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EASTERN MEDITERRANEAN UNIVERSITY
COMPUTER ENGINEERING DEPARTMENT
CMPE-242 Operating Systems
SPRING 2013-2014
MID I
16/04/2014
100 minutes, attempt all...
Student No
:…SOLUTION……………...........…
Name Surname
:…………………..........
Group:
:……………………………
(Instructors: [Gr. 1] Dr. Ahmet Ünveren, [Gr.2] Dr. Alexander Chefranov)
Grades Collected
Q1
Q2
Q3
Q4
Q5
Q6
Q7
TOTAL
CMPE242 Operating Systems
Q.1.(20 pts.)
a) (2.5 points) What are the two main goals of an operating system?
- interfacing with a user,
- resource management
b) (2.5 points) What services are provided by the kernel of an operating system?
Specify any three
-
Process management
Processor scheduling
Processes communication
c) (2.5 points) What are the benefits of multiprogramming? Specify any two
- Increase CPU utilization
- Decrease response time
d)
-
(2.5 points) What are the types of parallel systems? Specify any three
Centralized
Decentralized
symmetric
e) (2.5 points) What an interrupt is? What an interrupt vector is?
-
Interrupt is a signal (usually, externally generated) causing switching CPU to the
respective interruption service routine (ISR)
Interrupt vector contains ISR address
f) (2.5 points) How information is organized in magnetic disks? Explain notions of
surface, track, sector, cylinder
Information is kept on several magnetic material covered surfaces and on each surface is
organized in concentric circles (tracks) split into sectors, usually of 512 bytes. Same radius
tracks of different surfaces form a cylinder accessible by magnetic headers without moving
g) (2.5 points) What two registers are used for memory protection in multiprogramming
environment? What do they hold?
- Base register – contains minimal address of the memory allocated to a process
- Limit register – contains the size of the allocated memory
h) (2.5 points) Give an example of volatile and non-volatile storage devices
RAM – volatile, Hard disk – non-volatile memory
CMPE242 Operating Systems
Q.2. (10 pts.) Fill in the blanks of the following sentences:
a) ………OS………… is a program that acts as an intermediary between a user of a
computer and the computer hardware.
b) CPU Scheduling choose among several jobs ready to run. The system must know
their exact locations in memory.
c) In computer systems generally we have two communication models. With these
models communication may take place using either
Msg Passing or Shared Memory .
d) The long-term scheduler controls the degree . This corresponds to the number of
processes in memory.
e) CPU_bound process – spends more time doing computations; few I/O, very long CPU
bursts.
f) PCB holds the information associated with each process.
Q.3. (10 pts) Below there is a diagram of process states and transitions between them. Fill
in entries 1. - 11.
CMPE242 Operating Systems
Q.4. (15 pts.)
Consider Algorithm 1 – Strict Alternation Approach
Shared variables:
int turn;
initially turn = 0
turn = i  Pi can enter its critical section
Process Pi
Process Pj
do {
do {
while (turn != i) ;
critical section
turn = j;
reminder section
} while (true);
while (turn != j) ;
critical section
turn = i;
reminder section
} while (true);
This solution does not satisfy the progress requirement: If Pi wants to enter
its critical section, and turn is j, then it can not even if Pj may be executing in
its remainder section
a) (6 points) What the progress requirement is? Give its definition
Processes executing in their remainder sections shall not affect on the decision of entering a
critical section by the waiting it processes. Waiting of entering shall not be indefinite
b) (9 points) Give an example of scenario of the two processes execution
(what each process does in time) leading to the violation of the progress
requirement
Let the process P0 passes the critical section, and sets turn=1. Then the process P1 passes
the critical section, and sets turn=0 while the process P0 is still in its remainder section. Let
P1 quickly process its remainder section and tries to enter the critical section, but it can not
enter since turn=0, and P0 is still in its remainder section. Hence, P1 willing to enter the
critical section depends on P0 being in its remainder section, that violates the progress
requirement
CMPE242 Operating Systems
Q.5.( 15 pts) The processes A, B and C are given below.
Synchronize the processes in the following execution order:

produce1 , produce3, produce2, produce4, and produce5
Shared data:
Semaphore ...S1=1;S2=0;S3=0;S4=0;.S5=0;....................................................
Process A:
{
Process B:
{
Process C:
{
Wait(S1);
Wait(S2);
Wait(S3);
//produce1;
Signal(S3);
//produce2;
Signal(S4);
Wait(S4);
//produce4;
Signal(S5);
}
//produce3;
Signal(S2);
Wait(S5);
//produce5;
Signal(S1);
}
}
CMPE242 Operating Systems
Q.6. (15 pts) Consider the following set of processes:
Process ID
P1
P2
P3
Arrival Time
0
3
5
4
P4
CPU Burst Time (ms)
9
3
5
2
a) Draw Gantt chart illustrating the execution of these processes using the First Come First
Serve (FCFS) algorithm. (9 points)
P1
1
P4
P2
2
3
4
5
6
7
8
9
0
1
2
3
P3
4
5
6
7
8
9
b) Calculate mean waiting time, mean turn-around time, and CPU utilization. Show
details of your calculations (6 points)
W(P1)=0, W(P2)=6, W(P3)=9, W(P4)=8, W=(0+6+9+8)/4=5.75
Tot(P1)=9, Tot(P2)=9, Tot(P3)=14, Tot(P4)=10, Tot=(9+9+14+10)/4=10.5
U=19/19*100%=100%
CMPE242 Operating Systems
Q.7. (15 pts) Consider the following set of processes:
Process ID
P0
P1
P2
Arrival Time
2
1
0
CPU Burst Time
1
3
5
I/O Burst Time
1
7
2
CPU Burst Time
2
1
1
c) Draw Gantt charts illustrating the execution of these processes using the
Shortest Remaining Time First (SRTF) algorithm. (note: I/O device uses FCFS
scheduling)
P2| P1| P0| P1
0 1
2
3
| P0
5
| P0 | |
| P2| | P1|
7
| P2
11 12 13 14 15
P1
| P2
|
0
d) Based on your work above, fill in the table below giving both the waiting time (Wait) and turnaround
time (tat) for each process:
Scheduling Algorithm
Parameter
P0
SRTF
Wait
1
Tat
Process ID
P1
P2
1
6
12
15
5
c) Calculate the CPU utilization.
(13/15)x100
CMPE242 Operating Systems