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Computer Architecture A Quantitative Approach by John L. Hennessy
Experiment Findings · September 2020
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Assignment_01
Date: OCT 18, 2019
Course: ACA
Submitted by:
Student Name
Students Reg. ID
Email
Mohsan Ali
181484
181484@students.au.edu.pk
Ayesha Ejaz
181482
181482@students.au,edu.pk
Submitted To:
Dr. Shaharyar Kamal (Assistant Professor Au.edu.pk)
Advanced computer architecture-Assignment_01
2|Page
Contents
Case study 1:
1.
Chip Fabrication Cost ......................................................................................................................3
Question 1.1 .....................................................................................................................................................3
Statement: ...................................................................................................................................................3
Solution: ......................................................................................................................................................3
2.
Question 1.2 .....................................................................................................................................................4
Statement: ....................................................................................................................................................4
Solution: .......................................................................................................................................................4
Case Study 2: Power Consumption in Computer Systems .............................................................................................7
3.
Question 1.4 .....................................................................................................................................................7
Statement: .....................................................................................................................................................7
Solution:........................................................................................................................................................7
4.
Question 1.6 .....................................................................................................................................................8
Statement: ....................................................................................................................................................8
Solution: ........................................................................................................................................................9
Exercise ..........................................................................................................................................................................9
5.
Question 1.9 .....................................................................................................................................................9
Statement: .....................................................................................................................................................9
Solution:........................................................................................................................................................9
6.
Question 1.10 .................................................................................................................................................10
Statement: ..................................................................................................................................................10
Solution: ......................................................................................................................................................10
7.
Question 1.11 .................................................................................................................................................12
Statement: ...................................................................................................................................................12
Solution:......................................................................................................................................................12
8.
Question 1.13 .................................................................................................................................................14
Statement: ..................................................................................................................................................14
Solution: .....................................................................................................................................................14
9.
Question 1.15 .................................................................................................................................................15
Statement: ...................................................................................................................................................15
Solution:......................................................................................................................................................15
10.
Question 1.16 .............................................................................................................................................16
Statement: ....................................................................................................................................................16
Solution:.......................................................................................................................................................16
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Case study 1:
Chip Fabrication Cost
1. Question 1.1
Statement:
<Table 1> gives the relevant chip statistics that influence the cost of several
current chips. In the next few exercises, you will be exploring the effect of different
possible design decisions for the IBM Power5.
Table 1: Manufacturing cost factors for several modern processors
Chips
Die
size(mm2)
IBM Power5
Sun Niagara
AMD Opteron
389
380
199
Estimated
defect rate
(per cm2)
.30
.75
.75
Manufacturing
size (nm)
Transistors
(million)
130
90
90
276
279
233
Solution:
A. What is the yield for the IBM Power5?
Answer:
Given data:
Defects per area= 0.30cm2
Die size or die area= 389 mm2
N=4
Yield =0.36approximately
B. Why does the IBM Power5 have a lower defect rate than the Niagara and Opteron?
Answer:
There are many reasons can be considered to do this question but the company
itself matter a lot in production of IBM Power5. It is possible that the age of IBM
Power5 firm more than the age of Niagara and Opteron. The die age also matter a lot. As
the die age increase the chance of defects in the die decrease.
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2. Question 1.2
Statement:
It costs $1 billion to build a new fabrication facility. You will be selling a range of
chips from that factory, and you need to decide how much capacity to dedicate to each chip. Your
Woods chip will be 150 mm2 and will make a profit of $20 per defect-free chip. Your Markon chip will be
250 mm2 and will make a profit of $25 per defect-free chip. Your fabrication facility will be identical to
that for the Power5. Each wafer has a 300 mm diameter.
Solution:
A. How much profit do you make on each wafer of Woods chip?
Answer:
Given information:
Markon-profit per defect-free chip= $25
Woods-Profit per defect-free chip=$20
Each Wafer diameter=300mm
Woods Die Area= 1.5cm2
Markon Die Area=2.5cm2
The given below formula can be used to get required answer
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BY computing LCM
By re-arranging the equation
Profit of Phonix chip= Yield * dies per wafer * price of each wafer
Profit of each wafer= 0.65 * 416* 20$
Profit of each Woods Wafer= 5408$
B. How much profit do you make on each wafer of Markon chip?
Answer:
perform same task as above for the Markon chip
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BY computing LCM
By re-arranging the equation
Profit of Markon chip= Yield * dies per wafer * price of each wafer
Profit of each wafer of Markon= 0.50 * 240* 25$
Profit of each wafer of Markon = $3000
C. Which chip should you produce in this facility?
Answer: there are different parameters to produce the best chips. According to the defect
rate, yield and profit Woods chip is most suitable for more benefits.
D. What is the profit on each new Power5 chip? If your demand is 50,000 Woods chips
per month and 25,000 Markon chips per month, and your facility can fabricate 150
wafers a month, how many wafers should you make of each chip?
Answer:
We have two kind of chips one is Markon and one is Woods chips.
As earlier we know that the number of dies per wafer in Markon and Woods
Number of dies per wafer in Woods= 416
Number of dies per wafer in Markon= 240
Wafers needed for Woods:
= total chips per month/chips per wafer
= 50000/416 =120 wafers
Wafers needed for Markon:
= total chips per month/ dies per wafer
=25000/ 240= 104 wafers
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The above statistics shows that if Woods Company wants to make 50000
chips/dies a month then 416 dies/chips needed to make one wafer so total number
of wafers that can be made with 50k dies are 120. The same manner we can
explain the Markon Company.
Case Study 2: Power Consumption in Computer Systems
3. Question 1.4
Statement:
Figure 1.23 presents the power consumption of several computer system
components. In this exercise, we will explore how the hard drive affects power
consumption for the system.
Solution:
A. Assuming the maximum load for each component, and a power supply efficiency of
80%, what wattage must the server’s power supply deliver to a system with an Intel
Pentium 4 chip, 2 GB 240-pin Kingston DRAM, and one 7200 rpm hard drive?
Answer:
Given information:
Peak power of intel P4=66
Power Supply efficiency=80%
1gb Kingston D2N3 requires 240-pin and 2.3W
For 2GB Kingston D2N3 (2.3*2)
7200RPM hard drive needs 7.9W to run smoothly
Power efficiency of load= sum of all the loads
0.8x=(2.3*2)+66+7.9
x=4.6+66+7.9
x=98.125
This is the required power supply for all the three devices.
B. How much power will the 7200 rpm disk drive consume if it is idle roughly 60% of the
time?
Answer:
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The given information first:
7200rpm hard drive consumes 4.0W in Idle situation
Now find the power consumed by 7200rpm in idle and running condition:
60% when idle+ 40 % when busy= (0.6* 4W+ 0.4*7.9W)
=3.16+2.4
Total power Consumed=5.56
C. Given that the time to read data off a 7200 rpm disk drive will be roughly 75% of a
5400 rpm disk, at what idle time of the 7200 rpm disk will the power consumption be
equal, on average, for the two disks?
Answer:
Given Information:
The time to read data off a 7200 rpm disk drive will be roughly 75% of a 5400 rpm disk
seek7200 = .75 × seek5400
The total seek and idle 7200rpm
seek7200 + idle7200 = 100
The total seeks and idle 5400rpm
seek5400 + idle5400 = 100
At what idle time of the 7200rpm
seek7200 × 7.9W + idle7200 × 4W = seek5400 × 7W + idle5400 × 2.9W
idle7200 = 29.8%
4. Question 1.6
Statement:
Figure 1.24 gives a comparison of power and performance for several
benchmarks comparing two servers: Sun Fire T2000 (which uses Niagara) and IBM x346
(using Intel Xeon processors). This information was reported on a Sun Web site. There
are two pieces of information reported: power and speed on two benchmarks. For the
results shown, the Sun Fire T2000 is clearly superior. What other factors might be
Advanced computer architecture-Assignment_01
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important and thus cause someone to choose the IBM x346 if it were superior in those
areas?
Solution:
There are number of options available to evaluate the servers. Few of them is listed below





Reliability of server
Benchmark for comparison
Space for deployment
Physical structure
Price of server
If we were running applications that didn’t have any benchmark for characteristics then
IBMx346 might be faster. At the same time we don’t know those characteristics which
are unknown to us to decide about the IBMx346 is either good or bad.
Exercise
5. Question 1.9
Statement:
You are designing a system for a real-time application in which specific deadlines
must be met. Finishing the computation faster gains nothing. You find that your system can
execute the necessary code, in the worst case, twice as fast as necessary.
Solution:
A. How much energy do you save if you execute at the current speed and turn off
the system when the computation is complete?
Answer:
This is the natural phenomena that if after completing the task system
should be turned off to save the resources provide the benefits directly or indirectly.
According the statement in question:
Let’s suppose our system complete the whole task in half time because it is
twice faster. So you can save 50% of the energy in worst time. So, in best and
average time you also can save the 50% energy.
Energy Saving=50%
B. How much energy do you save if you set the voltage and frequency to be half as
much?
Answer: The original energy equation is given below
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We didn’t use the frequency term because changing the frequency does not
affect energy but only power.
So the new energy is
Now we evaluate the effect of power term
(
)
According to above equation Energy saved is ¼ of the old energy.
6. Question 1.10
Statement:
Server farms such as Google and Yahoo! provide enough compute capacity for the
highest request rate of the day. Imagine that most of the time these servers operate at only 60%
capacity. Assume further that the power does not scale linearly with the load; that is, when the
servers are operating at 60% capacity, they consume 90% of maximum power. The servers could
be turned off, but they would take too long to restart in response to more load. A new system has
been proposed that allows for a quick restart but requires 20% of the maximum power while in
this “barely alive” state.
Solution:
A. How much power savings would be achieved by turning off 60%
of the servers?
Answer:
The old power consumption is:
To find the old power let’s suppose power is represented by P and number
of available servers are N.
Then 90% of the servers were using the maximum power is
Old power=0.9NP
If we reduce the power consumption from 0.9NP to 0.4NP because 60% of servers
are at rest.
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Then 55% power approximately can be saved.
B. How much power savings would be achieved by placing 60% of the servers in
the “barely alive” state?
Answer:
The power consumption is given below:
The description of the variable used in this part is explained in a part of this
question.
C. How much power savings would be achieved by reducing the voltage by 20%
and frequency by 40%?
Answer:
The voltage is going to be decrease by 20% and remaining voltage is 80%
and the frequency is going to be decrease by 40% and remaining frequency will be
60% or 0.6.
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D. How much power savings would be achieved by placing 30% of the servers in
the “barely alive” state and 30% off?
Answer:
Let’s assume that other 40% the system have to operate with full capacity
and consume the 90% of maximum power
The power saving in new system can be calculated that 30%(0.3N) of servers are off
and 30% are in barely state. And second term shows that 40% of the systems are running with 90% of
maximum power.
0.3N*0.3P + 0.4N*0.9P= 0.45NP
Let’s power consumption suppose is equal to 0.9NP.
Power saving = (0.9NP – 0.45NP) / 0.9NP  0.309= 30.9%
7. Question 1.11
Statement:
Availability is the most important consideration for designing servers,
followed closely by scalability and throughput.
Solution:
A. We have a single processor with a failure in time (FIT) of 100. What is the
mean time to failure (MTTF) for this system?
Answer:
Given information:
Failure time= 100
MTTF=?
As a metric, MTTF represents how long a product can reasonably be expected
to perform in the field based on specific testing.
109 represent billion. Generally reported as failures per billion hours of
operation or Failure in Time (FIT)
MTTF=107
B. If it takes one day to get the system running again, what is the availability of the
system?
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Answer: the calculated mean time to failure is 107 then we can calculate the
availability of the system.
107/107 + 24 = 1 failure
Mean Time to Failure is a reliability measure; the reciprocal of MTTF is a rate of
failures. Generally reported as failures per billion hours of operation or Failure In
Time (FIT).
C. Imagine that the government, to cut costs, is going to build a supercomputer out
of inexpensive computers rather than expensive, reliable computers. What is the
MTTF for a system with 1000 processors? Assume that if one fails, they all fail.
Answer: MTTF have a complete focus on the hardware side rather than software and
applications of the system. It is compulsory that if you are going to develop the large
system on which a city or country is relying. You must focus on the mean time
failure. The availabity is the main thing in the development of any system.if a system
contains 1k processors then the system failure time will be less than the system
which has more nodes installed.the below graph shows the relationship between
them.
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8. Question 1.13
Statement:
Your company is trying to choose between purchasing the Opteron or Itanium 2. You
have analyzed your company’s applications, and 60% of the time it will be running
applications similar to wupwise, 20% of the time applications similar to ammp, and 20%
of the time applications similar to apsi. Fig. 1.17
Solution:
A. If you were choosing just based on overall SPEC performance, which would you
choose and why?
Answer: the wupwise application consumes less energy than the apsi and ammp.
Similarly there are other attributes on the basis which we can decide about Opteron
or Itanium 2. The SPEC table shown in this question tells us that the execution time
of Itanium2 is less than the Opteron. One more point is that the SpecRatio of
Opteron is (20.87) is less than the Itanium 2(27.8). So, Itanium 2 is the best option.
B. What is the weighted average of execution time ratios for this mix of
applications for the Opteron and Itanium 2?
Answer: keeping in view the main statement of question which is telling about 60%,
10% and 30% of different applications.
The formula to compute the weighted average of execution of time ratios for
this mix of applications is:
0.6*(
) + 0.1*(
0.6*(
)+0.1(
) + 0.3*
)+0.3(
)
The weighted average execution time ratio is = 0.649
Weighted time=64.9%
C. What is the speedup of the Opteron over the Itanium 2?
Answer: to calculate the speedup of Opteron over the Itanium we need the
weighted execution time of both applications.
The weighted execution time for the Opteron is:
Keep in view the main statement of question which is telling
about 60%, 10% and 30% of different applications.
The formula to compute the weighted average of execution of time ratios for
this mix of applications is:
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0.6*(
) + 0.1*(
) + 0.3*
0.6*(51.5)+0.1(136)+0.3(150)
The weighted average execution time for Opteron = 89.5 seconds
Similarly calculate weighted average time for the Itanium 2
0.6*(
) + 0.1*(
) + 0.3*
0.6*(56.1)+0.1(132)+0.3(230)
The weighted average execution time for Itanium = 116.16 seconds
Now,
Speed Up of Opteron Over the Itanium 2=
=
Speed Up of Opteron Over the Itanium 2= 1.30 Approxi.
9. Question 1.15
Statement:
Assume that we make an enhancement to a computer that improves some mode of
execution by a factor of 10. Enhanced mode is used 50% of the time, measured as a percentage of
the execution time when the enhanced mode is in use. Recall that Amdahl’s law depends on the
fraction of the original, unenhanced execution time that could make use of enhanced mode. Thus,
we cannot directly use this 50% measurement to compute speedup with Amdahl’s law.
Solution:
A. What is the speedup we have obtained from fast mode?
Answer: if we enhance the execution by 10 factors. This means
50% is enhanced mode will be 500%
Without enhancement, the accelerated phase would have taken just as long(50%) ,
but the accelerated phase would take 10 times as long i.e 500%. So the relative
execution time without the enhancement would be
50%+500%=550%
Therefore,
Overall speed up is:
=
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Overall speed up=5.5
B. What percentage of the original execution time has been converted to fast
mode?
Answer: the fast mode is also called accelerated.
Amdahl’s law can be used to construct equation for this problem
The 90.9% speedup is achieved by turning the accelerated on.
10.Question 1.16
Statement:
When making changes to optimize part of a processor, it is often the case that
speeding up one type of instruction comes at the cost of slowing down something else. For
example, if we put in a complicated fast floating point unit, that takes space, and something might
have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay
to reach that unit. The basic Amdahl’s law equation does not take into account this trade-off.
Solution:
A. What is the speedup If the new fast floating-point unit speeds up floating-point
operations by, on average, 2×, and floating-point operations take 20% of the
original program’s execution time, what is the overall speedup (ignoring the
penalty to any other instructions)?
Answer:
20% is original program’s execution time
Then remaining program execution is 80%
2 x speeds up for 20% is divide up
The overall speedup = 1 / (0.8 + 0.2 / 2) = 1/0.9
= 1.1 will be the speed up
B. What is the speedup Now assume that speeding up the floating-point unit slowed
down data cache accesses, resulting in a 1.5× slowdown (or 2/3 speedup). Data
Cache accesses consume 10% of the execution time. What is the overall speedup
now?
Answer:
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20% is original program’s execution time
Data cache access consumes 10% of the execution time
Then remaining program execution is 60%
2 x speeds up for 20% is divide up
The overall speedup = 1 / (0.6 + 0.2 / 2 + 0.1 × 1.5) = 1/0.9
= 1/(0.6+0.1+0.15)
=1/0.85
=1.17
C. After implementing the new floating-point operations, what percentage of
execution time is spent on floating-point operations? What percentage is spent
on data cache accesses?
Answer:
There are two questions in the above statement one is for floating-point operations
and one is for cache access operation.
First we compute:
The percentage of execution time spent on floating-point operations is
(0.3 / 2) / (0.6 + 0.3 / 2 + 0.1 × 1.5) = 0.167 = 16.7%
The percentage spent on data cache accesses is
(0.1 × 1.5) / (0.6 + 0.3 / 2 + 0.1 × 1.5) = 0.167 = 16.7%
Advanced computer architecture-Assignment_01
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