# Computer Organization and Systems Software - Coursework Example

Date:Â  2021-09-02 08:54:42
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There are three independent jobs. They need 10 s, 15 s and 20 s CPU time, respectively, and during their executions, each of them spend half a minute waiting for I/O. Compute the minimal overall runtime of these jobs when they are processed

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(a) In a uniprogrammed system: In a uniprogrammed system, Processor must wait for (blocked) I/O instruction to complete before preceding and also each job to finish at a time before the next starts.

Job 1 will take its time to finish that is 10s plus the half minute I/O waiting time of 30s:

Job 1 will take 10s + 30s = 40s to complete

Job 2 will take its time to finish that is 15s plus job 1 time taken to finish that is 40s then half minute I/O waiting time of 30s:

Job 2 will take 40 + 15 + 30 = 85s / 1 min 25s to complete

Job 3 will take its time to finish that is 20s plus job 2 time taken to finish that is 85s then half minute I/O waiting time of 30s:

Job 3 will take 85 + 20 + 30 = 135s / 2 min 15s to complete

(b) In multiprogrammed system: a multiprogrammed system utilizes time sharing hence when one job needs to wait for I/O, the processor can switch to the other job

Job 1 will take its time to finish that is 10s plus half minute I/O waiting time of 30s:

Job 1 will take 10 + 30 = 40s to complete

Job 2 will not wait for job 1 to finish hence will begin while job 1 is waiting for the I/O to finish. Hence it will take 15s plus the half minute I/O waiting time of 30s:

Job 2 will take 15 + 30 = 45s to complete

Job 3 will not wait for job 1, and 2 to finish hence will begin while job 1 and 2 are waiting for the I/O to finish. Hence it will take 20s plus the half minute I/O waiting time of 30s: Job 3 will take 20 + 30 = 50s to complete

2. A computer has a cache, main memory, and a hard disk. If a referenced word is in the cache, it takes 15 ns to access it. If it is in main memory but not in the cache, it takes 45 ns to load (the block containing) 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, it takes 10 ms to load (the block containing) it from the disk into main memory, and then the reference is started again. The cache hit ratio is 0.5. In the case of a cache miss, the probability that the word is in the main memory is 0.7. Compute the average time required to access a referenced word in this system.

Cache access time is 15 ns

Memory access time is the sum of cache access time plus memory load time:

15 ns + 45 ns = 60 ns

Disk access time is the sum of cache access time plus disk load time:

10 x 10 ^ 6 + 60 ns = 10000060 ns

Average access time: 0.5 x 15 ns + 0.1 x (0.7 x 60 ns + 0.4 x (10 x 10 ^ 6 + 60 ns))

Solution = 400014.1 ns

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