1. CS 704 Advanced Computer Architecture
Lecture 3
Quantitative Principles … Cont’d
Design for Performance
Prof. Dr. M. Ashraf Chughtai
Welcome to the third lecture of he series on Advanced Computer Architecture.
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Lecture 3 - Performance... Cont'd

2. Lecture 3 - Performance... Cont'd
Today’s Topics
Recap
I/O performance
Laws and Principles
Performance enhancement
Concluding: quantitative principles
Home work
Summary
After a quick review of the previous two lectures on the computer design we will continue with the discussion on the quantitative principles of computer design.
An introduction to the computer processor performance, which is the key to the computer design for performance, has been the theme of the second lecture. Today we will talk about:
I/O performance measures
Laws and principles of performance measure
Computer performance enhancement
Concluding quantitative principles of computer design
Homework
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

3. Lecture 3 - Performance... Cont'd
Recap: Lecture 1-2
Computer architecture verses organization
Technological Developments
Computer design cycle
Performance metrics: time verses throughput
Price-Performance design
Benchmarks: Performance evaluation
Distinguishing between architecture and organization of processors we concluded that ‘the architecture of the members of a processor family are same whereas organization of same architecture may differ between different members of the family’
Technological developments, from vacuum tubes to VLSI circuits, dynamic memory and network technology gave birth to four different generations of computers.
In the computer design cycle, the decisive factors for rapid changes in the computer development have been the performance enhancements, price reduction and functional improvements
The processor performance of two designs is often compared by the factor n, which determines how much lower execution time one machine takes as compared to the other or how much faster the other machine is than first.
Time is the key measurement of performance. However, the throughput - number of tasks completed in specified time cannot be ignored.
A desktop user may define the performance of his/her machine in terms of time taken by the machine to execute a program; whereas a computer center manager running a large server system may define the performance in terms of the number of jobs completed in a specified time.
Price-Performance Design: The relationship between cost and price is complex one; and computer designers must understand this relationship as it effects the selling of their design.
The cost is the total amount spends to produce a product and the price is the amount for which a finished good is sold and it is controlled by the die yield and volume.
Growth in Processor Performance: The supercomputers and mainframes, costing millions of dollars and occupying excessively large space, prevailing by early 1970’s have been replaced with very low-cost microprocessor-based desktop computing machines in the form of personal computer (PC) and workstation massively parallel processing machines.
Benchmark is a program developed to evaluate the performance of a computer. Good products created when have: proficient benchmarks and expert ways to summarize performance
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Lecture 3 - Performance... Cont'd

4. Lecture 3 - Performance... Cont'd
Computer I/O System
Producer-Server model
Producer: the device that generates request to be serviced
Queue: the area where the tasks accumulate waiting to be serviced
Server: the device performing the requested service
Response Time: the time a task takes from the moment it is placed in the buffer to the time server finishes the task
Server
I/O device/
controller
Producer
Queue
Arrivals
departures
An I/O system works on the principle of producer-server model, which comprises an area, called queue, where the tasks accumulate waiting to be serviced and the device performing the requested service, called server.
Producer creates tasks to be processed and place them in a FIFO buffer – queue. The server takes the task form buffer and perform them
The response time is the time task takes from the moment it arrives in the buffer to the time the server finishes the task
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Lecture 3 - Performance... Cont'd

5. I/O Performance Parameters
Diversity: Which I/O device can connect to the CPU
Capacity: How many I/O devices can connect to the CPU
Latency: Overall response time to complete a task
Bandwidth: Number of task completed in specified time - throughput
The parameters diversity that refers to which I/O device and capacity means how many I/O devices can connect to the CPU are the I/O performance measures having no counterpart in CPU performance metrics.
In addition, the latency (response time) and bandwidth (throughput) also apply to the I/O system.
An I/O system is said to be in equilibrium state when the rate at which the I/O requests from CPU arriving, at the input of I/O queue (buffer) equals the rate at which the requests departs the queue after being fulfilled by the I/O device.
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Lecture 3 - Performance... Cont'd

6. Lecture 3 - Performance... Cont'd
I/O Transaction Time
The interaction time or transaction time of a computer is sum of three times:
Entry Time: the time for user to enter a command – average sec; from keyboard 4.0 sec.
System Response Time: time between when user enters the command and system responds
Think Time: the time from reception of the command until the user enters the next command
The interaction or transaction time of a computer is sum of:
Entry Time: the time for user to enter a command – average sec; from keyboard 4.0 sec.
System Response Time: time between when user enters the command and system responds
Think Time: the time from reception of the command until the user enters the next command
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Lecture 3 - Performance... Cont'd

7. Throughput verses Response time: Performance Measures .. Cont’d
| | | | | |
0% 20% 40% 60% 80% 100%
200 _
150 _
100 _
50 _ 20 
% of maximum throughput - bandwidth
Response time – latency ms
The minimum response time achieves only 10% of the throughput
The response time of 100% throughput takes 7-8 times the minimum response time
The knee of the curve is the area where a little more throughput results in much longer response time
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Lecture 3 - Performance... Cont'd

8. Response time and throughput calculation
Arrivals
Departures
If the system is in steady state, then the number of tasks entering the system must be equal to the number of tasks leaving the system
Little’s Law:
Mean number of tasks in system =
Mean response time x Arrival rate
The interaction or transaction time of a computer is sum of:
Entry Time: the time for user to enter a command – average sec; from keyboard 4.0 sec.
System Response Time: time between when user enters the command and system responds
Think Time: the time from reception of the command until the user enters the next command
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Lecture 3 - Performance... Cont'd

9. Little’s Law – A Little queuing theory
Mean number of tasks in the system = (Time accumulated) / (Time observe)
Mean response time =
(Time accumulated) / (Number tasks)
Arrival rate λ =
(Number tasks) / (Time observe)
The expression for mean number of task may be written as:
Time accumulated Timeaccumulated x Number tasks =
Time observe Number tasks Time observe
Mean number of tasks = mean response time x Arrival rate
Assume that we observe a system for time (Time observe) minutes and found that the:
number of task are completed in this time (Number task) and
the sum of the times each task spends in the system is (Time accumulated),
the arrival rate (λ) is the average number of arriving tasks/second; and
the mean response time is the ratio of Timeaccumulated and
number of tasks completed (Number task) during Time observe.
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Lecture 3 - Performance... Cont'd

10. Amdahl's Law
Suppose that enhancement E accelerates a fraction F of the task by a factor S, and the remainder of the task is unaffected
Original
Execution
Time of Task
Time
after fraction F
Enhanced by factor S
Execution time of the Fraction Enhanced
Time for Fraction F to be Enhanced by factor S
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Lecture 3 - Performance... Cont'd

11. Lecture 3 - Performance... Cont'd
Amdahl's Law
Speedup due to enhancement E:
Ex Time without E
Speedup (E) =
Ex Time with E
Performance with E
= Performance without E
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Lecture 3 - Performance... Cont'd

12. Lecture 3 - Performance... Cont'd
Amdahl’s Law
Ex Time new = Ex Time old x (1 – Fraction enhanced) + Fraction enhanced
Speedup enhanced
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

13. Lecture 3 - Performance... Cont'd
Amdahl’s Law
ExTimenew = ExTimeold x (1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced 1
ExTimeold
ExTimenew
Speedupoverall = =
(1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

14. Lecture 3 - Performance... Cont'd
Amdahl’s Law
Floating point instructions improved to run 2X; but only 10% of actual instructions are FP
ExTimenew =
Speedupoverall =
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

15. Lecture 3 - Performance... Cont'd
Amdahl’s Law
Floating point instructions improved to run 2X; but only 10% of actual instructions are FP
ExTimenew = ExTimeold x ( /2) = 0.95 x ExTimeold 1
Speedupoverall = =
1.053
0.95
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

16. Lecture 3 - Performance... Cont'd
Amdahl’s Law
ExTimenew = ExTimeold x (1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced 1
ExTimeold
ExTimenew
Speedupoverall = =
(1 - Fractionenhanced) + Fractionenhanced
Speedupenhanced
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd

17. Lecture 3 - Performance... Cont'd
Amdahl’s Law
Solution
ExTimenew = ExTimeold x ( /2) = 0.95 x ExTimeold 1
Speedupoverall = =
1.053
0.95
MAC/VU-Advanced Computer Architecture
Lecture 3 - Performance... Cont'd