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Princess Sumaya Univ. Computer Engineering Dept. د. بســام كحـالــه Dr. Bassam Kahhaleh.

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Presentation on theme: "Princess Sumaya Univ. Computer Engineering Dept. د. بســام كحـالــه Dr. Bassam Kahhaleh."— Presentation transcript:

1 Princess Sumaya Univ. Computer Engineering Dept. د. بســام كحـالــه Dr. Bassam Kahhaleh

2 Princess Sumaya Univ. Computer Engineering Dept. Chapter 1:

3 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 2 / 16 Technology Year Technology Used in ComputersRelative Perf/Unit Cost 1951Vacuum tube1 1965Transistor35 1975Integrated circuit900 1995Very Large-Scale Integrated Circuit2,400,000 2005Ultra Large-Scale Integrated Circuit6,200,000,000

4 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 3 / 16 Technology 19852006 A computer with certain performance, main memory, disk storage $1,000,000$500

5 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 4 / 16 Classes of Computers 1960s Large Mainframes 1970s Minicomputer & Supercomputers 1980s Desktop Computers 1990s Servers (Internet & WWW)  Desktop ●Features, Price & Performance  Servers ●Reliability ●Scalability ●Efficient Throughput (Performance: TPM) Amazon.com (fall quarter of 2005) had an average revenue per hour of $1.35 million Computing capacity, Memory, Storage, & I/O Bandwidth

6 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 5 / 16 Classes of Computers  Desktop  Servers  Embedded Computers ●Price ●Real-time Performance ●Minimized Memory ●Minimized Power

7 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 6 / 16 Computer Performance  Response Time The time it takes to do a task  Execution Time  Throughput The total amount of work done in a given time  Difference? Task ATask B Time t Calculate Read File Calculate Save File Calculate Read File Calculate Save File 012345678910

8 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 7 / 16 Computer Performance  Response Time The time it takes to do a task  Execution Time  Throughput The total amount of work done in a given time  Difference? Task B Time t Calculate Read File Calculate Save File Calculate Read File Calculate Save File 012345678910 Task A

9 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 8 / 16 CPU Performance  Performance = 1 / Execution Time  Relative Performance: Performance Computer X Execution time Y ●n = ──────────────── = ─────────── Performance Computer Y Execution time X  CPU Performance ●CPU Execution time = CPU Clock Cycles × Clock Cycle time CPU Clock Cycles = ───────────── Clock Rate

10 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 9 / 16 CPU Performance Exercise: A program takes 10 seconds to run on a 2 GHz CPU. The same program on another CPU would take 20% extra clock cycles, yet it finishes in 6 seconds. What is the other CPU clock rate?

11 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 10 / 16 Instruction Performance  Clocks Per Instruction, CPI The average number of clock cycles each instruction takes to execute. Exercise: Which computer is faster? ComputerClock Cycle TimeCPI A250 picoseconds2 B500 picoseconds1.2 # of instructions in a program = CPU-A: CPU Execution Time = × × ps = ps CPU-B: CPU Execution Time = × × ps = ps Computer A is ( / ) = times faster than B

12 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 11 / 16 Instruction Performance Exercise: Given 3 groups of instructions: A, B and C, it takes different clock cycles to execute an instruction within each group. Given the shown instruction mix, which code sequence is faster to execute? Instruction Class ABC CPI123 Code Instruction Count ABC Seq1212 Seq2411 Seq1: CPU Execution Time = × + × + × = cycles Seq2: CPU Execution Time = × + × + × = cycles Seq2 is / = times faster than Seq1 Seq1 average CPI = = cycles / instruction Seq2 average CPI = = cycles / instruction

13 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 12 / 16 The Power Wall  Voltage, Capacitive Load & Frequency

14 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 13 / 16 Multi-Core Architectural & Organizational Improvements Technology Limitation in: Power Parallelism Memory

15 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 14 / 16 Chip Manufacturing

16 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 15 / 16 Die Cost Cost per Wafer Die Cost = ────────────── Dies per Wafer × Yield Wafer Area Dies per Wafer = ─────── Die Area 1 Yield = ───────────────────── Defects per Area × Die Area 2 ( 1+ ───────────────── ) 2

17 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. 16 / 16 Amdahl’s Law  When introducing an improvement, Execution Time is divided into 2 parts: ●Affected by the improvement ●Not affected Execution Time Execution Time Affected Execution After = ──────────────── + Time Improvement Amount of Improvement Unaffected Example: How much improvement is required for the multiply hardware to make the program run 5 times faster? AddMultiply Program Execution Time20 seconds80 seconds ─── = ─── +

18 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept.  Exercise 1.3  Exercise 1.4  Exercise 1.5  Exercise 1.6  Exercise 1.11  Exercise 1.13  Exercise 1.14  Exercise 1.15  Exercise 1.16

19 Princess Sumaya University 22540 – Computer Arch. & Org (2) Computer Engineering Dept. Chapter 1


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