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 CS 370: Operating Systems Department of Computer Science Colorado State University FALL 2015 URL: http://www.cs.colostate.edu/~cs370 INSTRUCTOR: Yashwant K Malaiya HW1 CPU Scheduling DUE DATE: Friday October 8 2015 5 PM. Submit PDF file using Canvas. Problem 1 (20 points) We had seen the exponential average expression used to predict the length of the
next CPU burst. What are the implications of assigning the following values to the parameters used by the
algorithm? 1. α = 0 and τ0 = 20 milliseconds
2. α = 0.99 and τ0 = 10 milliseconds
Answer: a) τn +1= α * tn + (1-­‐ α) * τn When α = 0 and τ0 = 20 milliseconds, the equation will be: τn +1= 0* tn + (1-­‐ 0) * 20 = 20 milliseconds When α= 0 and τ0 = 20 milliseconds, the formula always makes a prediction of 20 milliseconds for the next CPU burst. b) τn +1= α * tn + (1-­‐ α) * τn When α = 0.99 and τ0 = 10 milliseconds, the equation will be: τ1= 0.99* t0 + (1-­‐ 0.99) * 10 = [ 0.99 t0 + 0.1] milliseconds τn +1= 0.99* tn + (1-­‐ 0.99) * τn = [ 0.99 tn + 0.1 τn] milliseconds When α= 0.99 and τ0 = 10 milliseconds, the most recent behavior of the process is given much higher weight than the past history associated with the process. Consequently, the scheduling algorithm is almost memoryless, and simply predicts the length of the previous burst for the next quantum of CPU execution. CS 370: Operating Systems FALL 2015 Department of Computer Science URL: http://www.cs.colostate.edu/~cs370 Colorado State University INSTRUCTOR: Yashwant K Malaiya Problem 2 (20 points) A system is running ten I/O-bound tasks and one CPU-bound task. The I/O-bound
tasks issue an I/O operation once for every millisecond of CPU computing and that each I/O operation
takes 10 milliseconds to complete. Also assume that the context switching overhead is 0.1 millisecond
(which does not count as the CPU being utilized) and that all processes are long-running tasks. What is
the CPU utilization for a round-robin scheduler when: 1. The time quantum is 1 millisecond
2. The time quantum is 10 milliseconds
Answer: • The time quantum is 1 millisecond: Irrespective of which process is scheduled, the scheduler incurs
a 0.1 millisecond context-switching cost for every context-switch. This results in a CPU utilization of
1/1.1 * 100 = 91%. • The time quantum is 10 milliseconds: The I/O-bound tasks incur a context switch after using up only
1 millisecond of the time quantum. The time required to cycle through all the processes is therefore
10*1.1 + 10.1 (as each I/O-bound task executes for 1 millisecond and then incur the context switch
task, whereas the CPU-bound task executes for 10 milliseconds before incurring a context switch). The
CPU utilization is therefore 20/21.1 * 100 = 94%. CS 370: Operating Systems Department of Computer Science Colorado State University FALL 2015 URL: http://www.cs.colostate.edu/~cs370 INSTRUCTOR: Yashwant K Malaiya Problem 3 (60 points) The processes in the table below are being scheduled using round-robin scheduling
algorithm with preemption. Each process is has a numerical priority, with a higher number indicating a higher
priority. In addition to the processes listed below, the system also has an idle task (which consumes no CPU
resources and is identified as Pidle ). This task has priority 0 and is scheduled whenever the system has no other
available processes to run. The length of a time quantum is 10 units. If a process is preempted by a higherpriority process, the preempted process is placed at the end of the queue. a. Show the scheduling order of the processes using a Gantt chart. b. What is the turnaround time for each process? c. What is the waiting time for each process? d. What is the CPU utilization rate? Thread P1 P2 P3 P4 P5 P6 Answer: Priority 40 30 30 35 5 10 Burst 20 25 25 15 10 10 Arrival 0 25 30 60 100 105 b) P1=20-­‐0=20 , P2=80-­‐25=55 , P3=90-­‐30=60 , P4= 75-­‐60 =15, P5=120-­‐100=20 , P6=115-­‐105=10 c) P1=0 , P2=0+10+20=30 , P3=5+10+15+5=35 , P4=0 , P5=0+10=10 , P6=0 d) CPU util= (20+25+25+15+10+10)/120 =105/120=87.5 %