1. ICS 143 Principles of Operating Systems
Discussion Section Week 4

2. Today Announcements Programming Assignment Preview
Lifetime of a Process
Schedulers
Pop Quiz #1
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3. Announcements Homework #2 Mid-term exam New office hours
Due Thu, 30 April 2015, 11:55pm via EEE DropBox
Mid-term exam
Fri, 8 May 2015, 11:00-11:50am in EH 1200
New office hours
Ekin: Mons 2:00-3:00pm in ICS 468A
Michael: Tues 1:00-2:00pm in SBSG 2236
Prof V: Thus 10:00-11:00am in DBH 2086
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4. Programming Assignment Preview
Individual submission – everyone gets to code!
It’s ok to discuss ideas, but everyone must submit their own code
Implement several schedulers
Simulator will provide template / skeleton (Java)
Preemptive / non-preemptive
Short report (1 page max.)
Black-box testing
Public tests and expected output provided
Will test on public / private tests to make sure your code works
More details will be announced next week…
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5. Schedulers Overview Long Term Short Term Medium Term Creation
Terminated
Short Term
Ready
Running
Blocked
Medium Term
Suspended
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6. Lifetime of a Process
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7. Lifetime of a Process No PCB yet Must go through scheduler dispatch
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8. Levels of Scheduling
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9. CPU Scheduler
CPU scheduler selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them.
Non-preemptive Scheduling
Once CPU has been allocated to a process, the process keeps the CPU until
Process exits OR
Process switches to waiting state
Preemptive Scheduling
Process can be interrupted and must release the CPU.
Need to coordinate access to shared data
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10. Dispatch Module
Dispatcher module gives control of the CPU to the process selected by the short-term scheduler. This involves:
switching context
switching to user mode
jumping to the proper location in the user program to restart that program
Dispatch Latency:
time it takes for the dispatcher to stop one process and start another running.
Dispatcher must be fast.
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11. Scheduling Criteria CPU Utilization Throughput Turnaround time
Keep the CPU and other resources as busy as possible
Throughput
# of processes that complete their execution per time unit.
Turnaround time
amount of time to execute a particular process from its entry time.
Waiting time
amount of time a process has been waiting in the ready queue.
Response Time (in a time-sharing environment)
amount of time it takes from when a request was submitted until the first response is produced, NOT output.
MAX
MAX
MIN
MIN
MIN
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12. CPU Burst Distribution
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13. Determining Length of Next CPU Burst
One can only estimate the length of burst.
Use the length of previous CPU bursts and perform exponential averaging.
tn = actual length of nth burst
n+1 =predicted value for the next CPU burst
 = 0, 0    1
Define: n+1 =  tn + (1- ) n
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14. Exponential Averaging(cont.)
 = 0
n+1 = n; Recent history does not count
= 1
n+1 = tn; Only the actual last CPU burst counts.
Similarly, expanding the formula:
n+1 = tn + (1-) tn-1 + … (1-)^j tn-j + … (1-)^(n+1) 0
Each successive term has less weight than its predecessor. j
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15. First Come First Serve (FCFS) Scheduling
Policy: Process that requests the CPU FIRST is allocated the CPU FIRST.
Commonly preemptive or non-preemptive?
Non-preemptive
Implementation - using FIFO queues
incoming process is added to the tail of the queue.
Process selected for execution is taken from head of queue.
Performance metric?
Average waiting time in queue.
Advantage?
Simple, no starvation
Disadvantages?
Convoy effect: Short processes at the back of the queue may have to wait a long time
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16. First-Come, First-Served(FCFS) Scheduling
Example
Suppose the arrival order for the processes is
P1, P2, P3
Waiting time
P1 = 0;
P2 = 24;
P3 = 27;
Average waiting time
( )/3 = 17
Gantt Chart for Schedule 24 P1 P2 P3 27 30
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17. Shortest-Job-First(SJF) Scheduling
Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time.
Two Schemes:
Scheme 1: Non-preemptive
Once CPU is given to the process it cannot be preempted until it completes its CPU burst.
Scheme 2: Preemptive
If a new CPU process arrives with CPU burst length less than remaining time of current executing process, preempt. Also called Shortest-Remaining-Time- First (SRTF).
Advantage?
SJF is optimal - gives minimum average waiting time for a given set of processes.
Disadvantage?
May lead to starvation
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18. Non-Preemptive SJF Scheduling
Example
Gantt Chart for Schedule P1 7 P3 P2 P4
Average waiting time =
( )/4 = 4 8 12 16
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19. Preemptive SJF Scheduling(SRTF)
Example
Waiting time:
P1 = 9
Waited from t=2 to t=11
P2 = 1
Waited from t=4 until t=5
P3 = 0
P4 = 2
Waited from t=5 until t=7
Gantt Chart for Schedule P1 2 P2 4 P3 5 7 P2 P4 11 16 P1
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20. Priority Scheduling
A priority value (integer) is associated with each process. Can be based on
Cost to user
Importance to user
Aging
%CPU time used in last X hours.
CPU is allocated to process with the highest priority.
Preemptive
Non-preemptive
Advantage?
Prioritize important jobs
Disadvantage?
May result in starvation
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21. Priority Scheduling Example (non-preemptive) Highest priority = 1
Waiting time:
P1 = 0
Came in at t=0, processed right away
P2 = 6.8
Came in at t=0.2, had to wait until t=7
P3 = 11
Came in at t=4, had to wait until t=15
P4 = 6
Came in at t=5, had to wait until t=11
Gantt Chart for Schedule P1 7 P2 11 15 P4 16 P3
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22. Round Robin (RR) Each process gets a small unit of CPU time
Time quantum usually milliseconds.
After this time has elapsed, the process is preempted and added to the end of the ready queue.
n processes, time quantum = q
Each process gets 1/n CPU time in chunks of at most q time units at a time.
No process waits more than (n-1)q time units (plus context swtich time).
Advantages?
No starvation, fair time-sharing, good response time.
Disadvantages?
Typically higher average turnaround time than SRTF, might be complicated to implement
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23. Round Robin Example Waiting time: Time Quantum = 20
Assume arrival time = 0 for all processes
Neglect context switch time for simplicity
Waiting time:
P1 = = 81
from t=20 to t=77, from t=97 to t=121
P2 = 20
from t=0 to t=20
P3 = = 94
From t=0 to t=37, from t=57 to t=97, from t=117 to t=134
P4 = = 97
From t=0 to t=57, from t=77 to t=117
Gantt Chart for Schedule P1 20 P2 37 P3 57 P4 77 P1 97 P3
117 P4
121 P1
134 P3
154 P3
162
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24. Real-Time Scheduling Hard Real-time Computing -
required to complete a critical task within a guaranteed amount of time.
Soft Real-time Computing -
requires that critical processes receive priority over less fortunate ones.
Types of real-time Schedulers
Periodic Schedulers - Fixed Arrival Rate
Demand-Driven Schedulers - Variable Arrival Rate
Deadline Schedulers - Priority determined by deadline
…..
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25. Issues in Real-time Scheduling
Dispatch Latency
Problem - Need to keep dispatch latency small, OS may enforce process to wait for system call or I/O to complete.
Solution - Make system calls preemptible, determine safe criteria such that kernel can be interrupted.
Priority Inversion and Inheritance
Problem: Priority Inversion
Higher Priority process needs kernel resource currently being used by another lower priority process... higher priority process must wait.
Solution: Priority Inheritance
Low priority process now inherits high priority until it has completed use of the resource in question.
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26. Pop Quiz #1
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