COMP SCI 3SH3: Operating System Concepts (Term 2 Winter 2006) Test 2 February 27, 2006; Time: 50 Minutes ;. Questions Instructor: Dr.

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1 COMP SCI 3SH3: Operating System Concepts (Term 2 Winter 2006) Test 2 February 27, 2006; Time: 50 Minutes ;. Questions Instructor: Dr. Kamran Sartipi Name: Student ID: Question 1 (Disk Block Allocation): [12 marks] For each of the following disk allocation techniques specify its characteristics in terms of: 1) Ease of growth of file size; 2) How the free disk blocks are assigned; 3) Ease of random access to a location in a file: 4) Advantages and disadvantages. Contiguous allocation: 1) Files can not grow, the max size of the file is set from the beginning 2) Contiguous free disk blocks are assigned to a file 3) Random access is easily performed 4) Advantage: scheduling is easy but external fragmentation wastes disk space and needs de-fragmentation. Linked allocation: 1) File grow easily by adding a new free block at the end of the linked free blocks 2) The blocks connected by linking a new block to the last block, where a small portion of each block is used for the address of the next block. In FAT technique a linked list of block numbers are kept in the memory. 3) Random access is not easy and needs to traverse the link of addresses. 4) Advantage: easy to grow or shrink, no external fragmentation. Disadvantage: random access is time consuming. Indexed allocation 1) Files grow easily and using indirect index tables a file can grow almost unlimited. 2) Each file has an index-block which is a regular disk block and the address of a free disk block is assigned to the next location in this index block. 3) Random access is easily performed with very low overhead. 4) Advantage: data blocks and index blocks are regular disk blocks so the method is very flexible, and does not need a FAT table to be kept in memory. Disadvantage: for small size files the index block space is not used efficiently; however in Unix a small number of indexes are kept in the FCB to accelerate the address mapping and eliminate the need to read index-block from file. Question 2 (File system): [12 marks] Figure below illustrates the block diagram of a file system and the tables system wide open-file table and per-process open-file table. Two processes perform open file commands to open two instances of files foo and bar and perform operations on these two files. Page 1 of 8

2 Answer to the following questions: 1. Which of the above shown pieces are in hard-disk and which ones are in memory? Only the File system is in hard disk and the rest are in memory. 2. What is the purpose of Super block and what are its contents? Page 414 text book: A super block in Unix File System (extra: also known as partition control block and Master File Table in NTFS) contains partition details, such as the number of blocks in the partition, size of the blocks, free-block count and free-block pointers, and free FCB count and FCB pointers. 3. By drawing arrows and labeling the arrows show how the values of four file descriptors fd1, fd2, fd3, and fd4 are determined, and write the values below. fd1: 4, fd2: 5, fd3: 3, fd4: 4 4. What are the contents of entries in System-wide open-file table and per-process open-file table? Contents of System wide open-file table (SWOFT): class slide 5 for chapter 12 and pages 414 to 416 of text book Four of six answers gets full mark - File permissions (access controls) - File dates (create, access, write) - File owner, group, others - File size Page 2 of 8

3 - (pointers to) File data blocks. - File count (number of open file instances of a file). For the above example file count is 2 for entries 7 and 8 in SWOFT. Contents of per-process open-file table (PPOFT): - File permissions or access controls (more specific than those in SWOFT) - Pointer to the appropriate entry in the SWOFT. In the above example, number 7 or 8. - Pointer to the current location for read or write in the file. 5. Suppose Process 1 is a producer and Process 2 is a consumer. Specify the sequence of instructions and use the values you obtained for file descriptors such that Process 1 passes 300 bytes to Process 2 (in Figure rc is read-count and wc is write-count). Process 1: Put 300 bytes from address space of Process 1 into Buffer starting from the first entry of the Buffer. fd2 = open(/../../bar, Read-and-write); // fd2=5 wc = write(fd2, address of first entry of Buffer, 300); // wc should be 300 if wc!= 300 then Error( couldn t write 300 bytes into file bar ) close(fd2) /// this line is optional for this question.. Process 2: fd3 = open(/../../bar, Read-only); // fd3 = 3 rc = read(fd3, address of first entry of Buffer, 300); // rc should be 300 if rc!= 300 then Error ( couldn t read 300 bytes from file bar into Buffer) Get 300 bytes from Buffer starting from the first entry of Buffer and move them into the address space of Process 2. close(fd3); /// this line is optional for this question 6. Using Buffer and file system blocks explain briefly how data would be transferred between Buffer and disk blocks, and how many disk blocks are involved in this transfer? Process 1 puts 300 bytes from the address space of Process 1 into the Buffer, starting from first entry of the Buffer. The OS allocates 3 free blocks (2 x 128 =256 < 300 < 3 x 128 = 384) to put 300 bytes from Buffer. Page 3 of 8

4 The OS assigns these three blocks as the file blocks of file bar. Depending on the file allocation mechanism of OS, this can be contiguous allocation, Indexed allocation, or linked allocation. Question 3 (CPU scheduling): [20 marks] 2.1) Consider the following set of processes P1 to P5, with the length of CPU-burst time and arrival time given in milliseconds. Fill in the provided empty Gantt chars below to illustrate the execution of these processes using Round Robin (RR with quantum size = 2), First Come First Serve (FCFS), and a preemptive Shortest Remaining Time First (SRTF) scheduling. Process Burst time Arrival Time P P2 1 1 P3 3 4 P4 2 8 P Page 4 of 8

5 2.2) Compute the average waiting time for the FCFS and SRTF scheduling and write your calculation below. [10 marks] FCFS: Waiting time for: P1: 0; P2: 9; P3: 7; P4: 6; P5: 5 Total waiting time: ( ) = 27 Average waiting time: 27 / 5 = 5.4 SRTF: P1: 6; P2: 0; P3: 0; P4: 0; P5: 5 Average waiting time: 11/ 5 = 2.2 Page 5 of 8

6 Question 4 (Critical Section): [20 marks] Answer True or False to the following statements. 1. Critical section is a global variable that two or more processes can access and modify. FALSE 2. Critical section is a part of a program where two or more processes assess a shared resource. TRUE 3. Any solution to the critical section problem must satisfy mutual exclusion, progress and bounded waiting conditions. TRUE 4. Semaphore is a means to provide both mutual exclusion and synchronization of the concurrent processes. TRUE Question 5 (Semaphore): [20 marks] Below two processes access a shared variable counter. The content of counter before execution of these pieces of code is 6. Process 1 Process 2. register1 = counter register2 = counter register1 = register1 + 1 register2 = register2-1 counter = register1 counter = register2. 1. Specify two interleaving of these machine language instructions such that at the end the value of register counter becomes: counter = 5: register1 = counter register1 = register1 + 1 register2 = counter register2 = register2-1 counter = register1 counter = register2 counter = 7: register1 = counter register1 = register1 + 1 register2 = counter register2 = register2-1 counter = register2 counter = register1 Page 6 of 8

7 2. Use semaphore S and initialize it to a proper value; then use two instructions wait(s) and signal(s) in the above processes to protect the critical section of the processes. Semaphore S S := 1 Process 1 Process 2.. wait(s) wait(s) register1 = counter register2 = counter register1 = register1 + 1 register2 = register2-1 counter = register1 counter = register2 signal(s) signal(s). 3. Explain using the values of S and the blocking processes on how semaphore S provides mutual exclusion for the critical section. S is initialized to 1 in order to be used for process synchronization. All operations on semaphore S are performed atomically (i.e., they cannot be interrupted). The first process that performs wait(s) the value of S is decreased to 0 and that process can proceed to its critical section and manipulate value of counter exclusively. The process performs signal(s) when it exits from its critical section and S becomes 1. During the time that a process is in its critical section (S=0) if another process performs wait(s) then the process is blocked an is put in the waiting queue for semaphore S until the first process exits form its critical section and performs signal(s) that causes OS to wake-up the process in the head of the queue for semaphore. Now the latter process can enter its critical section. Page 7 of 8

8 Question 6 (Thread): [20 marks] Answer True or False to the following statements about multi-trading modes. 1. Threading (or light weight processes) is a mechanism of multiplexing the execution time of a heavy-weight process into several threads of execution to reduce the high overhead for context switching. TRUE 2. In one-to-one model, many user-level threads are mapped onto a single kernel thread, and threads can run on a multi-processor system. FALSE 3. In many-to-one model, many kernel-level threads are mapped to one single userlevel thread and thread can run on distributed systems. FALSE 4. In many-to-many model, many user-level threads are mapped to many kernel threads, and this is the most flexible mode. TRUE Page 8 of 8