Question

Operating Systems Questions (Please help if you can)

1. A computer has cache, main memory, and a disk used for virtual memory. If a

referenced word is in the cache, 20ns are required to access it. If it is in main memory but not

in the cache, 60ns are required to load it into the cache (this includes the time to originally

check the cache), and then, the reference is started again. If the word is not in main memory,

12ms are required to fetch the word from disk, followed by 60ns to copy it to the cache, and

then, the reference is started again. The cache hit ratio is 0.9 and the main memory hit ratio

is 0.6. What is the average time in ns required to access a referenced word on this system?

2. [6 pt] An i/o-bound process is one that, if run alone, would spend more time waiting for i/o

than using the cpu. A cpu-bound process is the opposite. Suppose a short-term scheduling

algorithm favors those processes that have used little cpu time in the recent past. Explain

why this algorithm favors i/o-bound processes yet does not permanently deny cpu time to

cpu-bound processes.

3. What is an instruction trace? What is an interleaved instruction trace?

4. You have executed the following C program:

#include <stdio.h>

#include <sys/types.h>

#include <unistd.h>

int main()

{

pid_t pid = fork();

printf ( "%d\n", pid );

return ( 0 );

}

What are the possible outputs, assuming the fork succeeded?

Answer 1

Answer:------------------
1.
There are three cases to consider:---------------------

Location of referenced ------------------ word Probability ------------- Total time for access in ns In cache ------------------------------------------- 0.9 ----------------------------------------- 20 Not in cache, but in main memory ------ (0.1)(0.6) = 0.06 -------------------------- 60 + 20 = 80 Not in cache or main memory ----------- (0.1)(0.4) = 0.04 ---------------------12ms + 60 + 20 = 12,000,080

So the average access time would be:
Avg = (0.9)(20) + (0.06)(80) + (0.04)(12000080) = 480026 ns

2.
I/O-bound programs use shorter processor time so the algorithm favors them. But if a processor-bound process is denied processor time for a longer time, the same algorithm will grant the processor to that process since it has not used the processor at all in the recent past. Therefore this algorithm favors I/O-bound programs and yet does not permanently deny processor time to processor-bound programs

3.
An instruction trace is a list of the sequence of instructions that are executed by an individual process.