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| Subject: cs501 ASSIGNMENT SOLUTION Wed Jul 07, 2010 12:56 pm | |
| · Elapsed time = CPU time + I/O time· This gives us the I/O time = 500 – 280 = 220 seconds at the beginning, which is 44 % of the elapsed time. · Out of the total 457 instructions executed to print a line, 65x4=260 are required for polling.· For a 20MIPS processor, the execution of the remaining 197 instructions takes 197 / (20x106) = 9.85 sec.· Since the printing of 65 characters takes 195ms.· (195-0.00985) = 194.99msec is spent in the polling loop before the next 65 characters can be printed. · This is 194.99/195=99.99% of the total time. | |
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