CS501 Current Assginment Solution

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GMT - 11 Hours CS501 Current Assginment Solution

Post by Admin on Tue Jun 14, 2011 10:22 pm

1. What percentage of time will a 20 MIPS processor spend in the busy wait
loop of 65-character line printer when it takes 3 m-sec to print a character
and a total of 457 instructions need to be executed to print 65 character
lines? Assume that 4 instructions are executed in the polling loop?
Solution:
Out of the total 457 instructions executed to print a line, 65x4=260 is required for polling.
For a 20MIPS processor, the execution of the remaining 197 instructions takes 197/
(20x10
6
) = 9.85 sec.
Since the printing of 65 characters takes 65*3 msec, (195 – 0.00985) = 194.99 msec 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
2. Suppose that a certain program takes 500 seconds of elapsed time to execute.
Out of these 500 seconds, 280seconds is the CPU time and the rest is I/O time.
What will be the elapsed time?
Solution:
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.

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