Engineering High School

Suppose we have four jobs in a computer system, in the order JOB1, JOB2, JOB3 and JOB4. JOB1 requires 8 s of CPU time and 8 s of I/O time; JOB2 requires 4 s of CPU time and 14 s of disk time; JOB3 requires 6 s of CPU time; and, JOB4 requires 4 s of CPU time and 16 s of printer time. Define the following quantities for system utilization: • 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 (not waiting) Compute these quantities (with illustrations if needed) in each of the following systems: a. A uniprogramming system, whereby each job executes to completion before the next job can start its execution. b. A multiprogramming system that follows a simple round-robin scheduling. Each process gets 2 s of CPU time turn-wise in a circular manner.

Answers

Answer 1

Answer:

Explanation:

a. Uniprogramming System:

Turnaround Time: For JOB1, the turnaround time is 16 s (8 s CPU time + 8 s I/O time). For JOB2, the turnaround time is 18 s (4 s CPU time + 14 s disk time). For JOB3, the turnaround time is 6 s (6 s CPU time). For JOB4, the turnaround time is 20 s (4 s CPU time + 16 s printer time).

Throughput: In a uniprogramming system, the throughput is equal to 1 job per time period, as only one job can be processed at a time.

Processor Utilization: For JOB1, the processor utilization is 8/16 = 50%. For JOB2, the processor utilization is 4/18 = 22.2%. For JOB3, the processor utilization is 6/6 = 100%. For JOB4, the processor utilization is 4/20 = 20%.

b. Multiprogramming System:

Turnaround Time: For JOB1, the turnaround time is 24 s (8 s CPU time + 8 s I/O time + 8 s waiting). For JOB2, the turnaround time is 24 s (4 s CPU time + 14 s disk time + 6 s waiting). For JOB3, the turnaround time is 16 s (6 s CPU time + 10 s waiting). For JOB4, the turnaround time is 32 s (4 s CPU time + 16 s printer time + 12 s waiting).

Throughput: In a multiprogramming system, the throughput is equal to 4 jobs per time period, as all four jobs can be processed concurrently.

Processor Utilization: For JOB1, the processor utilization is 8/24 = 33.3%. For JOB2, the processor utilization is 4/24 = 16.7%. For JOB3, the processor utilization is 6/16 = 37.5%. For JOB4, the processor utilization is 4/32 = 12.5%.

Answer 2

Here are the turnaround time, throughput, and processor utilization for the four jobs in a uniprogramming and round-robin system:

Uniprogramming

Turnaround time:

JOB1: 16 s

JOB2: 22 s

JOB3: 12 s

JOB4: 20 s

Throughput: 1 job/16 s

Processor utilization: 50%

Round-Robin

Turnaround time:

JOB1: 18 s

JOB2: 14 s

JOB3: 10 s

JOB4: 18 s

Throughput: 2 jobs/16 s

Processor utilization: 62.5%

The difference in turnaround time between the two systems is due to the fact that in a uniprogramming system, the CPU is idle while a job is waiting for I/O. In a round-robin system, the CPU is always running, even if it is only running a short burst of code for another job.

The difference in throughput between the two systems is also due to the fact that in a round-robin system, the CPU is always running, so more jobs can be completed in a given amount of time.

The processor utilization is higher in a round-robin system because the CPU is not idle as often. However, the turnaround time is also higher because the jobs have to share the CPU.

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Related Questions

how many grams of water can be heated from 10 degrees to 45 degrees using 20,500 joules of energy

Answers

Therefore, 20,500 Joules of energy can heat 1,104.3 grams of water from 10 degrees Celsius to 45 degrees Celsius.

Specific heat capacity

Calculate the energy (in Joules) required to heat 1 gram of water from 10 degrees Celsius to 45 degrees Celsius:

Q = m * c * ΔT

Q = 1 gram * 4.184 Joules/gram-degree Celsius * (45 - 10) degrees Celsius

Q = 18.47 Joules

Calculate the amount of water (in grams) that can be heated with 20,500 Joules of energy:

20,500 Joules / 18.47 Joules = 1,104.3 grams

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