Operating Systems
Lecture 16
Scheduling II
Operating System Concepts
6.1
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Recall Scheduling Criteria
 CPU utilization –fraction of time the CPU is busy.
 CPU efficiency – fraction of time the CPU is executing user code.
 Throughput – # of processes completed per unit time
 Average Turnaround time – average delay between job submission
and job completion.
 Normalized turnaround time – Ratio of turnaround time to service
time per process. Indicates the relative delay experienced by a
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 was
submitted until the first response is produced.
Operating System Concepts
6.2
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example: Computing Criteria
Suppose two processes arrive at time zero.
ti = service time for process i to complete.
P1: t1 = 7
P2: t2 = 11
Assume 1 unit of time for dispatch of process.
Assume non-preemptive scheduling.
a) Draw the Gantt diagram for these processes.
Operating System Concepts
6.3
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example: Computing Criteria
(continued)
Suppose two processes arrive at time zero.
P1: t1 = 7
P2: t2 = 11
Assume 1 unit of time for dispatch of process.
Assume non-preemptive scheduling.
b) Compute the following:
Avg. service time:
Throughput:
CPU utilization:
CPU efficiency:
Avg. Turnaround:
Avg. Wait time:
(Note: Avg wait time = avg. turnaround - avg service - avg dispatch)
Normalized turnaround for P1 :
Normalized turnaround for P2 :
c) Suppose P2 arrives at time 4. What statistics change?
What are their new values?
Operating System Concepts
6.4
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Assumptions in next examples
 Assume dispatch time = 0 (unless otherwise noted)
 CPU efficiency = CPU utilization
 Assume each process has one and only one CPU
burst (unless otherwise noted).
 Variable definitions:
 w = time spent in system (wait time, blocked time,
execution time).
 e = time spent in execution thus far.
 s = total service time required, including e.
Operating System Concepts
6.5
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example with FCFS scheduling
FCFS (first come first served)
selection function: max(w)
Example:
P1 : t1 = 60 (arrival at 0 - e)
P2 : t2 = 10 (arrival at 0)
P3 : t3 = 30 (arrival at 0 + e)
P4 : t4 = 20 (arrival at 0 + 2*e)
e = an extremely small unit of time.
a) Draw the Gantt diagram.
Operating System Concepts
6.6
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example with FCFS continued
Example:
P1 : t1 = 60 (arrival at 0 - e)
P2 : t2 = 10 (arrival at 0)
P3 : t3 = 30 (arrival at 0 + e)
P4 : t4 = 20 (arrival at 0 + 2*e)
e = an extremely small unit of time.
b) Compute the following quantities:
Avg. service time:
Throughput:
CPU utilization:
CPU efficiency:
Avg turnaround time:
Avg wait time:
Normalized turnaroundtime for each process:
Operating System Concepts
6.7
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example with FCFS continued
Suppose order changes to: P2 P3 P4 P1
Example:
P1 : t1 = 60 (arrival at 0 + 2*e)
P2 : t2 = 10 (arrival at 0- e)
P3 : t3 = 30 (arrival at 0 )
P4 : t4 = 20 (arrival at 0 + e)
e = an extremely small unit of time.
a) Draw the Gantt diagram.
b) What values change? What are the new values?
Operating System Concepts
6.8
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Notes on FCFS
 FCFS is unfair to processes with short CPU
bursts. (They may have long wait times
compared to their service requirements).
 Question: Is starvation possible with FCFS?
 FCFS is generally used only with batch
systems. (May be used as part of another
system).
Operating System Concepts
6.9
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
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) Selection criterion: min(s)
P1
0
3
P3
7
P2
8
P4
12
16
 Average waiting time = (0 + 3 + 6 + 7)/4 = 4
 Avg service time =
 Throughput =
 Avg turnaround =
 Check consistency. Wait = (Turnaround - Service - Dispatch)
Operating System Concepts
6.10
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Notes on SJF
 SJF is a priority algorithm
 Priority is based on the predicted next CPU burst
 Question: Is starvation possible with SJF?
Operating System Concepts
6.11
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example of Preemptive SJF
Process
Arrival Time
Burst Time
P1
0.0
7
P2
2.0
4
P3
4.0
1
P4
5.0
4
 SJF (preemptive) Selection criterion: min(s - e)
P1
0
P2
2
P3
4
P2
5
P4
7
P1
11
16
 Average waiting time = (9 + 1 + 0 + 2)/4 = 3
 What statistics are different from non-preemptive SJF?
What are the values of these stats?
Operating System Concepts
6.12
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Determining Length of Next CPU Burst
 Can be done by using the length of previous CPU bursts,
using exponential averaging.
1. tn  actual lenght of nthCPU burst
2.  n1  predicted value for the next CPU burst
3.  , 0    1
4. Define :
 n 1   t n  1   n .

If we expand the formula, we get:
n+1 =  tn+(1 - )  tn -1 + …
+(1 -  )j  tn -j + …
+(1 -  )n+1 0
Operating System Concepts
6.13
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Priority Scheduling
 A priority number (integer) is associated with each process.
 The CPU is allocated to the process with the highest priority
 Some systems have a high number represent high priority.
 Other systems have a low number represent high priority.
 Text uses a low number to represent high priority.
 Priority scheduling may be preemptive or nonpreemptive.
Operating System Concepts
6.14
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Assigning Priorities
 SJF is a priority scheduling where priority is the predicted
next CPU burst time.
 Other bases for assigning priority:
 Memory requirements
 Number of open files
 Avg I/O burst / Avg CPU burst
 External requirements (amount of money paid, political
factors, etc).
 Problem: Starvation -- low priority processes may never
execute.
 Solution: Aging -- as time progresses increase the priority
of the process.
Operating System Concepts
6.15
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Round Robin Scheduling
 Each process gets a small unit of CPU time (time
quantum).
 A time quantum is usually 10-100 milliseconds.
 After this time has elapsed, the process is preempted
and added to the end of the ready queue.
 If there are n processes in the ready queue and the
time quantum is q,
 then each process gets 1/n of the CPU time in
chunks of at most q time units at once.
 No process waits more than (n-1)q time units.
Operating System Concepts
6.16
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
Example of RR, time quantum = 20
Process
P1
P2
P3
P4
 The Gantt chart is:
P1
0
P2
20
37
P3
Burst Time
53
17
68
24
P4
57
P1
77
P3
97 117
P4
P1
P3
P3
121 134 154 162
 Compute: Avg service time, Throughput, avg turnaround, avg wait:
 Typically, higher average turnaround than SJF, but better response.
 Suppose P1 arrives at 0, P2 at 19, P3 at 23 and P4 at 25. What
changes? What are the new values?
Operating System Concepts
6.17
Silberschatz, Galvin and Gagne 2002
Modified for CSCI 399, Royden, 2005
RR Performance
 Performance varies with the size of the time slice, but not in a simple way.
 Short time slice leads to faster interactive response.
 Problem: Adds lots of context switches. High Overhead.
 Longer time slice leads to better system throughput (lower overhead), but
response time is worse.
 If time slice is too long, RR becomes just like FCFS.
 Time slice vs process switch time:
 If time slice = 20 msec and process switch time = 5 msec, then 5/25 =
20% of CPU time spent on overhead.
 If time slice = 500 msec, then only 1% of CPU used for overhead.
 The time slice should be large compared to the process switch time.
 A typical time slice is 1 sec (4.3 BSD UNIX)
 RR makes the implicit assumption that all processes are equally
important.
 Cannot use RR is you want different processes to have different
priorities.
Silberschatz, Galvin and Gagne 2002
Operating System Concepts
6.18
Modified for CSCI 399, Royden, 2005