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CS 471
Operating Systems Spring 2010Final 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 6 0 8 6 2
Working set size
PageReplacement Algorithm
1 2 10
Optimal 10 7 6Random 10 7|8|9 6FIFO 10 7 6LRU 10 9 6LFU 10 8 6MFU 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 completedNonblocking - 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
Servers
1 2 3
1 EC PD EC
2 PD DD EC
3 EC EC NC
ServerX.cintmain(void){
int len, listenfd, connfd;struct sockaddr_in servaddr, cliaddr;char buff[512];int 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.cintmain(int argc, char **argv){
int sockfd;struct sockaddr_in servaddr;struct hostent *hp, *gethostbyname();char buffer[512];int 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
FCFS SSTF C-LOOK
Total Head Movement
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 whereBit [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 D2
F1 F2 F3 D3
F4 F5
B. What is the output of the following two statements:
% cat /home/cs471w/public_html/spring10/final/Q5/D2/D3/F5Answer:
HI% cat /home/cs471w/public_html/spring10/final/Q5/D2/D3/F4/F2
Answer:
BY
C. Consider the following three programs:Q5R.cintmain(int argc, char **argv){ int fd; fd = open("Q5file", O_RDONLY); read_lock(fd); printf("Q5R"); while (1) sleep(20); unlock(fd);}Q5S.cintmain(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.cintmain(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: 0123456789Fill 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 Q5W2nd raw: execute Q5S then Q5R, Q5S and Q5W3rd raw: execute Q5W then Q5R, Q5S and Q5W
If no out is produced, enter: nothing
Q5R Q5S Q5W
Q5R Q5R Q5R Q5S56789 nothingQ5S Q5S56789 Q5R Q5S56789 nothingQ5W Q5W nothing nothing nothing
