CS604-Operating Systems Solved MCQs Final Term
Lectures Included 23 to 40
1…………….. Is caused due to unavailability of resources
Ans: Starvation Lec#23 p#112
2 The type of semaphore, whose value can not be greater than 1, is……………
Ans: Binary Semaphore Lec#24 p#114
3…………. Separates logical memory from physical memory
Ans: Virtual Memory Lec#37 p# 185
4……………. Register is used to point the smallest memory address of a process and its size.
Ans: Limit Register Lec# 31 p#158
5 There is no dynamic address translation in case of …………………….
Ans: MFT (Multi-Programming with Fixed Tasks) Lec#31 p#159
6 If HR is hit ratio and MR is miss ratio then effective access time is given by…………..
Teffective = HR (TTLB+ Tmem) + MR (TTLB + 2Tmem) Lec#33 p#167
7 Variation of fork ( ) system call used by UNIX is………….
Ans: vfork ( ) Virtual fork Lec#38 p# 193
8 FIFO page replacements has lowest page fault rates
Ans: False Correct Optimal Algorithm Lec#38 p#199
9 Correct about LRU is …………….
Ans: Replacement with least recently used page Lec#39 p#199
10 Physical Memory is broken into fixed sized blocks called……………..
Ans: Frames Lec#32 p#162
11 In UNIX ……………… can be used to request the operating system to memory map an opened file
Ans: mmap ( ) Lec#38 p# 195
12………….. Does not use copy-on-write
Ans: vfork ( ) Lec#38 p#193
The victim page that is selected depends upon the…………..
Ans: Algorithm used Lec#38 p#192
13……………… is the division of every segment into fixed sized pages
Ans: Paged Segmentation Lec#36 p#179
14……………. Is the finest operating system
Ans: MULTICS Lec#36 p#180
15 Segmentation may cause …………………….
Ans: External Fragmentation Lec#35 p#177
16 ……………….. Is a logical unit such as main program, procedure, function, method or object etc
Ans: Segment Lec#35 p#175
17 Each segment has …………….. And…………….
Ans: Name and Length Lec#35 p#175
18 Semaphore requires………………
Ans: Busy Waiting Lec#23 p#109,110
19 Semaphore in which the process spins while waiting for the process is called ………
Ans: Spinlock Lec#23 p#110
20 Spinlock is useful in…………….. Systems
Ans: multiprocessor systems Lec#23 p#110
21 Context switching can be………………
Ans: time consuming Lec#23 p#110
22 you can use semaphore for……………… of cooperating processes
Ans: synchronization Lec#23 p#111
23 Semaphore is ……………… solution for critical section problem
Ans: Hardware based Lec# 23, 22 p#107,108
24 Two operations related with semaphore are ………….
Ans: Wait and Signal Lec#22, 23 p# 108, 112
25 Problem with semaphore is if one process fails then………….
Ans: Entire system of cooperating processes fails Lec#23 p# 112
26 A process is said to be in ………………… state if every process is waiting for an event that can be caused only by another process
Ans: Deadlock state Lec#23 p#112
27 One way bridge crossing is an example of……………..
Ans: Deadlock Lec#23 p#112
28 condition of infinite blocking is called……………..
Ans: Starvation Lec#23 p#112
29 A typical deadlock situation is …………….
Ans: Starvation Lec#23 p#113
30 The problem in which principle of mutual exclusion is violated is……………..
Ans: Violation of mutual exclusion Lec#23 p#113
31 Violation of mutual exclusion problem occurs due to ……………..
Programming errors Lec#23 p#113
32 There are………….. Types of Semaphore
Ans: Two Lec#24 p#114
33 Two types of semaphore are………….. And…………….
Ans: Counting and Binary Semaphore Lec#24 P#114
34 The type of semaphore whose integer value can range over an unrestricted integer is……………
Ans: Counting Semaphore Lec#24 p#114
35 There are……………. Classical problems of synchronization
Ans: Three Lec#24 p#115
36 Classical problems of synchronization are named as…………………. And …………… And ………………………
Ans: Bounded Buffer, Reader and Writer and Dinning Philosophers problems
Lec#24 p#115
37 in Bounded buffer problem solution, semaphore empty is initialized to…………. And semaphore full is initialized to…………….
Ans: n and 0 Lec#24 p#115
38 in Reader and writer problem some process are called…………… and other are called as ………………….
Ans: Reader and Writer Lec#24 p#116
39 Simplest form of reader and writer problem is……………
Ans: First reader and writer Lec#24 P#116
40 If the writer is waiting to access the object, no new reader may start reading this condition represents of………………
.
Ans: Second reader writer problem Lec#24 p#116
41 For Dinning Philosopher problem use an ……………………….. Solution
Ans: Asymmetric solution Lec#24 p#120
42 There will be no deadlock if there does not exist…………………
Ans: Circular waiting Lec#24 p# 120
43 Critical region provides…………………
Ans: Encapsulation Lec#24 p#121
44 A………… is characterized by local data and set of programmer defined operators
Ans: Monitor Lec# 25 p#121
Condition variable is associated with……………….
45 Ans: Monitor Lec# 25 p#122
46 In case of monitor, if no process is suspended then signal operation has……………
Ans: No effect Lec# 25 p#123
47 ……………. Operation resumes exactly one suspended process
Ans: Signal operation Lec# 25 p#123
48 monitors with condition variables can solve…………………. Problems than monitor alone
Ans: More Lec# 25 p#123
49 Only……………….. Process can be active within a monitor and other may be waiting.
Ans: one Lec# 25 p#123
50A process must be holding at least …………… resource while waiting to acquire additional resources
Ans: One Lec# 26 p#128
51 ……………………. Can not be preemptive
Ans: Resources Lec# 26 p#128
52 Deadlocks can be described precisely through……………..
Ans: Resource allocation graph Lec #26 p#129
53 If process contains no cycle then it means that there is no …………………
Ans: Deadlock Lec #26 p#130
54 if there is only one instance per resource type then……………… deadlocks can exist
Only one Lec #26 p#130
55If there are several instances per resource type then……………… deadlocks can exist
Ans: Possibility of deadlock Lec #26 p#130
56 …………………Methods prevent deadlocks by constraining how process can request for resources
Ans: Deadlock prevention Lec#27 p#132
57 ………………………. Algorithm is used in Deadlock avoidance.
Ans: Banker’s Algorithm Lec#28 p#137
58 There are …………….. Ways to deal with deadlocks
Ans: three Lec #27 p# 132
59…………….. State can not have deadlocks
Ans: Safe Lec# 27 p# 134
60The system is in …………………. If it can allocate the resources to each process in same order and avoid a deadlock
Ans: Safe state Lec#27 p#134
61 An ………………….. State is not a deadlock state
Ans: Unsafe State Lec# 27 p#134
62 …………………… ensures that the system never enters in deadlock state
Ans: Deadlock avoidance Lec# 27 p#134
63 When a new process enters in the system it must declare the maximum number of instances of each resource type it may need it is ……………….
Ans: Banker’s Algorithm Lec#28 p#137
64 RRA stands for……………………
Ans: Resource Request Algorithm Lec#28 p# 138
65 Variation of resource allocation graph is………………………..
Ans: Wait for graph Lec#29 p#144
66 Most common solution for starvation is……………………
Ans: Rollbacks Lec# 29 p#149
67 Purpose of ………………… is efficient use of memory
Ans: Memory management Lec# 30 p# 150
68 In case of……………………………. Process can not be moved around in memory at run time
Ans: Compile and load time binding Lec#30 p#151
69 An address generated by CPU is referred as………………..
Ans: Logical Address Lec# 30 p#152
70 An address seen by memory unit loaded into the memory address register of the memory is called …………………..
Ans: Physical address Lec#30 p# 152
71 MMU stands for…………………………
Ans: Memory management unit Lec#30 p# 152
72 We refer logical address as…………………
Ans: Virtual address Lec#30 p#152
73 Set of all logical addresses generated by a program form…………………
Ans: Logical address space Lec#30 p#152
74 Set of all physical addresses corresponding to logical addresses is…………………
Ans: Physical address space Lec#30 p#152
75 Total size of physical address space is equal to total size of…………………
Ans: Main memory Lec# 30 p#152
76 For better memory space utilization we can use……………..
Ans: Dynamic loading Lec# 30 p#154
77 Gcc compilers invokes………………. linking by default
Ans: Dynamic Lec#30 p# 155
78 ……………….. Allows static linking
Ans: -static option Lec#30 p#155
79 If we want to enable a process larger than the memory allocated then we can make use of………………..
Ans: Overlays Lec#31 p#156
80 …………………. Register points to the size of a process
Ans: Limit Register Lec#31 p#158
81 ……………………… register points to the smallest memory address of process
Ans: Base Register Lec#31 p#158
82 …………………refers to the situation when free memory space exists to load a process in the memory but the space is not contiguous.
Ans: External Fragmentation Lec#32 p#162
83 Main Memory is……………….. Memory
:
Ans: Volatile (Ref: RAM is also called the main memory which is a volatile memory
.
84 Overlays are implemented by the ………………………..
Ans: Programmer Lec#31p157
85……………………..keep in memory only those instructions and data that are needed at any given time.
Ans: Overlays Lec#31p157
86 Secondary Storage memory devices have ……………………memory.
Ans: Permanent and non-volatile Lec#30 p151
87 If we want to enable a process larger than the memory allocated then we can make use of……………………..
Overlays Lec#31 p#156
88 Wasted space inside a region refers to………………………
Ans: Internal fragmentation Lec# 31 p#159
89 Inability to use free memory refers to…………………
Ans: Fragmentation Lec#31 p#159
90 MFT stands for………………………..
Multi-Programming with Fixed Tasks Lec#31 p# 159
91 MFT can have……………. Internal fragmentation
Ans: More lec#31 p#159
92 Base Register is also called ………………….
Ans: Reallocations register Lec#31 p#159
93 Base register is a part of……………………. Set
Ans: Programmer visible register set Lec#31 p#159
94 In MFT an alternative protection method is ………………
Ans: Base/Limit Register Lec#31 p#159
95 Advantage of Base/limit register is……………..
Ans: Easy movement of job Lec#31 p#159
96 MFT with multiple queues involves………………
Ans: Load Time binding Lec#31 p#160
97 MVT stands for…………………….
Ans: Multiprogramming with variable tasks Lec#32 p#161
98 MVT eliminates…………………. But introduces………………….
Ans: Internal and External Fragmentation Lec#32 p#161
99 Logical Memory is divided into blocks of same size called…………….
Ans: Pages Lec#32 p#162
100 in order to keep track of pages in the main memory……………….. Is used
Ans: Page Table Lec#32 p#162
- TLB stands for……………..
Translation look-aside buffer Lec#33 p#166
2. TLB has ……………… entries
Ans: 64-1024 Lec#33 p#166
3. Effective Memory access Time on hit is equal to ……………..
Ans: (Tmem + TTLB) Lec#33 p# 166
4. Effective Memory access Time on miss is equal to ……………..
Ans: (2Tmem + TTLB) Lec#33 p# 166
5 If MR is mean ratio and HR is hit ratio then Effective Memory access Time is equal to ……………..
Ans: Teffevtive = HR (TTLB + Tmem) + MR (TTLB + 2Tmem) Lec#33 p# 166
6. Protection bit is associated with each page in …………………
Ans: Paging Lec#34 p#169
7. A logical address space can be ………………….. Large depending upon page size
Ans: 32 or 64 bit large Lec#34 p#170
8. Modern Computers support ……………… logical address space
Ans: Large (usually 32 or 64 bit) Lec#34 p#170
9. Level of paging having 10 bit page number and 10 bit page offset is called…………….
Ans: Two Levels Paging Lec# 34 p# 170
10. Hashed page table approach is used to handle logical address space larger than……………
Ans: 32 Bits Lec#34 p#172
11. For less amount of physical memory to be consumed use……………….
Ans: Inverted page table Lec#34 p#172
12. Paging allows……………
Ans: Sharing of common code Lec#35 p#174
13. Logical address space is collection of ……………….
Ans: Segments Lec#35 p#175
14.……………….. Is a logical unit
Ans: Segment Lec#35 p#175
15. …………………. Indicates no of segments used by a program
Ans: Segment Table Length Register Lec#35 p#175
16………………. Indicates table location in the memory
Ans: Segment Table Base Register Lec#35 p#175
17 Segmentation is associated with the protection of……………….
Ans: Segments Lec#35 p#177
18 If validation bit is 0 it indicates……………….
Ans: illegal segment Lec#35 p#177
19. We divide every segment in fixed sized pages in………………
Ans: Paged segmentation Lec#36 p#179
20………………… is the finest operating system
Ans: MULTICS Lec#36 p#180
21. Physical address is of………………….
Ans: 20 bits Lec#36 p# 183
22. Effective address is of…………….
Ans: 16 bits Lec#36 p# 183
23………………… is separation of logical memory from physical memory
Ans: Virtual memory Lec# 37 p# 185
24…………….. Never swaps a page into memory unless the page will be needed
Ans: Lazy Swapper Lec#37 p#186
25. Access to page marked invalid causes ……………..
Ans: Page Fault trap Lec#37 p#188
26. Effective access time in demand paging is calculated as…………
Ans: EAT = (l-p)* ma + p* page fault time Lec#37 p# 189
27 when there is no free frame available…………………. Is required
Ans: Page replacement Lec#38 p# 192
28 ……………… is used when files may include binary executable
Demand paging Lec#38 p# 193
29 Fork ( ) system call uses…………….
Ans: Copy on write Lec#38 p# 193
30. Only that pages that may be modified are marked………
Ans: Copy on write Lec#38 p# 193
31. A variation of fork system call is………………….
Ans: Vfork ( ) Lec#38 p#193
32………………… does a not use copy on write
Ans: Vfork ( ) Lec#38 p#193
33 In LINUX shared pages are marked read only after ………………………
Ans: Fork ( ) Lec#38 p#194
34 in UNIX………………….. is used to request an operating system to memory map an opened file
Ans: Mmap ( ) Lec#38 p#195
35 We can reduce overhead in page replacement by using…………………
Ans: Modify bit Lec#39 p#197
36 Modify bit is also called …………………..
Ans: Dirty bit Lec#39 p#197
37 FIFO page replacement suffers from…………………..
Ans: Balady Anomaly Lec#39 p# 198
38 as the no of frames available increases, page faults………………
Ans: Decreases Lec#39 p# 198
39 Simplest page replacement algorithm is ……………..
Ans: FIFO replacement algorithm Lec#39 p#198
40………………. Algorithm has lowest page fault rates
Ans: Optimal algorithm Lec#39 p# 199
41 Optimal algorithm is mostly used for……………………
Ans: Comparison Lec#39 p#199
42 LRU stands for…………………..
Ans: Least recently used algorithm Lec#39 p#199
43 ……………… types of implementations are feasible with LRU algorithm
Ans: Two (Counter based & stack based implementations) Lec#39 p# 200
44 in Belady Anomaly as the no of frames allocated increases, page fault rate
………….
Ans: increase Lec#40 p# 202
45 ………………Algorithm is based on locality of reference concept
Ans: Least frequently used algorithm Lec#40 p#203
46 In ………………. Frames are equally divided among processes
Ans: Fixed allocation Lec#4 p# 204
47 In………………. no of frames is proportional to its size
Ans: Proportional allocation Lec#40 p#204
48 Priority based proportional allocation is……………..
Priority allocation Lec#40 p#205
49 The condition in which a process spends more time in paging than executing is known as……………….
Ans: Thrashing Lec#40 p#2065