Review question: process  (Page 4/4)

In a threaded program, which resources are shared by the entire process and which are associated with each thread?

The process address space, structures used for interprocess communication, files, and I/O resources are associated with theentire process. Each thread gets its own execution state, context (registers), stack, and local variables.

Consider the following code:

1. for (i = 0; i<10; i++)
2. for (j = 0; j<10; j++)
3. a[i] = a[i]* j
   • Give one example of the spatial locality in the code.

Spatial locality occurs when the array a is accessed sequentially.

• Give one example of the temporal locality in the code.

Temporal locality occurs when i is used repeatedly with the same value in the second loop.

Consider a memory system with the following parameters:

• Tc = 100 ns
• Tm = 1200 ns
• Cc = 0.01 cents/bit
• Cm = 0.001 cents/bit

1. What is the cost of 1 MByte of main memory?

Cost = Cm x 8 x 220 = $83.89

1. What is the cost of 1 Mbyte of main memory using cache memory technology?

Cost = Cc x 8 x 220 = $838.86

1. If the effective access time is 10% greater than the cache access time, what is the hit ratio H?

1.1 x Tc = Tc + (1 - H)Tm110 = 100 + (1 - H) x 12001200 x H = 1190H = 1190/1200 = 99.2%

A computer has a cache, main memory, and a disk used for virtual memory. if a referenced word is in the cache, 20 ns are required to access it.If it is in main memory but not in the cache, 60 ns are needed to load it into the cache (this includes the time to originally check the cache), and then theReference is started again. if the word is not in main memory, 12 ms are required to fetch the word from disk, followed by 60 ns 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 aReferenced word on this system?

There are three cases to consider:

So the average access time would be:

Avg = (0.9)(20) + (0.06)(80) + (0.04)(12000080) = 480,026 ns

Suppose that we have a multi-programmed computer in which each job has identical characteristics. in one computation period, t, for a job, half theTime is spent in i/o and the other half in processor activity. each job runs for a total of n periods. assume that simple round-robin scheduling is used, andThat i/o operations can overlap with processor operation. define the following quantities:

• Turnaround time = actual time to complete a job.
• Throughput = average number of jobs completed per time period T
• Processor utilization = percentage of time that the processor is active (waiting)

Compute these quantities for one, two, and four simultaneous jobs, assuming that the period T is distributed in each of thefollowing ways:

1. I/O first half, processor second half

When there is one job, it can do I/O or run on the processor whenever it wants. So the quantities are:

• Turnaround Time = N*T
• Throughput = 1/N
• Processor Utilization = 50%

When there are two jobs, one starts right away and does I/O. When it switches to run on the CPU, the second can start its I/O. Thisdelays the second job for 1/2*N, but otherwise they alternate between I/O and CPU. Assume the jobs are long, so the extra 1/2 a cycle is insignificant.Then:

• Turnaround Time = N*T
• Throughput = 2/N
• Processor Utilization = 100%

When there are 4 jobs, the CPU is round-robin among the four, as is the I/O. This means the jobs are interleaved as:

Job1: I/O CPU I/O CPU

Job2: I/O CPU I/O CPU

Job3: I/O CPU I/O CPU

Job4: I/O CPU I/O CPU

A job can execute for one cycle T, then it must wait for T before doing another cycle. Again assume the jobs are long so that anyinitial wait is insignificant. Then:

• Turnaround Time = (2N-1)*T
• Throughput = 2/N
• Processor Utilization = 100%
• I/O first and fourth quarters, processor second and third quarter

The answers for this part are the same as the first. This is easy to see for the case of 1 job and 2 jobs. When there are 4 jobs, theCPU is round-robin among the four, as is the I/O. This means the jobs are interleaved as:

Job1: I CP O I CP O

Job2: I CP O I CP O

Job3: I CP O I CP O

Job4: I CP O I CP O

An i/o-bound program is one that, if run alone, would spend more time waiting for i/o than using the processor. a processor-bound program is theOpposite. suppose a short-term scheduling algorithm favors those programs that have used little processor time in the recent past.

• Explain why this algorithm favors I/O-bound programs.

The algorithm favors I/O bound processes because these will be much less likely to have used the processor recently. As soon asthey are done with an I/O burst, they will get to execute on the processor.

• Explain why this algorithm does not permanently deny processor time to processor-bound programs.

This algorithm does not permanently deny processor time to processor-bound programs, because once they have been prevented fromusing the processor for some time they will be favored by the scheduling algorithm.