CS 3530 - Operating Systems
Test_1 - Sample Exercise
Example: The following is the FCFS execution diagram given in the book for 5
processes P1 through P5, that have the following CPU Burst times in milliseconds
respectively: 135, 102, 56, 148, 125. All processes arrive at 0ms.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Problem:
 Now you have just 3 processes, P1, P2 and P3, and they arrive at 1, 15 and
23 milliseconds respectively, and they each have two CPU Burst as
follows (there are no I/O Bursts).
P1:
P2:
P3:
10
5
7
3
12
2
For FCFS scheduling
(a) complete the execution diagram,
(b) calculate total waiting time for every process, and
(c) calculate the average waiting time for all processes.
(d) for RR scheduling with a time quantum of 5 milliseconds,
(d) complete the execution diagram,
(e) calculate total waiting time for every process, and
(f) calculate the average waiting time for all processes.
Example: The following is the SRT (Shortest Remaining Time) execution diagram
given in the book for 6 processes P1 through P6, that have the following CPU Burst
times respectively: 135, 102, 56, 148, 125, 65. All processes except P6 arrive at 0ms,
and P6 arrives at 200ms.
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Problem:
 Now you have 5 processes, P1 through P5, and they arrive at 1, 2, 4, 4 and 5
milliseconds respectively, and they have the following CPU Burst times respectively:
9, 8, 3, 10, 4.
For SRT scheduling,
(a) complete the execution diagram,
(b) calculate total waiting time for every process,
(c) calculate the average waiting time for all processes. Assume negligible context
switch time.
d) 1) Explain how priorities can be used in non-preemptive and preemptive scheduling.
(2) List two examples.
Problem:
 For the Batch System Model with Multiprogramming there are Processes P1, P2,
and P3 that arrive at 2, 11, and 16 milliseconds respectively. The degree of
multiprogramming is three.
Each process has two CPU burst times as given in milliseconds:
P1: 4
5
P2: 5
6
P3: 6
5
Each process has two I/O burst times as given in milliseconds:
P1: 4
4
P2: 8
5
P3: 3
3
Assume that there is no delay time caused by context switching. Calculate the
TOTAL wait time for P1, P2 and P3. If two processes are waiting at the same
time, P1 takes precedence over P2 and P3, and P2 take precedence over P3.
Extra credit: Calculate the total turnaround time required for each of the three
processes.
 Using SRT (Shortest Remaining Time) scheduling, you have 5 processes, P1 through
P5, that arrive at 0, 2, 4, 6 and 8 milliseconds respectively, and they have the following
CPU Burst times respectively: 15, 10, 4, 9, 2. There are no I/O bursts.
For SRT scheduling, (a) complete the execution diagram, (b) calculate total waiting time
for every process, and c) calculate the average waiting time for all processes. Assume
negligible context switch time.
Calculate the total turnaround time required for each of the five processes.
Deadlock –Banker’s Algorithm
1. A system has three resource types {A, B, C} and five processes {P1, P2, P3, P4, P5}. The total units of
system resources are: (6, 10, 13). The maximum demands for every process are: P1(2,2,1), P2(3,7,4),
P3(6,6,5), P4(4,3,5), P5(1,6,4). The current allocation is: P1 (0,0,1), P2 (2,0,0), P3 (1,0,3), P4 (2,3,5),
and P5 (0,3,3). All this information is shown in the table below.
(a) Fill in the NEED column shown below. (b) Is this state a safe state? Explain your answer by
identifying a successful future sequence of processes that makes the state safe, or by explaining which
processes are part of the problem that makes the state unsafe?
Answer
(3a)
PROCESS
P1
P2
P3
P4
P5
(3b)
ALLOCATION
A B C
0 0 1
2 0 0
1 0 3
2 3 5
0 3 3
MAX
A B C
2 2 1
3 7 4
6 6 5
4 3 5
1 6 4
NEED
A B C
2 2 0 Subtract (Allocation-Max) = need
1 7 4
5 6 2
2 0 0
1 3 1
Starting allocation is (5 6 12) subtract ( - ) Total Unit is (6, 10, 13)
which leaves (1 4 1) as the available resources.
It is a safe state because the following order of processes can work:
The total system units are (6,10,13) or 6 resources for A, 10 for B, and 13 for C. And the current
Allocation for P1 through P5 show a total current allocation of 5 for A, 6 for B and 12 for C:
A B C
0 0 1
2 0 0
1 0 3
2 3 5
0 3 3
------5 6 12
If you subtract (5,6,12) from total units of system resources (6,10,13) you get (1,4,1);
Note: use (AVAILABLE := AVAILABLE – Request)
Of all the processes, P1-P5, only P5 can use (1,4,1) to get to its MAX need, because P5 is at
Allocated (0,3,3) and its MAX need is (1,6,4) and all it needs (Need Allocation – MAX) is (1,3,1) to get
there, so it takes the (1,3,1) and (by subtracting available resources of (1,4,1) and (need) (1,3,1) is
leaves (0,1,0). And so P5 goes first and then when it is totally finished, it releases the max
resources it used, which was (1,6,4) and you add that to (0,1,0) and (by adding Allocation :=
Allocation + Request ti), Which is (1,6,4) +(0,1,0) your get (1,7,4).
Then next P2 uses (2,0,0) and needs to go to (3,7,4) so it uses all the resources (1,7,4) and frees its
(3,7,4) resources when done, leaving (3,7,4) available since it used all the resources.
Then P4 or P1 can go because P4 needs (2,0,0) and (3,7,4) is available and P1 can go because it
needs (2,2,0) and (3,7,4) is available. So in the case where P4 goes next, it uses (2,0,0) making it
go to its max of (4,3,5) and leaving (1,7,4), and then when it is done, it frees its (4,3,5) and that
adds to the (1,7,4) making (5,10,9) available for the next process, and so on.
<P5, P2,P4,P1,P3> or <P5, P2,P4,P3,P1> or <P5, P2,P1,P4,P3>
P5:
P2:
P4:
P1:
P3:
Uses (1 3 1) and when done leaves (1 7 4) as the available resources
Uses (1 7 4) and when done leaves (3 7 4) as the available resources
Uses (2 0 0) and when done leaves (5 10 9) as the available resources
Uses (2 2 0) and when done leaves (5 10 10) as the available resources
Uses (5 6 2) and when done leaves (6 10 13) as the available resources
Or:
P5:
P2:
P4:
P3:
P1:
Uses (1 3 1) and when done leaves (1 7 4) as the available resources
Uses (1 7 4) and when done leaves (3 7 4) as the available resources
Uses (2 0 0) and when done leaves (5 10 9) as the available resources
Uses (5 6 2) and when done leaves (6 10 12) as the available resources
Uses (2 2 0) and when done leaves (6 10 13) as the available resources
Or:
P5:
P2:
P1:
P4:
P3:
Uses (1 3 1) and when done leaves (1 7 4) as the available resources
Uses (1 7 4) and when done leaves (3 7 4) as the available resources
Uses (2 2 0) and when done leaves (3 7 5) as the available resources
Uses (2 0 0) and when done leaves (5 10 10) as the available resources
Uses (5 6 2) and when done leaves (6 10 13) as the available resources
CS 3530 – Operating Systems
Test_1 - Review Topics
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Definition of an Operating System
Linux commands been used with cs3
Stochastic Model
Deterministic Model
Processes Vs. Threads
User Vs. Kernel Level Resources
Types of Performance Models
Performance Metrics
Resource Capacity and Bottleneck
Simple Batch Model
Multiprogramming
Multiprogramming Batch Model
CPU and I/O Bursts
Interrupts
Context Switching
Queues: Input, Ready, I/O
Scheduling Policies
Types of Schedulers
Role of the Dispatcher
Execution Diagrams
Coordination between Processes / Processor Synchronization
Definition of Critical Section
Different types of Semaphores / Implementation Code
Classical Synchronization Problems including Dining Philosophers
Process Deadlock / Handling Deadlock
Resource Allocation Graphs
File Management
Differences in File Systems
 Disk Scheduling Algorithms