1. CGS 3763 Operating Systems Concepts Spring 2013
Dan C. Marinescu
Office: HEC 304
Office hours: M-Wd 11: :30 AM

2. Lecture 24 – Wednesday, March 13, 2013
Last time:
CPU Scheduling
Today:
CPU scheduling
Next time
Process synchronization
Reading assignments
Chapter 6 of the textbook
Lecture 24

3. Comparison of FCFS, SJF, and RR
We have three processes A, B, and C.
Each process is characterized by
Release time (R) – the time the process arrives in the system
Start time (S) – the time the scheduler allows it to run
Wait time (W) – the time till it has to wait until it is started: W = S - R
Work (Wrk) – the amount of time it needs to use the CPU
Finish time (F) – the time the process finishes: F=S+Wrk
Time in system (T) – the time from when the process arrives until it finishes: T = F- R.
We compare
The average time in system
The average waiting time
for three scheduling policies: FCFS, SJF, and RR.
Lecture 24

4. Proc Release time Work Start time Finish time Wait time till start
Time in system A 3 B 1 5
3 + 5 = 8
3 – 1 = 2
8 – 1 = 7 C 2 8
8 + 2 = 10
8 – 3 = 5
10 – 3 = 7 A 3 B 1 5
5 + 5 = 10 4
10 – 1 = 9 C 2
3 + 2 = 5
5 – 3 = 2 A 3 6
6 – 0 = 6 B 1 5 10
1 – 1 = 0
10 – 1 = 9 C 2 8
5 – 3 = 2
8 – 3 = 5
Lecture 24

5. Average waiting time till the job started Average time in system
Scheduling policy
Average waiting time till the job started
Average time in system
FCFS
7/3
17/3
SJF
4/3
14/3 RR
3/3
20/3
Lecture 24

6. Priority scheduling
Each thread/process has a priority and the one with the highest priority (smallest integer  highest priority) is scheduled next.
Preemptive
Non-preemptive
SJF is a priority scheduling where priority is the predicted next CPU burst time
Problem  Starvation – low priority threads/processes may never execute
Solution to starvation  Aging – as time progresses increase the priority of the thread/process
Priority my be computed dynamically
Lecture 24

7. Priority inversion other types of software.
A lower priority thread/process prevents a higher priority one from running.
T3 has the highest priority, T1 has the lowest priority; T1 and T3 share a lock.
T1 acquires the lock, then it is suspended when T3 starts.
Eventually T3 requests the lock and it is suspended waiting for T1 to release the lock.
T2 has higher priority than T1 and runs; neither T3 nor T1 can run; T1 due to its low priority, T3 because it needs the lock help by T1.
Allow a low priority thread holding a lock to run with the higher priority of the thread which requests the lock
other types of software.
Lecture 24