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07-07-2010, 01:32 PM
1. 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?
2. 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?
Solutions
1. 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.
2. Solution:
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.
2. 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?
Solutions
1. 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.
2. Solution:
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.