l1_handouts_4up - Hennessy and Patterson ECE 3140/CS 3420...

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1 ECE 3140/CS 3420 Computer Organization Spring 2009 MIPS Instruction Set Architecture ECE3140/CS3420 Hennessy and Patterson • Read Chapter 1 – 1.1-1.9 ead Chapter 2 Read Chapter 2 – 2.1 through 2.7 – 2.8 through 2.14 (for Next Tuesday) – Skim B.1,B.2, B.10 (for Next Tuesday) ECE3140/CS3420 2 Programming Process swap(int v[], int k) {int temp; mp = v[k]; C Language swap: Assembly Language Compiler temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; } muli $2,$5,4 add $2, $4, $2 lw $15, 0($2) lw $16, 4($2) sw $16, 0($2) sw $15, 4($2) jr $31 00000000101000010000000000011000 0000000000110000001100000100001 Binary Machine Language Assembler ECE3140/CS3420 3 00000000000110000001100000100001 10001100011000100000000000000000 10001100111100100000000000000100 10101100111100100000000000000000 10101100011000100000000000000100 00000011111000000000000000001000 How Well do Computers Work? • Performance is critical to both end users and hardware designers – I.e., faster is better! • What to measure? – Execution time – the total time from start to finish hroughput e total amount of work done in ECE3140/CS3420 Throughput the total amount of work done in a give time • Execution time is most important to us at the moment (ignoring system overhead) 4
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2 Performance • We define performance in terms of execution time. Performance X = 1/Execution Time X • We say that “X is n times faster than Y” if Performance X /Performance Y = n • Equivalently, ECE3140/CS3420 Execution time Y /Execution time X = n 5 Performance • Execution time is defined in terms of the system clock xecution Time = Clock Cycles x Clock Cycle Time Execution Time = Clock Cycles x Clock Cycle Time • Where Clock Cycles = Instruction Count x CPI • Where CPI = Average number of clock cycles per instruction • Classic CPU performance equation
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l1_handouts_4up - Hennessy and Patterson ECE 3140/CS 3420...

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