COMP3370A1.docx - COMP 3370 Assignment 1 Jingyi Huang 7787399 performancex = 1 according to the definition of performance a b 1 executionTime x Prefm1 =

# COMP3370A1.docx - COMP 3370 Assignment 1 Jingyi Huang...

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COMP 3370 Assignment 1 Jingyi Huang 7787399 1. according to the definition of performance: performance x = 1 executionTime x ; a) Pref m1 = 1/5 ; Pref m2 = 1/30; So, the performance of M 1 is 0.2; the performance of M 2 is 1/30. b) pref m 1 pref m 2 = 1 5 1 30 = 30 5 = 6. So M 1 is 6 times faster than M 2 . 2. 4 minutes = 240 seconds for M 2 a) M 1 is 8 times faster than M 2 , then pref m 1 pref m 2 = 1 n 1 240 = 240 n = 8 , therefore n = 240/8 = 30 sec, so the execution time on M 1 is 30 sec. b) M 3 is 4 times slower than M 2 , then pref m 2 pref m 3 = 1 240 1 n = n 240 = 4 , therefore n = 240*4 = 960 sec, it is 16 minutes. 3. Since seconds program = cycles program seconds cycle , clock rate = 1 seconds cycle = 2.8*10 9 cycles/sec; a) Clock cycles = CPU time / cycle time = CPU time*clock rate = 9*2.8*10 9 = 2.52*10 10 cycles. b) We want to a 1.2 times speed up, then the CPU time became: 9/1.2 = 7.5 sec; then, clock rate = 1 clockcycletime = 1 CPU time clockcycles = clockcycles CPU time = 2.52 × 10 10 7.5 = 3.36 × 10 9 sec . So, the clock rate we need is 3.36 GHz. 4. For M 1 , clock rate = 2*10 9 cycles/sec, clock cycles = 2*10 8 cycles/program. Then, the CPU time
of M 1 is: CPU time = clockcycles clock rate = 2 × 10 8 2 × 10 9 = 0.2 sec. For M 2 , clock rate = 2.5*10 9 cycles/sec, clock cycles = 2*10 8 *1.3 = 2.6*10 8 cycles/program. Then, the CPU time of M 2 is: CPU time = clockcycles clock rate = 2.6 × 10 8 2.5 × 10 9 = 0.104 sec . a) Since Tm 1 = 0.2 sec, Tm 2 = 0.104 sec, machine M 2 is much faster than machine M 1 . b) Since performance x = 1 executionTime x ; Pref m1 = 1 0.2 ; Pref m2 = 1 0.104 ; then, Pref m 2 Pref m 1 = 1 0.104 1 0.2 = 0.2 0.104 = 1.92 . so, machine M 2 is 1.92 times faster than machine M 1 . 5. CPU time for M 1 is: 10 sec, clock rate = 2*10 9 cycles/sec. Now, for a better implementation, we need CPU time for M 2 is: 4 sec; for the M 1 ’s clock cycles, we know: CPU time = clockcycles clock rate , so clock cycles for M 1 = CPU time ×clock rate = 10 × 2 × 10 9 = 2 × 10 10 cycles/program. Now, we required 10% more clock cycles than on M 1 , so M 2 ’s clock cycles = 2 × 10 10 × 1.1 = 2.2 × 10 10 cycles/program. M 2 ’s clock rate = clockcycles CPU time = 2.2 × 10 10 4 = 5.5 × 10 9 cycles/sec = 5.5 GHz. So, M 2 ’s clock rate has to be 5.5 GHz. 6. Since performance of a machine for a program = 1 executionTime x .

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