 Therefore, the average CPI= 410/ 100 =4.1 The average instruction execution time = Average CPI * clock cycle Clock cycle = 1/f =1/ 2* 108 = 5* 10-10 So, IET = 4.1 * 5* 10-10 = 2.05 * 10-10 ns For the pipelined CPU; The average instruction execution time = clock cycle + overhead =5* 10-10 + 0.1 = 0.1 ns Speed up = average IET (non-pipelined) /average IET (pipelined) RUNNING HEAD = 2.05 * 10-10 / 0.1 = 2.05 * 10-9 Question 2: RAW-type dependencies are: t1 of I2 to I5* t0 of I4 to I3# t0 of I4 to I3* t2 of I6 to I5# t2 of I6 to I5* t2 of I7 to I5# t2 of I7 to I6# It is impossible to solve by reordering. Question 3: AMAT = time for a hit + (miss rate x miss penalty) For level 3 cache, [H1*T1] + [(1-H1) *H2*T2] + [(1-H1) (1-H2) *H3*T3] + [(1-H1) (1-H2) (1-H3) *Hm*Tm] Where: H1- Hit rate L1 H2- Hit rate L2 H3- Hit rate L3 Hm- Hit rate main memory 3 RUNNING HEAD 4 T1- Hit time L1 T2- Hit time L2 T3- ...
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