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(COMP4611)[2011](f)midterm~dli^_78798.pdf
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COMP 4611 Design and Analysis of Computer Architectures
Fall Semester 2011
Midterm Examination
Date: 20 Oct. 2011 Time: 9:00 am - 10:20 am
Name:_____________________ Student ID: _____________Email:____________________
Instructions
1. Please write your name, student ID and email address on this page.

2. Please answer all the questions within the space provided on the examination paper. You may use the back of the pages for your rough work.

3. This paper consists of 5 questions and 7 pages.

4. Please read each question very carefully and answer the questions carefully and clearly to the point. Make sure your answers are neatly written, legible and readable.

5. Show all the steps used in deriving your answer, wherever appropriate.



Question
Points
Score

1
10


2
20


3
20


4
20


5
30


Total
100







1. (10 points) Amdahl's Law. A given benchmark has a floating-point computation phase and an integer computation phase. To speed up the execution of this benchmark the floating-point performance of the processor has been enhanced by a factor of 10. With this enhancement, the benchmark spends 50% of its total execution time in its floating-point phase.

a) Calculate the speedup achieved by using the processor's floating-point enhancement.


Let T be the execution time taken for the benchmark execution with the enhancement
T/2 is the time the floating point unit is used in the enhanced version
Since the speed-up S for floating point unit is 10
Original execution time taken for the floating-point operations = T/2*S = T/2*10 = 5T
Original execution time taken by the benchmark = T/2 + 5T = 11T/2

Overall speed-up = Original Execution time / Enhanced execution time
= 11T/2 / T = 11/2 = 5.5









b) Calculate the original percentage of the benchmark's floating-point computation.




Original percentage of floating point computation = 100*5T / (11T/2)
= 1000/11
= 90.91%








2. (20 points.) A 1 GHz pipelined processor takes 1 cycle for an ALU instruction but all other instructions take two cycles. A certain benchmark executes 900 million instructions out of which 44% are ALU instructions.

a) Calculate the MIPS rating of the processor for this benchmark and the total execution time.


ALU Instructions = 0.44*900 = 396 million
non-ALU instructions = 504 million
MIPS Rating:
CPI = 1 * 0.44 + 2 * 0.56 = 1.56
MIPS Rating = 109 / (1.56 * 106) = 641.0256


Total Execution Time:
Execution Time = IC * CPI / Clock rate = 900*106*1.56 / 109 = 1.404 s



b) An optimized compiler has been used for the above benchmark and it managed to reduce the number of executed ALU instructions by 25% and the number of the other instructions by 12.5%. Repeat the calculations in a) for this case.


New ALU instructions = 0.75 * 396 = 297 M
New non-ALU in