chap2Solns - COD

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SOLUTION OF SAMPLE EXERCISES IN CHAPTER 2 2.4 We are given the number of cycles per second and the number of seconds, so we can calculate the number of required cycles for each machine. If we divide this by the CPI we’ll get the number of instructions. For M1, we have 3 seconds × 200 × 10 6 cycles/second = 600 × 10 cycles per program / 10 cycles per instruction = 60 × 10 6 instructions per program. For M2, we have 4 second × 300 × 10 6 cycles/second = 1200 × 10 6 cycles per program / 9.4 cycles per instruction = 127.7 × 10 6 instructions per program. 2.9 We do this problem by finding the amount of time that program 2 can be run in an hour and using that for executions per second, the throughput measure. seconds seconds 3600 200 hour Execution of P1 Executions of P2 per hour= seconds Execution of P2 seconds 3600 200 100 1600 hour Executions of P2 per hour on M1= 53 33 −× = = seconds 3600 200 5 2600 hour Executions of P2 per hour on M2= 650 44 == With performance measured by throughput for program 2, machine M2 is 650 533 =1.2 times faster than M1. The cost-effectiveness of the machines is to be measured in units
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This note was uploaded on 09/06/2009 for the course ECE 1212 taught by Professor Jame during the Spring '09 term at Punjab Engineering College.

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chap2Solns - COD

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