CS 471
Operating Systems
Spring 2010
Final Exam
Time 2 & 1/2 hours
Open Book & Notes
Name: wahab
Unix Login: wahab
Each Question is 20 points.
Question 1
A. Fill in the number of page faults in the following matrix, assuming the following page reference
string:
5 6 1 0 1
Working set
6 0 8 6
1
2
2
10
size
Page
Replacement
Algorithm
Optimal
10
7
6
Random
10
7|8|9
6
FIFO
10
7
6
LRU
10
9
6
LFU
10
8
6
MFU
10
9
6
Question 2
A. Explain the meaning of s in the following listing of the passwd file:
% cd /usr/bin
% ls -l passwd
-r-sr-sr-x
1 root
sys
27228 Aug 16
2007 passwd
Answer:
So that any user can be the root during the execution of
this command
B. Consider the following access matrix.
1. Is it possible for D1 to access the laser printer? Explain.
Answer:
Yes.
Switch to D2.
2. Is it possible for D1 to access F2? Explain.
Answer:
Yes.
Switch to D2 then Switch to D3.
Question 3
In class we discussed the concepts of Blocking and Nonblocking I/O:
Blocking - process suspended until I/O completed
Nonblocking - Returns quickly with count of bytes read or written
To set a file descriptor fd to Nonblocking we use the following system calls:
val = fcntl (fd, F_GETFL, 0);
fcntl (fd, F_SETFL, val | O_NONBLOCK);
Consider the following 6 programs:
ServerX and ClientX, where X = 1, 2, 3.
Fill in the following matrix with one of the following choices:

DD = Definite Deadlock, the client and server are deadlocked

PD = Possible Deadlock, the client and server may reach a deadlock state.

NC = Normal Chatting, a message typed by one “immediately” is seen by the other.

EC = Erratic Chatting, a message typed by one “eventually” will be seen by the other.
Clients
1
2
3
Servers
1
EC
PD
EC
2
PD
DD
EC
3
EC
EC
NC
ServerX.c
int
main(void)
{
int
len, listenfd, connfd;
struct sockaddr_in servaddr, cliaddr;
char
int
buff[512];
nread;
listenfd = socket(AF_INET, SOCK_STREAM, 0);
bzero(&servaddr, sizeof(servaddr));
servaddr.sin_family = AF_INET;
servaddr.sin_addr.s_addr = htonl(INADDR_ANY);
servaddr.sin_port = htons(10123);
bind(listenfd, (SA *) & servaddr, sizeof(servaddr));
listen(listenfd, 0);
len = sizeof(cliaddr);
connfd = accept(listenfd, (SA *) & cliaddr, &len);
ChatX(connfd);
}
ClientX.c
int
main(int argc, char **argv)
{
int
sockfd;
struct sockaddr_in servaddr;
struct hostent *hp, *gethostbyname();
char
int
buffer[512];
nread;
sockfd = socket(AF_INET, SOCK_STREAM, 0);
bzero(&servaddr, sizeof(servaddr));
hp = gethostbyname(argv[1]);
bcopy(hp->h_addr, &(servaddr.sin_addr.s_addr), hp->h_length);
servaddr.sin_family = AF_INET;
servaddr.sin_port = htons(10123);
if (connect(sockfd, (SA *) & servaddr, sizeof(servaddr)) < 0) {
perror("connect error");
exit(-1);
}
ChatX (sockfd);
}
ChatFunctions.c:
Void Chat1(int sockfd)
{
int
nread;
char
buffer[512];
for (;;) {
nread = read(0, buffer, 512);
if (nread == 0) exit(0);
write(sockfd, buffer, nread);
nread = read(sockfd, buffer, 512);
if (nread == 0) exit(0);
write(1, buffer, nread);
}
}
void Chat2(int sockfd)
{
int
nread;
char
buffer[512];
for (;;) {
nread = read(sockfd, buffer, 512);
if (nread == 0) exit(0);
write(1, buffer, nread);
nread = read(0, buffer, 512);
if (nread == 0) exit(0);
write(sockfd, buffer, nread);
}
}
void Chat3(int sockfd)
{
int
val, nread;
char
buffer[512];
val = fcntl (sockfd, F_GETFL, 0);
fcntl (sockfd, F_SETFL, val | O_NONBLOCK);
val = fcntl (0, F_GETFL, 0);
fcntl (0, F_SETFL, val | O_NONBLOCK);
for (;;) {
nread = read(0, buffer, 512);
if (nread == 0) exit(0);
write(sockfd, buffer, nread);
nread = read(sockfd, buffer, 512);
if (nread == 0) exit(0);
write(1, buffer, nread);
usleep(1000);
}
}
Question 4
A. Calculate the total head movement for servicing the following request Queue according to the
specified disk I/O scheduling algorithms.
Request Queue (cylinder range 0-100): 83 37 14 24 65
Head pointer: cylinder 0
Disk Scheduling
Algorithm
Total Head
Movement
FCFS
SSTF
C-LOOK
203
83
83
B. The following is the typical information kept about a file in UNIX.
Assume the following:
 A data block pointer = 10 bytes.
 Number of direct blocks = 10
 A block size = 1000 bytes.
Calculate the maximum possible file size in blocks.
Answer:
10+100+100**2+100**3
C. To keep track of the free disk blocks, one method is to use a Bit vector where
Bit [i] = 0 means block i is occupied.
Assume a word length of 32 bits and the prefix of the bit map is: 003
What is the index number of the first empty block?
Answer:
32+32+30 = 94
Question 5
The following is an example of a general directory structure.
A. Draw the director structure rooted at Q5 after executing the followingcommands:
% pwd
/home/cs471w/public_html/spring10/final
% mkdir Q5
% cd Q5
% mkdir D1 D2
% cd D1
% echo HI > F1
% echo BY > F2
% cd ../D2
% touch F3
% mkdir D3
% cd D3
% ln –s /home/cs471w/public_html/spring10/final/Q5/D1 F4
% ln F4/F1 F5
Answer:
Q5
D1
F1 F2
D2
F3
D3
F4
F5
B. What is the output of the following two statements:
% cat /home/cs471w/public_html/spring10/final/Q5/D2/D3/F5
Answer:
HI
% cat /home/cs471w/public_html/spring10/final/Q5/D2/D3/F4/F2
Answer:
BY
C. Consider the following three programs:
Q5R.c
int
main(int argc, char **argv)
{
int
fd;
fd = open("Q5file", O_RDONLY);
read_lock(fd);
printf("Q5R");
while (1) sleep(20);
unlock(fd);
}
Q5S.c
int
main(int argc, char **argv)
{
int
fd;
fd = open("Q5file", O_RDONLY);
read_lock(fd);
printf("Q5S");
lseek(fd, 5, SEEK_SET);
read(fd, buf, 5);
write(1, buf, 5);
while (1) sleep(20);
unlock(fd);
}
Q5W.c
int
main(int argc, char **argv)
{
int
fd;
fd = open("Q5file", O_WTONLY);
write_lock(fd);
printf("Q5W");
while (1) sleep(20);
unlock(fd);
}
Assume the contents of Q5file is: 0123456789
Fill in the following matrix with the output of each program, assuming we have 4 open windows and in
each window we execute one program as follows:
1st raw: execute Q5R then Q5R, Q5S and Q5W
2nd raw: execute Q5S then Q5R, Q5S and Q5W
3rd raw: execute Q5W then Q5R, Q5S and Q5W
If no out is produced, enter: nothing
Q5R
Q5S
Q5W
Q5R Q5R
Q5S Q5S56789
Q5W Q5W
Q5R
Q5R
nothing
Q5S56789
Q5S56789
nothing
nothing
nothing
nothing