Transcript Module 6: CPU Scheduling

Chapter 5: CPU Scheduling
Chapter 5: CPU Scheduling
 Basic Concepts
 Scheduling Criteria
 Scheduling Algorithms
 Multiple-Processor Scheduling
 Real-Time Scheduling
 Thread Scheduling
 Operating Systems Examples
 Java Thread Scheduling
 Algorithm Evaluation
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Basic Concepts
 Maximum CPU utilization obtained with
multiprogramming
 CPU–I/O Burst Cycle – Process execution
consists of a cycle of CPU execution and I/O wait
 CPU burst distribution: a large number of short
CPU bursts and a small number of large CPU
bursts
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Alternating Sequence of CPU And I/O Bursts
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Histogram of CPU-burst Times
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CPU Scheduler
 Selects from among the processes in memory that are ready
to execute, and allocates the CPU to one of them
 CPU scheduling decisions may take place when a process:
1. Switches from running to waiting state
2. Switches from running to ready state
3. Switches from waiting to ready
4. Terminates
 Scheduling only under conditions 1 and 4 is nonpreemptive
or cooperative.
 Scheduling under conditions 2 and 3 is preemptive
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CPU Scheduler
 CPU scheduling conditions:
1. Process switches from running to waiting state
2. Process switches from running to ready state
3. Process switches from waiting to ready
4. Process terminates
 Nonpreemptive or cooperative scheduling (1 and 4): Under
this type of scheduling, once the CPU has been allocated to
a process, the process keeps the CPU until it releases the
CPU either by terminating or by switching to the weighting
state (Windows 3.x).
 Preemptive Scheduling (2 and 3): (Windows- 95, NT, 2000,
XP, Vista; Mac OS X, UNIX)
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Dispatcher
 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
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Scheduling Criteria
 CPU utilization – keep the CPU as busy as possible
 Throughput – # of processes that complete their
execution per time unit
 Turnaround time – amount of time to execute a
particular process
 Waiting time – amount of time a process has been
waiting in the ready queue
 Response time – amount of time it takes from when
a request (I/O )was submitted until the first response
is produced
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Optimization Criteria
 Max CPU utilization
 Max throughput
 Min turnaround time
 Min waiting time
 Min response time
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Scheduling Algorithms
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First-Come, First-Served (FCFS) Scheduling
Process
Burst Time
P1
24
P2
3
P3
3
 Suppose that the processes arrive in the order: P1 , P2 , P3
The Gantt Chart for the schedule is:
P1
P2
0
24
P3
27
30
 Waiting time for P1 = 0; P2 = 24; P3 = 27
 Average waiting time: (0 + 24 + 27)/3 = 17
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FCFS Scheduling (Cont.)
Suppose that the processes arrive in the order
P2 , P3 , P1
 The Gantt chart for the schedule is:
P2
0
P3
3
P1
6
30
 Waiting time for P1 = 6; P2 = 0; P3 = 3
 Average waiting time: (6 + 0 + 3)/3 = 3
 Much better than previous case
 Convoy effect (one big CPU-bound process may force other
processes to wait for completion of its CPU burst if this big process
arrives first)
 Hence, execution of short processes prior to a long process is
desirable
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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 (the smallest next CPU burst).
 Two schemes:

nonpreemptive – once CPU given to the process it
cannot be preempted until completes its CPU burst

preemptive – if a new process arrives with CPU burst
length less than remaining time of current executing
process, it can preempt the current process. This
scheme is known as the
Shortest-Remaining-Time-First (SRTF)
 SJF is optimal if it gives minimum average waiting time for
a given set of processes
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Example of Non-Preemptive SJF
Process
Arrival Time
Burst Time
P1
0.0
7
P2
2.0
4
P3
4.0
1
P4
5.0
4
 SJF (non-preemptive)
P1
0
3
P3
7
P2
8
P4
12
16
 Average waiting time = (0 + 6 + 3 + 7)/4 = 4
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Example of Preemptive SJF
Shortest-Remaining-Time-First (SRTF)
Process
Arrival Time
Burst Time
P1
0.0
7
P2
2.0
4
P3
4.0
1
P4
5.0
4
 SJF (preemptive)
P1
0
P2
2
P3
4
P2
5
P4
P1
11
7
16
 Average waiting time = (9 + 1 + 0 +2)/4 = 3
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Determining Length of Next CPU Burst
 Can only estimate the length
 Can be done by using the length of previous CPU bursts, using
exponential averaging
1. tn  actual length of nth CPU burst
2.  n stores the past history
3.  n 1  predicted value for the next CPU burst
4.  , 0    1
5. Define:

tn
n 1
  tn  1    n
contains the most recent information,
n
contains the past history
α is the relative weight of recent and past history; more commonly it is
taken α=1/2, so recent history and past history are equally weighted
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Philosophy of Exponential Averaging
  =0
n+1 = n
 Recent history does not count
  =1
 n+1 =  tn
 Only the actual last CPU burst counts
 If we expand the formula, we get:
n+1 =  tn+(1 - ) tn -1 + …
+(1 -  )j  tn -j + …
+(1 -  )n +1 0

 Since both  and (1 - ) are less than or equal to 1, each
successive term has less weight than its predecessor
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Prediction of the Length of the Next CPU
Burst
 n1   tn  1    n
 0  10;   1/2; t0  6
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