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1 INTERNAL ASSESSMENT TEST II Date: 04/04/2018 Max Marks: 40 Subject & Code: Operating Systems 15CS64 Semester: VI (A & B) Name of the faculty: Mrs.Sharmila Banu.A Time: 8.30 am am Answer any FIVE full questions, choosing one full question from each part PART I Q.No Questions Marks 1 a) What is race condition? Explain with an example. 4 count++ could be implemented as register1 := count register1 := register1 + 1 count := register1 count-- could be implemented as register2 := count register2 := register2-1 count := register2 Consider this execution interleaving with count = 5 initially: T0: producer execute register1 := count {register1 = 5 T1: producer execute register1 := register1 + 1 {register1 = 6 T2: consumer execute register2 := count {register2 = 5 T3: consumer execute register2 := register2-1 {register2 = 4 T4: producer execute count := register1 {count = 6 T5: consumer execute count := register2 {count = 4 several processes access and manipulate the same data concurrently and the outcome of the execution depends on the particular order in which the access takes place, is called a race condition. b) Define busy waiting. Re-implement these two semaphore functions that eliminate the busy waiting loop. wait (S) { signal (S) { S++; while (S <= 0); S--; 4 Busy waiting means that a process maintains the CPU while it is waiting for some I/O operation to complete (rather than giving up the CPU). This is generally undesirable because the processor could be doing useful work otherwise. Solution without Busy Waiting: wait(s) {

2 value--; if (value <0) ( add process to wait list; block; signal(s) { value++; if (value<=0) { remove P from wait list wakeup(p); 2 a) Define monitor. Explain dining philosopher's solution using monitor. 6 A higher-level abstraction that provides a convenient and effective mechanism for process synchronization Key Idea: Only one process may be active within the monitor at a time monitor DP { enum { THINKING; HUNGRY, EATING) state [5] ; condition self [5]; void pickup (int i) { state[i] = HUNGRY; test(i); if (state[i]!= EATING) self [i].wait; void test (int i) { if ( (state[(i + 4) % 5]!= EATING) && (state[i] == HUNGRY) && (state[(i + 1) % 5]!= EATING) ) { state[i] = EATING ; self[i].signal () ; void putdown (int i) { state[i] = THINKING; // test left and right neighbors test((i + 4) % 5); test((i + 1) % 5); initialization_code() { for (int i = 0; i < 5; i++) state[i] = THINKING; Each philosopher I invokes the operations pickup() and putdown() in the following sequence: dp.pickup (i) EAT dp.putdown (i) 2

3 b) How does the signal() operation associated with monitors differ from the corresponding 2 operation defined for semaphores? The signal() operations associated with monitors is not persistent in the following sense: if a signal is performed and if there are no waiting threads, then the signal is simply ignored and the system does not remember the fact that the signal took place. If a subsequent wait operation is performed, then the corresponding thread simply blocks. In semaphores, on the other hand, every signal results in a corresponding increment of the semaphore value even if there are no waiting threads. A future wait operation would immediately succeed because of the earlier increment. PART II 3 a) Define deadlock. What are the necessary conditions for deadlock to occur? 3 A set of processes is in a deadlock state when every process in the set is waiting for an event that can be caused only by another process in the set. i. Mutual Exclusion - at least one thread must hold a resource in a non-sharable mode. ii. Hold and Wait - at least one thread holds a resource and is waiting for other resource(s) to become available. iii. No Preemption - A thread can only release a resource voluntarily; another thread or the OS cannot force the thread to release the resource. iv. Circular wait - A set of waiting threads {t1; : : : ; tn where ti is waiting on ti+1 (i = 1 to n) and tn is waiting on t1. b) Explain the deadlock prevention methods in detail. 5 Restrain the ways request can be made Mutual Exclusion not required for sharable resources; must hold for nonsharable resources Hold and Wait must guarantee that whenever a process requests a resource, it does not hold any other resources o Require process to request and be allocated all its resources before it begins execution o Allow process to request resources only when the process has none o Low resource utilization; starvation possible No Preemption o If a process that is holding some resources requests another resource that cannot be immediately allocated to it, then all resources currently being held are released o Preempted resources are added to the list of resources for which the process is waiting o Process will be restarted only when it can regain its old resources, as well as the new ones that it is requesting. This protocol is often applied to resources whose state can be easily saved and restored later, such as CPU registers and memory space. It cannot generally be applied to such resources as printers and tape drives.

4 Circular Wait impose a total ordering of all resource types, and require that each process requests resources in an increasing order of enumeration we define a one-to-one function F: R > N, where N is the set of natural numbers. a process can initially request any number of instances of a resource type say, Ri. After that, the process can request instances of resource type Rj if and only if F(Rj) > F(Ri). 4 a) Consider the following snapshot of a system: Allocation Max Available A B C D A B C D A B C D P P P P P Answer the following questions using the banker s algorithm: a. What is the content of the matrix Need? b. Is the system in a safe state? c. If a request from process P1 arrives for (0,4,2,0), can the request be granted immediately? If it can be granted, please find and show the safe sequence. 8 a. What is the content of the matrix Need? The values of Need for processes P0 through P4 respectively are (0, 0, 0, 0), (0, 7, 5, 0), (1, 0, 0, 2), (0, 0, 2, 0), and (0, 6, 4, 2). b. Is the system in a safe state? Yes. With Available being equal to (1, 5, 2, 0), either process P0 or P3 could run. Once process P3 runs, it releases its resources which allow all other existing processes to run. c. If a request from process P1 arrives for (0,4,2,0), can the request be granted immediately? Yes it can. This results in the value of Available being (1, 1, 0, 0). One ordering of processes that can finish is P0, P2, P3, P1, and P4. PART III 5 a) Define paging. How does it eliminate external fragmentation? Explain paging hardware in detail. Logical address space of a process can be noncontiguous; process is allocated physical memory whenever the latter is available. Thus it avoids external fragmentation. Concept of Paging: Divide physical memory into fixed-sized blocks called frames (size is power of 2, between 512 bytes and 8,192 bytes) Divide logical memory into blocks of same size called pages Keep track of all free frames, To run a program of size n pages, need to find n free frames and load program Set up a page table to translate logical to physical addresses. Address generated by CPU is divided into: Page number (p) used as an index into a page table which contains base address of each page in physical memory Page offset (d) combined with base address to define the physical memory address that is sent to the memory unit as shown in fig. For given logical address space 2 m and page size 2 n 2 6

5 Example: b) On a single paging system with 2 32 bytes of physical memory, 2 12 pages of logical memory address space and page size of 512 bytes, i)how many bits are in a physical address? ii) how many bits are in a logical address? 2 i) 32 bits ii) 21 bits

6 6 a) Explain the concept of segmentation with a neat diagram. 5 Memory-management scheme that supports user view of memory A program is a collection of segments A segment is a logical unit such as: main program procedure function Logical address consists of a two tuple: <segment-number, offset>, Segment table maps two-dimensional physical addresses; each table entry has: base contains the starting physical address where the segments reside in memory limit specifies the length of the segment Segment-table base register (STBR) points to the segment table s location in memory Segment-table length register (STLR) indicates number of segments used by a program; The segment number is used as an index to the segment table. The offset d of the logical address must be between 0 and the segment limit. If it is not, we trap to the operating system (logical addressing attempt beyond, end of segment). When an offset is legal, it is added to the segment base to produce the address in physical memory of the desired byte. The segment table is thus essentially an array of base-limit register pairs. b) Consider a paging system with the page table stored in memory. If a memory reference takes 200 nanoseconds how long does a paged memory reference take? If we add a TLB and 75 percent of all page-table references are TLB hits, what is the effective memory reference time? (Assume that finding a page-table entry in the TLB takes zero time, if the entry is there.) 3 2

7 a. 400 nanoseconds; 200 nanoseconds to access the page table and 200 nanoseconds to access the word in memory. b. Effective access time = 0.75 (200 nanoseconds) (400 nanoseconds) = 250 nanoseconds. PART IV 7 a) Suppose a computer lab contains two printers, three tape drives, and four crystal ROM writers. Assume program 1 is currently using a printer and two crystal ROM writers, program 2 is using a tape drive, and program 3 is using two crystal ROM writers, a tape drive, and a printer. Suppose now that program 1 requires a tape drive, program 2 requires a printer, and program 3 requires one more crystal ROM writer. Draw the corresponding resource allocation graph. Is the system in a deadlocked state? Why or why not? 4 No. P1's request is satisfied immediately. Assuming that no more resource requests are made, P1 will be able to eventually complete, freeing the resources required by the other processes. b) A paging system uses 16-bit address and 4K pages. The following shows the page tables of two running processes, Process 1 and Process 2. Translate the logical address 16,000 of Process 1 and the logical address 9,000 of Process 2 to their physical addresses. Fill your answers into the table below. 4

8 Process 1 16, ,712 11,904 Process 2 9, ,480 8 a) What is a critical section problem? What requirements should a solution to critical section 8 problem satisfy? State Peterson's solution and indicate how it satisfies the above requirements. Critical-Section Problem Consider a system consisting of n processes {P0, P1,..., Pn. Each process has a segment of code, called a critical section, in which the process may be changing common variables, updating a table, writing a file. No two processes are executing in their critical sections at the same time. The critical-section problem is to design a protocol that the processes can use to cooperate. The section of code implementing this request is the entry section. The critical section may be followed by an exit section. The remaining code is the remainder section. General Structure of process Pi Three requirements to be satisfied: 1. Mutual Exclusion No two processes can execute simultaneously in their critical section. 2. Progress - If no process is executing in its critical section and some processes wish to enter their critical sections, then only those processes that are not executing in their remainder sections can participate in the decision on which will enter its critical section next, and this selection cannot be postponed indefinitely. 3. Bounded Waiting - There exists a bound, or limit, on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted. Peterson s Solution Two process solution Assume that the LOAD and STORE instructions are atomic. The two processes(pi and Pj) share two variables: int turn; Boolean flag[2] Initially: flag[0]=flag[1]= false The variable turn indicates whose turn it is to enter the critical section. if turn == i, then process Pi is allowed to execute in its critical section The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process P i is ready. 2

9 Algorithm for Process P i do { flag[i] = TRUE; turn = j; while ( flag[j] && turn == j); CRITICAL SECTION flag[i] = FALSE; REMAINDER SECTION while(true); To prove property 1, we note that each P; enters its critical section only if either flag[j] == false or turn -- i. Also note that, if both processes can be executing in their critical sections at the same time, then flag [0] == flag [1] == true. These two observations imply that Po and Pi could not have successfully executed their while statements at about the same time, since the value of turn can be either 0 or 1 but cannot be both. Hence, one of the processes say Pj must have successfully executed the while statement, whereas P, had to execute at least one additional statement ("turn == j"). However, since, at that time, f lag[j] == true, and turn == j, and this condition will persist as long as Pj is in its critical section, the result follows: Mutual exclusion is preserved. To prove properties 2 and 3, we note that a process P, can be prevented from entering the critical section only if it is stuck in the while loop with the condition flag [j] == true and turn == j; this loop is the only one possible. If P; is not ready to enter the critical section, then flag [j] == false, and P; can enter its critical section. If Pj has set flag [j ] to true and is also executing in its while statement, then either turn == i or turn == j. If turn == i, then P, will enter the critical section. If turn == j, then Pj will enter the critical section. However, once P; exits its critical section, it will reset f lag[j] to false, allowing P, to enter its critical section. If Pj resets flag [j ] to true, it must also set turn to i. Thus, since P, does not change the value of the variable turn while executing the while statement, P,- will enter the critical section (progress) after at most one entry by P/ (bounded waiting). PART V 9 a) Define Readers - Writers problem. Explain in detail how it can be solved using 8 semaphores. A data set is shared among a number of concurrent processes o Readers only read the data set; do not perform any updates o Writers can both read and write. Problem o Allow multiple readers to read at the same time. o Only one writer can access the shared data at the same time. first readers-writers problem, requires that no reader will be kept waiting unless a writer has already obtained permission to use the shared object.

10 second readers-writers problem requires that, once a writer is ready, that writer performs its write as soon as possible. That is no new readers may start reading. Shared Data o Semaphore mutex initialized to 1. o Semaphore wrt initialized to 1. o Integer readcount initialized to 0. The semaphore wrt is common to both reader and writer processes. The mutex semaphore is used to ensure mutual exclusion when the variable readcount is updated. The readcount variable keeps track of how many processes are currently reading the object. The structure of a writer process while (true) { wait (wrt) ; // writing is performed signal (wrt) ; if a writer is in the critical section and n readers are waiting, then one reader is queued on wrt, and n-1 readers are queued on mutex. When a writer executes signal (wrt), we may resume the execution of either the waiting readers or a single waiting writer. The selection is made by the scheduler. The structure of a reader process while (true) { wait (mutex) ; readcount ++ ; if (readcount == 1) wait (wrt) ; signal (mutex) // reading is performed wait (mutex) ; readcount - - ; if (readcount == 0) signal (wrt) ; signal (mutex) ; The semaphore wrt functions as a mutual-exclusion semaphore for the writers. It is also used by the first or last reader that enters or exits the critical section. It is not used by readers who enter or exit while other readers are in their critical sections. 10 a) Explain hierarchical paging. 4 Computer systems support a large logical address space (2 32 to 2 64 ). In such an environment, the page table itself becomes excessively large. we would not want to allocate the page table contiguously in main memory. One simple solution to this problem is to divide the page table into smaller pieces. Break up the logical address space into multiple page tables A simple technique is a two-level page table A logical address (on 32-bit machine with 4K page size) is divided into: a page number consisting of 20 bits a page offset consisting of 12 bits Since the page table is paged, the page number is further divided into: a 10-bit page number 2

11 a 10-bit page offset Address-Translation Scheme Thus, a logical address is as follows: where p i is an index into the outer page table, and p 2 is the displacement within the page of the outer page table b) Show a hardware design for a dynamic memory relocation scheme. What is the responsibility of the OS in managing this hardware? 4 Every time a context switch is made, the OS must reload the relocation and the limit registers with values appropriate for the new process.