## Memory-management

 Question 1
The Operating System of a computer may periodically collect all the free memory space to form contiguous block of free space. This is called:
 A Concatenation B Garbage collection C Collision D Dynamic Memory Allocation
Operating-Systems       Memory-Management       ISRO-2018
Question 1 Explanation:
→ The Operating System of a computer may periodically collect all the free memory space to form a contiguous block of free space. This is called garbage collection
→ We can also use compaction to minimize the probability of external fragmentation.
→ In compaction, all the free partitions are made contiguous and all the loaded partitions are brought together.
 Question 2
A computer has 1000 K of main memory. The jobs arrive and finish in the sequence Job 1 requiring 200 K arrives Job 2 requiring 350 K arrives Job 3 requiring 300 K arrives Job 1 finishes Job 4 requiring 120 K arrives Job 5 requiring 150 K arrives Job 6 requiring 80 K arrives Among the best fit and first fit, which performs better for this sequence?
 A First fit B Best fit C Both perform the same D None of the above
Operating-Systems       Memory-Management       ISRO-2018
Question 2 Explanation:
Main memory = 1000K
Job 1 requiring 200 K arrives
Job 2 requiring 350 K arrives
Job 3 requiring 300 K arrives and assuming continuous allocation:
Free memory = 1000 − 850(200 + 350 + 300) = 150 K (till these jobs first fit and best fit are same)
Since, job 1 finishes, Free memory = 200 K and 150 K
Case 1: First fit
Job 4 requiring 120 K arrives
Since 200 K will be the first slot, so Job 4 will acquire this slot only. Remaining memory = 200 – 120 = 80 K
Job 5 requiring 150 K arrives
It will acquire 150 K slot
Job 6 requiring 80 K arrives
It will occupy 80 K slot, so, all jobs will be allocated successfully.
Case 2: Best fit
Job 4 requiring 120 K arrives
It will occupy best fit slot which is 150 K. So, remaining memory = 150 − 120 = 30 K
Job 5 requiring 150 K arrives
It will occupy 200 K slot. So, free space = 200 − 150 = 50 K
Job 6 requiring 80 K arrives
There is no continuous 80 K memory free. So, it will not be able to allocate.
So, first fit is better.
 Question 3
Consider a logical address space of 8 pages of 1024 words mapped into memory of 32 frames. How many bits are there in the logical address?
 A 13 bits B 15 bits C 14 bits D 12 bits
Operating-Systems       Memory-management       ISRO CS 2008
Question 3 Explanation:
logical address space = 8 pages of 1024 words
number of bits in logical address space = p (page bits) + d (offset bits)
number of bits = log28 + log21024 = 3 + 10 = 13 bits
 Question 4
Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. If the one-page fault is generated for every 106 memory accesses, what is the effective access time for the memory?
 A 21.4 ns B 29.9 ns C 23.5 ns D 35.1 ns
Operating-Systems       Memory-Management       ISRO-2016
Question 4 Explanation: Question 5
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation lookaside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is:
 A 11 bits B 13 bits C 15 bits D 20 bits
Operating-Systems       Memory-Management       ISRO-2016
Question 5 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
 Question 6
If the page size in a 32-bit machine is 4K bytes then the size of the page table is
 A 1 M bytes B 2 M bytes C 4 M bytes D 4 K bytes
Operating-Systems       Memory-Management       ISRO CS 2011
Question 6 Explanation:
→Page size is total space taken up by page and Page table entry size is memory taken for indexing the Page in Page Table
→Size of logical address = 32 bits
→Page size = 4K =22210=212 Bytes
→Number of pages = logical address space/ size of each page = 232/ 212= 220
→Page table size = number of pages * size of a page table entry
= 220 * 22
= 222
 Question 7
In a system using a single processor, a new process arrives at the rate of six processes per minute and each such process requires seven seconds of service time. What is the CPU utilization?
 A 70% B 30% C 60% D 64%
Operating-Systems       Memory-Management       ISRO CS 2011
Question 7 Explanation:
From the given question,
The number of new processes will arrive per minute = 6
Each process require to complete its task = 7 secs
CPU utilization time within a minute = 6*7 = 42 secs
The percentage of CPU utilization = time which is spent for utilization / total time * 100
= (42/60) * 100
= 70%
 Question 8
Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%, and it takes 20 nanosecond to search the TLB and 100 nanoseconds to access the main memory what is effective memory access time in nanoseconds?
 A 126 B 128 C 122 D 120
Operating-Systems       Memory-Management       ISRO CS 2011
Question 8 Explanation:
Hit ratio to TLB(H) is 98%
Searching time of TLB(T) is 20ns
Access time(M) is 100ns and four level paging scheme is used.
Effective Memory access Time, EAT = H* T+ (1 - H)[ T+ 4*M] + M]
EAT = (0.98 *20) + 0.02(20 + 400) + 100
= 19.6 + 8.4 + 100 = 128 ns
 Question 9
Consider a logical address space of 8 pages of 1024 words each, mapped onto a physical memory of 32 frames. How many bits are there in the physical address and logical address respectively?
 A 5, 3 B 10, 10 C 15, 13 D 15, 15
Operating-Systems       Memory-Management       ISRO CS 2013
Question 9 Explanation:
→Number of pages= 8= 23=(3 bits)
→Each page consists of 1024 words =210(10 bits)
→Logical address space consists of 8 pages of 1024 words each,
→Then the number of bits required for logical address is 3+10=13 bits.
→Total number of frames =32=25(5 bits).
→The logical memory is mapped to physical memory which means mapping should done between pages and frames.
→Physical address = 5(number of bits for frames) + 10 (number of bits for pages)= 15 bits
 Question 10
In a 64-bit machine, with 2 GB RAM, and 8 KB page size, how many entries will be there in the page table if it is inverted?
 A 218 B 220 C 233 D 251
Operating-Systems       Memory-Management       ISRO CS 2013
Question 10 Explanation:
Given data is
Memory size = 2 GB = 231
Page size = 8 KB = 213
Number of entries in inverted page table = physical address space / page size = 231/213 = 218
 Question 11
Consider the following segment table in the segmentation scheme: What happens if the logical address requested is Segment ID 2 and offset 1000?
 A Fetches the entry at the physical address 2527 for segment Id2 B A trap is generated C Deadlock D Fetches the entry at offset 27 in Segment Id 3
Operating-Systems       Memory-Management       ISRO CS 2014
Question 11 Explanation:
From the question we need to find the logical address for segment id-2.
From given table,
Segment-2 has a base address = 1527 ‘
Process can access memory from the location 1527 to 2025(1527+498)
If the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.
In computing and operating systems, a trap, also known as an exception or a fault, is typically a type of synchronous interrupt caused by an exceptional condition (e.g., breakpoint, division by zero, invalid memory access)
 Question 12
Dirty bit is used to indicate which of the following?
 A A page fault has occurred B A page has corrupted data C A page has been modified after being loaded into cache D An illegal access of page
Operating-Systems       Memory-Management       ISRO CS 2014
Question 12 Explanation:
→ The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device
→ When a page is modified inside the cache and the changes need to be stored back in the main memory, the valid bit is set to 1 so as to maintain the record of modified pages.
 Question 13
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512
 A 211 B 215 C 219 D 220
Operating-Systems       Memory-Management       ISRO CS 2014
Question 13 Explanation:
Size of Physical Address = Paging bits + Offset bits
Paging bits = 11 – 1 = 10 (As 1 valid bit is also included)
Offset bits = log2 (page size) =log2 (512) =9
Size of Physical Address = 10 + 9 = 19 bits
 Question 14
Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory is four frames. -
 A 25 B 6169 C 2073 D 4121
Operating-Systems       Memory-Management       ISRO CS 2014
Question 14 Explanation:
Given data,
Page size=2048 words
= 211 bytes
Step-1: Total number of pages = 216/211
= 25
Step-2: The physical address is nothing but [number of frames * size of each frame] Physical address= 4*211
= 213
Step-3: Given physical address (25)10 = (0000000011001)2 in 13 bits
The 13 bits address, we are representing into Question 15

In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is

 A 105 B 68 C 75 D 78
Operating-Systems       Memory-Management       UGC-NET JUNE Paper-2
Question 15 Explanation:
Average time to access a page = page hit ratio(time required to access a page in primary memory) + page miss ratio(time required to access a page in primary memory + time required to access a page in secondary memory)
Average time to access a page = 0.40(15) + 0.60(120)
Average time to access a page = 6 + 72
Average time to access a page = 78
 Question 16

In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is

 A e-15 B 1 - e-15 C 1 - e-20 D e-20
Operating-Systems       Memory-Management       UGC-NET JUNE Paper-2
Question 16 Explanation:
→ In probability theory, a Poisson process is a stochastic process that counts the number of events and the time points at which these events occur in a given time interval.
→ The time between each pair of consecutive events has an exponential distribution with parameter λ and each of these inter-arrival times is assumed to be independent of other inter-arrival times.
→ λ = (40*30)/60
= 20
P(T > 40min) = 1 - P(T ≤ 40 min)
= 1 - (1 - e-40min/λ)
= 1 - (1 - e-40min/20)
= e-20
 Question 17

Which of the following statements are true ?

(a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous.
(b) Memory Fragmentation can be internal as well as external.
(c) One solution to external Fragmentation is compaction.

 A (a) and (b) only B (a) and (c) only C (b) and (c) only D (a), (b) and (c)
Operating-Systems       Memory-Management       UGC-NET JUNE Paper-2
Question 17 Explanation:
External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is not contiguous.
Yes, it is true that memory Fragmentation can be internal as well as external.
Yes, compaction is a solution to external Fragmentation.
 Question 18

Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are

 A Page Access information B Virtual Page number C Page Frame number D Both virtual page number and Page Frame Number
Operating-Systems       Memory-Management       UGC-NET JUNE Paper-2
Question 18 Explanation:
→ For every page table it contains page frame number.
→ Virtual page number can represents index in the page table to get the page frame number.
 Question 19

Consider a virtual page reference string 1, 2, 3, 2, 4, 2, 5, 2, 3, 4. Suppose LRU page replacement algorithm is implemented with 3 page frames in main memory. Then the number of page faults are

 A 5 B 7 C 9 D 10
Operating-Systems       Memory-Management       UGC-NET JUNE Paper-2
Question 19 Explanation: So, total number of page faults are 7.
 Question 20
If there are 32 segments, each of size 1 K byte, then the logical address should have
 A 13 bits B 14 bits C 15 bits D 16 bits
Operating-Systems       Memory-Management       Nielit Scientist-C 2016 march
Question 20 Explanation:
There are 32 segments which is equal to 2​5
Each segment size 1K byte =2​ 10
Then total number of bits that logical address consists is 15 bits.
 Question 21
How many wires are threaded through the cores in a coincided-current core memory?
 A 2 B 3 C 4 D 6
Operating-Systems       Memory-Management       Nielit Scientist-B CS 22-07-2017
Question 21 Explanation:
The most common form of core memory, X/Y line coincident-current, used for the main memory of a computer, consists of a large number of small toroidal ferrimagnetic ceramic ferrites (cores) held together in a grid structure (organized as a "stack" of layers called planes), with wires woven through the holes in the cores' centers.
In early systems there were four wires: X, Y, Sense, and Inhibit, but later cores combined the latter two wires into one Sense/Inhibit line. Each toroid stored one bit (0 or 1).
 Question 22
Which access method is used for obtaining a record from cassette tape?
 A Direct B Sequential C Random D Parallel
Operating-Systems       Memory-Management       Nielit Scientist-B CS 22-07-2017
Question 22 Explanation: Question 23
A CPU generates 32 bit virtual addresses. The page size is 4KB. The processor has a Translation Lookaside Buffer(TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is
 A 11 bits B 13 bits C 15 bits D 20 bits
Operating-Systems       Memory-Management       Nielit Scientist-B CS 22-07-2017
Question 23 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
 Question 24
Computer uses 46-bit virtual address, 32 bit physical address, and a three level paged page table organization. The page table base register stores the base address of the first level table(T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second level table (T2). Each entry of T2 stores the base address of a page of the third level table(T3). Each entry of T3 stores a page table entry(PTE). The PTE is 32 bits in size, The processor used in the computer has a 1MB 16 way set associative virtually indexed physically tagged cache. the cache block size is 64 bytes. What is the size of a page in KB in this computer?
 A 2 B 4 C 8 D 16
Operating-Systems       Memory-Management       Nielit Scientist-B CS 22-07-2017
Question 24 Explanation:
Architecture of physically indexed cache: VIPT cache and aliasing effect and synonym.
Synonym: Different virtual addresses mapped to same physical address (for data sharing).
So these synonyms should be in same set to avoid write-update problems.
In our problem VA = 46bits We are using 16 bits for indexing into cache.
To have two synonym is same set we need to have same 16 bits index for PA & VA.
Assume that physical pages are colored and each set should have pages of same color so that any synonyms are in same set.
Since page size = 8KB ⇒ 13 bits
These 13bits are not translated during VA→PA. So 13 bits are same out of 16 Index bits, 13 are same we need to make 3 bits (16-13) same now.
3 bits can produce, 23 = 8 combinations which can be mapped on the different sets, so we need 8 different colors to color our pages. >br> In physically indexed cache indexing is done via physical address bits, but in virtual indexed cache, cache is indexed from (offset + set) bits. In physical Index cache indexing is done one to one (1 index maps to one page in one block of cache). In VIPT we have more/ extra bits, so mapping is not one-one. Hence these extra bits have to be taken care, such that if two virtual address refers to same page in cache block of different sets then they have to be assumed same i.e., we say of same color and put same color page in one set to avoid write update problems.
 Question 25

Consider a disk pack with 32 surfaces, 64 tracks and 512 sectors per pack. 256 bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is

 A 19 B 20 C 18 D 22
Operating-Systems       Memory-Management       UGC-NET DEC Paper-2
Question 25 Explanation:
There are 32(25) surfaces, each surface have 64(26) tracks and each surface have 512(29) sectors.
So, to identify each sector uniquely,
5+6+9 = 20-bits are needed.
 Question 26
If there are 32 segments, each size 1K bytes, then the logical address should have
 A 13 bits B 14 bits C 15 bits D 16 bits
Operating-Systems       Memory-Management       ISRO CS 2015
Question 26 Explanation:
Given data, total 32 segments
Each segment size is 1K bytes.
32 segments=25
Segment size=210
⇒ 25*210
⇒ 215
So, 15 bits are required.
 Question 27
Increasing the RAM of a computer typically improves performance because:
 A Virtual Memory increases B Larger RAMs are faster C Fewer page faults occur D Fewer segmentation faults occur
Operating-Systems       Memory-Management       ISRO CS 2015
Question 27 Explanation:
→ When page frames increases, then no. of page faults decreases.
→ Such as if RAM size increases, then no. of page entries increases, then no. of page faults decreases.
 Question 28
Dirty bit for a page in a page table
 A helps avoid unnecessary writes on a paging device B helps maintain LRU information C allows only read on a page D None of the above
Operating-Systems       Memory-Management       ISRO CS 2015
Question 28 Explanation:
→ The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device.
 Question 29
In a computer system, memory mapped access takes 100 nanoseconds when a page is found in TLB. In case the page is not TLB, it takes 400 nanoseconds to access. Assuming a hit ratio of 80%, the effective access time is:
 A 120ns B 160ns C 200ns D 500ns
Operating-Systems       Memory-Management       KVS 22-12-2018 Part-B
Question 29 Explanation:
EAT := TLB_miss_time * (1- hit_ratio) + TLB_hit_time * hit_ratio
TLB time = 100ns,
Memory time = 400ns
Hit Ratio= 80%
E.A.T. = (0.80)*(500)+0.20*900
0.8*100+.2*400=160
 Question 30
The following diagram depicts a______cell. A Storage B Mobile C Memory D Register
Operating-Systems       Memory-Management       KVS DEC-2013
Question 30 Explanation:
● Computer memory is the storage space in the computer, where data is to be processed and instructions required for processing are stored.
● In the computer memory , we will perform read and write operations.
 Question 31
What is coalescing?
 A It is a second strategy for allocating kernel memory B The buddy system allocates memory from a fixed size segment consistency of physically contiguous pages C Kernel memory is often allocated from a free memory pool different from the list used to satisfy ordinary user mode processes D An advantage of the buddy system is how quickly adjacent buddies can be combined to form larger segments using this technique
Operating-Systems       Memory-Management       KVS DEC-2013
Question 31 Explanation:
● coalescing is the act of merging two adjacent free blocks of memory.
● When an application frees memory, gaps can fall in the memory segment that the application uses.
● Among other techniques, coalescing is used to reduce external fragmentation, but is not totally effective.
● Coalescing can be done as soon as blocks are freed, or it can be deferred until some time later (known as deferred coalescing), or it might not be done at all.
 Question 32
Each process is contained in a single section of memory that is contiguous to the section containing the next process is called
 A Contiguous memory​ ​ protection B Contiguous path name C Definite path name D Indefinite path name
Operating-Systems       Memory-Management       KVS DEC-2013
Question 32 Explanation:
Contiguous memory allocation is a classical memory allocation model that assigns a process consecutive memory blocks (that is, memory blocks having consecutive addresses).
 Question 33
Both the first fit and best fit strategies for memory allocation suffer from
 A External fragmentation B Internal fragmentation C 50-percent rule D segmentation
Operating-Systems       Memory-Management       KVS DEC-2013
Question 33 Explanation:
→ Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
→ External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→ Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
 Question 34
The simplest, but most expensive approach to introductory redundancy is duplicate to every disk. This technique is called
 A Swap space B Mirroring C Page slots D None of these
Operating-Systems       Memory-Management       KVS DEC-2013
Question 34 Explanation:
● Mirroring copies identical data onto more than one drive.
● Striping partitions each drive's storage space into units ranging from a sector (512 bytes) up to several megabytes.
●The stripes of all the disks are interleaved and addressed in order.
 Question 35
Copying a process from memory to disk to allow space for other processes is called___
 A Demand paging B Deadlock C page fault D Swapping
Operating-Systems       Memory-Management       KVS DEC-2017
Question 35 Explanation:
→ Swapping is a mechanism in which a process can be swapped temporarily out of main memory (or move) to secondary storage (disk) and make that memory available to other processes. At some later time, the system swaps back the process from the secondary storage to main memory.
→ The performance is usually affected by swapping process but it helps in running multiple and big processes in parallel and that's the reason Swapping is also known as a technique for memory compaction. Question 36

The segmentation memory management scheme suffers from:

 A External fragmentation B Internal fragmentation C Starvation D Ageing
Operating-Systems       Memory-Management       JT(IT) 2016 PART-B Computer Science
Question 36 Explanation:
• Segmentation and paging avoids external fragmentation but still it suffers from internal fragmentation.
• Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
• External fragmentation is the various free spaced holes that are generated in either your memory or disk space. External fragmented blocks are available for allocation, but may be too small to be of any use.
• Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work. Starvation may be caused by errors in a scheduling or mutual exclusion algorithm.
• Ageing is a scheduling technique used to avoid starvation.
 Question 37
Which of the following technique allows execution of programs larger than the size of physical memory?
 A Thrashing B DMA C Buffering D Demand Paging
Operating-Systems       Memory-Management       KVS DEC-2017
Question 37 Explanation:
Virtual memory technique allows execution of programs larger than the size of physical memory. Demand paging we are using for virtual memory technique.
 Question 38
An address in the memory is called
Operating-Systems       Memory-Management       KVS 30-12-2018 Part B
Question 38 Explanation:
A physical address (also real address, or binary address), is a memory address that is represented in the form of a binary number on the address bus circuitry in order to enable the data bus to access a particular storage cell of main memory, or a register of memory mapped I/O device.
 Question 39
The mechanism that brings a page memory only when it is needed in___
 A page replacement B segmentation C fragmentation D demand paging
Operating-Systems       Memory-Management       KVS DEC-2017
Question 39 Explanation:
→ Demand paging follows that pages should only be brought into memory if the executing process demands them. This is often referred to as lazy evaluation as only those pages demanded by the process are swapped from secondary storage to main memory. Contrast this to pure swapping, where all memory for a process is swapped from secondary storage to main memory during the process startup.
→ Demand paging is a method of virtual memory management. In a system that uses demand paging, the operating system copies a disk page into physical memory only if an attempt is made to access it and that page is not already in memory (i.e., if a page fault occurs).
→ It follows that a process begins execution with none of its pages in physical memory, and many page faults will occur until most of a process working set of pages are located in physical memory. This is an example of a lazy loading technique.
 Question 40
Consider the following statements
S1: a small page size causes large page tables
S2: Internal fragmentation is increase with small pages
S3: I/O transfers are more efficient with large pages
Which of the following is true?
 A S1 is true and S3 is false B S1 and S2 are true C S2 and S3 are true D S1 is true and S2 is false
Operating-Systems       Memory-Management       KVS DEC-2017
Question 40 Explanation:
S1: True:A small page size causes large page tables
S2: True:Internal fragmentation is increase with small pages. Actually Internal fragmentation is decrease with small pages
S3:True:I/O transfers are more efficient with large pages
 Question 41
First fit and best fit strategies for memory allocation suffer from ____ and _____ fragmentation, respectively.
 A Internal,internal B Internal,external C External,external D external,internal
Operating-Systems       Memory-Management       KVS 30-12-2018 Part B
Question 41 Explanation:
→Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
→External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
 Question 42
In a paging system, it takes 30 ns to search translation Lookaside Buffer (TLB) and 90 ns to access the main memory. If the TLB hit ratio is 70%, the effective memory access time is :
 A 48ns B 147ns C 120ns D 84ns
Operating-Systems       Memory-Management       UGC NET CS 2017 Jan -paper-2
Question 42 Explanation:
Effective memory access(EMA)=Hit ratio*(TLB access time + Main memory access time) +(1–hit ratio) * (TLB access time + 2 * main memory time)
EAM=0.7*(30+90)+0.3(30+(2*90))
=0.7*120 + 0.3(30+(180))
=0.7*120 + 0.3*210
= 84 + 63
= 147
 Question 43
In which of the following storage replacement strategies, is a program placed in the largest available hole in the memory ?
 A Best fit B First fit C Worst fit D Buddy
Operating-Systems       Memory-Management       UGC NET CS 2004 Dec-Paper-2
Question 43 Explanation:
First fit:​ Allocate the first hole that is big enough. Searching can start either at the beginning of the set of holes or at the location where the previous first-fit search ended. We can stop searching as soon as we find a free hole that is large enough.
Best fit: ​ Allocate the smallest hole that is big enough. We must search the entire list, unless the list is ordered by size. This strategy produces the smallest leftover hole.
Worst fit: ​ Allocate the largest hole. Again, we must search the entire list, unless it is sorted by size. This strategy produces the largest leftover hole, which may be more useful than the smaller leftover hole from a best-fit approach.
 Question 44
Moving Process from main memory to disk is called :
 A Caching B Termination C Swapping D Interruption
Operating-Systems       Memory-Management       UGC NET CS 2005 june-paper-2
Question 44 Explanation:
Swapping is a mechanism in which a process can be swapped/moved temporarily out of main memory to a backing store , and then brought back into memory for continued execution.
 Question 45
Loading operating system from secondary memory to primary memory is called ____________ .
 A Compiling B Booting C Refreshing D Reassembling
Operating-Systems       Memory-Management       UGC NET CS 2006 Dec-paper-2
Question 45 Explanation:
Loading operating system from secondary memory to primary memory is called booting.
 Question 46
A page fault __________ .
 A is an error in specific page B is an access to the page not currently in main memory C occurs when a page program accesses a page of memory D is reference to the page which belongs to another program
Operating-Systems       Memory-Management       UGC NET CS 2006 June-Paper-2
Question 46 Explanation:
A page fault is an access to the page not currently in main memory. A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process.
 Question 47
The memory allocation scheme subjected to ​ external​ fragmentation is :
 A Segmentation B Swapping C Demand paging D Multiple contiguous fixed partitions
Operating-Systems       Memory-Management       UGC NET CS 2006 June-Paper-2
Question 47 Explanation:
To avoid external fragmentation we have two methods
1. Paging
2. Segmentation
But both are still suffer in internal fragmentation.
 Question 48
A specific editor has 200 K of program text, 15 K of initial stack, 50 K of initialized data, and 70 K of bootstrap code. If five editors are started simultaneously, how much physical memory is needed if shared text is used ?
 A 1135 K B 335 K C 1065 K D 320 K
Operating-Systems       Memory-Management       UGC NET CS 2014 Dec-Paper-2
Question 48 Explanation:
Given data,
-- Program text=200 K
-- Initial stack=15 K
-- Initialized data=50 K
-- Bootstrap code=70 K
-- Physical memory needed=?
Step-1: Here, given constraint that, all five editors are started simultaneously.
So, all editors to perform all above operations. It need physical memory is
= Program text + Initial stack + Initialized data + Bootstrap code
= 200 K + 15 K + 50 K + 70 K
= 335 K
 Question 49
For the implementation of a paging scheme, suppose the average process size be ‘x’ bytes, the page size be ‘y’ bytes, and each page entry requires ‘z’ bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by
 A x/2 B xz/2 C √2xz D √ xz/ 2
Operating-Systems       Memory-management       UGC NET CS 2014 Dec-Paper-2
Question 49 Explanation:
Since the average number of pages required per process will be x/y and the amount of space required by the page table will be (x/y)*z. The amount of space lost due to internal fragmentation is y/2. So total space wastage is
Loss(L)=(x/y)*e + y/2
To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero. (dL/dy) =0
y=√(2xz)
 Question 50
​To overcome difficulties in Readers-Writers problem, which of the following statement/s is/are true?
Choose the correct answer from the code given below:
 A 1 only B Both 2 and 3 C 2 only D 3 only
Operating-Systems       Memory-Management       UGC NET CS 2018-DEC Paper-2
Question 50 Explanation:
 Question 51
A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?
 A 7,7,18 B 18,7,7 C 7,6,18 D 6,7,18
Operating-Systems       Memory-Management       UGC NET CS 2018-DEC Paper-2
Question 51 Explanation:
An instruction size is given as 32-bits.
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(2​ 6​ ) so to identify each register uniquely 6-bits are needed.
Address part : 256K(2​ 18​ ) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Operation code:
Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)
Size of Operation code= 7-bits Question 52
Consider a system with 2 level cache. Access times of Level 1, Level 2 cache and main memory are 0.5 ns, 5 ns and 100 ns respectively. The hit rates of Level1 and Level2 caches are 0.7 and 0.8 respectively. What is the average access time of the system ignoring the search time within cache?
 A 20.75 ns B 7.55 ns C 24.35 ns D 35.20 ns
Operating-Systems       Memory-Management       UGC NET CS 2018-DEC Paper-2
Question 52 Explanation:
Average access time = level 1 hit rate( level 1 access time)+ (level1 miss rate)(level 2 hit rate(level 2 access time)+ (level 1 miss rate)( level 2 miss rate) (main memory access time)
Average access time = 0.7(0.5)+ 0.3(0.8)(5)+ 0.3(0.2)(100)
Average access time = 7.55 ns
 Question 53
The hit ratio of a Translation Lookaside Buffer (TLAB) is 80%. It takes 20 nanoseconds (ns) to search TLAB and 100 ns to access main memory. The effective memory access time is ______.
 A 36 ns B 140 ns C 122 ns D 40 ns
Operating-Systems       Memory-Management       UGC NET CS 2013 Sep-paper-2
Question 53 Explanation:
Given data,
-- hit ratio=80% it is equivalent to 0.8
-- search time=20 ns
-- access memory=100 ns
-- miss ratio= 1-Hit ratio
= 1-0.8
= 0.2
-- Effective access memory=?
Step-1: Effective access memory= Hit ratio*(search time+access memory) + Miss ratio*(search time+2*access memory)
= 0.8*(120)+0.2*(220) ns
= 140 ns
 Question 54
In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is .
 A 105 B 68 C 75 D 78
Operating-Systems       Memory-Management       UGC NET CS 2018 JUNE Paper-2
Question 54 Explanation:
Average time to access a page = page hit ratio(time required to access a page in primary memory)+ page miss ratio(time required to access a page in primary memory+time required to access a page in secondary memory)
Average time to access a page = 0.40(15)+ 0.60(120)
Average time to access a page = 6+72
Average time to access a page = 78
 Question 55
Which of the following statements are true ?
(a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous.
(b) Memory Fragmentation can be internal as well as external.
(c) One solution to external Fragmentation is compaction.
 A (a) and (b) only B (a) and (c) only C (b) and (c) only D (a), (b) and (c)
Operating-Systems       Memory-Management       UGC NET CS 2018 JUNE Paper-2
Question 55 Explanation:
External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is ​ not ​ contiguous.
Yes, it is true that memory Fragmentation can be internal as well as external.
Yes, compaction is a solution to external Fragmentation.
 Question 56
Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are .
 A Page Access information B Virtual Page number C Page Frame number D Both virtual page number and Page Frame Number
Operating-Systems       Memory-Management       UGC NET CS 2018 JUNE Paper-2
Question 56 Explanation:
→ For every page table it contains page frame number.
→ Virtual page number can represents index in the page table to get the page frame number.
 Question 57
Given memory partitions of 100 K, 500 K, 200 K, 300 K and 600 K (in order) and processes of 212 K, 417 K,112 K, and 426 K (in order), using the first-fit algorithm, in which partition would the process requiring 426 K be placed ?
 A 500 K B 200 K C 300 K D 600 K E None of the above
Operating-Systems       Memory-Management       UGC NET CS 2012 Dec-Paper-2
Question 57 Explanation:
First fit: Allocate the first hole that is big enough. Searching can start either at the beginning of the set of holes or at the location where the previous first-fit search ended. We can stop searching as soon as we find a free hole that is large enough. Note: Given options are wrong. Excluded for evaluation.
 Question 58
Which of the following memory allocation scheme suffers from external fragmentation ?
 A Segmentation B Pure demand paging C Swapping D Paging
Operating-Systems       Memory-Management       UGC NET CS 2012 Dec-Paper-2
 Question 59
The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4-way set associative (i.e. with 4-cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is
 A 12 bits B 15 bits C 16 bits D 20 bits
Operating-Systems       Memory-Management       UGC NET CS 2013 Dec-paper-2
Question 59 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 64 page table entries with 4-way set associative
=64/4
=16
=24
→ 4 bits are needed to address a set.
→ The size of TLB tag = 20-4
= 16 bits
 Question 60
Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?
 A 9 bits B 11 bits C 13 bits D 15 bits
Operating-Systems       Memory-Management       UGC NET CS 2011 Dec-Paper-2
Question 60 Explanation:
Since we know page and frame both have the same size and since page size is given as 1024 words, it means frame size is of 1024 i.e. 210. Hence to uniquely identify each word inside a frame 10-bits are needed.
→ Number of frames is given as 32 i.e 25. So each frame can be uniquely identified using 5-bits.
→ Hence the total number of bits needed to identify a word inside memory is 5+10 = 15 bits. Question 61
Let the page fault service time be 10 millisecond(ms) in a computer with average memory access time being 20 nanosecond(ns). If one page fault is generated for every 106 memory accesses, what is the effective access time for memory ?
 A 21 ns B 23 ns C 30 ns D 35 ns
Operating-Systems       Memory-Management       UGC NET CS 2013 June-paper-2
Question 61 Explanation:
P=page fault rate
EA = p*page fault service time + (1-p) * Memory access time
=1/106*10*106+(1-1/106)*20
≅ 29.9 ns
 Question 62
Assume N segments in memory and a page size of P bytes. The wastage on account of internal fragmentation is :
 A NP/2 bytes B P/2 Bytes C N/2 Bytes D NP Bytes
Operating-Systems       Memory-Management       UGC NET CS 2009-June-Paper-2
Question 62 Explanation:
→ The wastage on account of internal fragmentation is NP/2 bytes.
where,
Segments in memory=N
Page size= P bytes.
 Question 63
Assertion (A) :  Bit maps are not often used in memory management.
Reason (R) :Searching a bitmap for a run of given length is a slow operation.
 A Both (A) and (R) are true and (R) is correct explanation for (A) B Both (A) and (R) are true but (R) is not correct explanation for (A) C (A) is true (R) is false D (A) is false (R) is true
Operating-Systems       Memory-Management       UGC NET CS 2009-June-Paper-2
Question 63 Explanation:
→ Bit maps are not often used in memory management because searching a bitmap for a run of given length is a slow operation.
 Question 64
Suppose it takes 100 ns to access a page table and 20 ns to access associative memory with a 90% hit rate, the average access time equals :
 A 20 ns B 28 ns C 90 ns D 100 ns
Operating-Systems       Memory-Management       UGC NET CS 2009-June-Paper-2
Question 64 Explanation:
Given data,
-- Access page table time=100 ns
-- Associate memory=20 ns
-- hit ratio=90% = 0.9
-- Miss ratio=1-hit ration
= 10% =0.1
-- Average Access Time=?
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= 0.9*20 + 0.1*100
= 28 ns
 Question 65
Variable partition memory management technique with compaction results in :
 A Reduction of fragmentation B Minimal wastage C Segment sharing D None of the above
Operating-Systems       Memory-Management       UGC NET CS 2009-June-Paper-2
Question 65 Explanation:
Variable partition memory management technique with compaction results in reduction of fragmentation
 Question 66
A page fault
 A is an error specific page. B is an access to the page not currently in memory. C occur when a page program occur in a page memory. D page used in the previous page reference.
Operating-Systems       Memory-Management       UGC NET CS 2009 Dec-Paper-2
Question 66 Explanation:
A page fault is the page is not in main memory.
 Question 67
A program is located in the smallest available hole in the memory is _________
 A best – fit B first – bit C worst – fit D buddy
Operating-Systems       Memory-Management       UGC NET CS 2009 Dec-Paper-2
Question 67 Explanation:
First fit: Allocate the first hole that is big enough. Searching can start either at the beginning of the set of holes or at the location where the previous first-fit search ended. We can stop searching as soon as we find a free hole that is large enough.
Best fit: Allocate the smallest hole that is big enough. We must search the entire list, unless the list is ordered by size. This strategy produces the smallest leftover hole.
Worst fit: Allocate the largest hole. Again, we must search the entire list, unless it is sorted by size. This strategy produces the largest leftover hole, which may be more useful than the smaller leftover hole from a best-fit approach.
 Question 68
Suppose it takes 100 ns to access page table and 20 ns to access associative memory. If the average access time is 28 ns, the corresponding hit rate is :
 A 100 percent B 90 percent C 80 percent D 70 percent
Operating-Systems       Memory-Management       UGC NET CS 2008 Dec-Paper-2
Question 68 Explanation:
Given data,
-- Access page table time=100 ns
-- Associate memory=20 ns
-- hit ratio=X
-- Miss ratio=1-X
-- Average Access Time=28
Step-1: AAT= Hit Ratio*Access page table + Miss Ratio*Associate memory
= X*20 + (1-X*100) [Note: If X=0.9 then we are getting exact AAT]
= 28 ns
(or)
28=X*20+(1-X*100)
X=0.9
 Question 69
An example of a memory management system call in UNIX is :
 A fork. B mmap. C sigaction. D execve.
Operating-Systems       Memory-Management       UGC NET CS 2008-june-Paper-2
Question 69 Explanation:
Fork() → It will create child process of already created process.
mmap() → It is memory management system call and implements demand paging.
sigaction() → Examine and change a signal action
execve() → Executes the program pointed to by filename.
 Question 70
With 64 bit virtual addresses, a 4KB page and 256 MB of RAM, an inverted page table requires :
 A 8192 entries. B 16384 entries. C 32768 entries. D 65536 entries.
Operating-Systems       Memory-Management       UGC NET CS 2008-june-Paper-2
Question 70 Explanation:
Given data,
-- Virtual addresses size= 64 bit
-- Page size = 4 KB
-- RAM size = 256 MB
-- Inverted page table requires = ? Question 71
A program has five virtual pages, numbered from 0 to 4. If the pages are referenced in the order 012301401234, with three page frames, the total number of page faults with FIFO will be equal to :
 A 0 B 4 C 6 D 9
Operating-Systems       Memory-Management       UGC NET CS 2007-Dec-Paper-2
Question 71 Explanation:    Question 72
Average process size = s bytes. Each page entry requires e bytes. The optimum page size is given by :
 A √(se) B √(2se) C s D e
Operating-Systems       Memory-Management       UGC NET CS 2007-Dec-Paper-2
Question 72 Explanation: Question 73
Moving process from main memory to disk is called :
 A Caching B Termination C Swapping D Interruption
Operating-Systems       Memory-Management       UGC NET CS 2007 June-Paper-2
Question 73 Explanation:
→ Page making process from main memory to disk is called swapping.
→ Swapping is a mechanism in which a process can be swapped/moved temporarily out of main memory to a backing store , and then brought back into memory for continued execution.
 Question 74
Part of a program where the shared memory is accessed and which should be executed indivisibly, is called :
 A Semaphores B Directory C Critical section D Mutual exclusion
Operating-Systems       Memory-Management       UGC NET CS 2007 June-Paper-2
Question 74 Explanation:
→ Consider a system consisting of n processes {p0,p1,...,pn-1}. Each process has a segment of code, called a critical section, in which the process my changing changing common variable,updating a table, writing a file, and so on.
→ Part of a program where the shared memory is accessed and which should be executed indivisibly, is called critical section.
 Question 75
A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively:
 A 14 and 15 B 14 and 29 C 15 and 14 D 16 and 32
Operating-Systems       Memory-Management       UGC NET CS 2017 Jan- paper-3
Question 75 Explanation:
Given data,
-- Total number of pages=64
-- Page size=512
-- Page frames=32
Step-1: Logical address=Total number of pages*Page size
= 26*29
= 215
Step-2: Physical address= Page size*Page frames
= 29*25
= 214
 Question 76
Match the following with respect to various memory management algorithms: A (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) B (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv) C (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) D (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Operating-Systems       Memory-Management       UGC NET CS 2015 Dec - paper-3
Question 76 Explanation:
Working set is used to provide frames according to the dynamically changing requirements of a process in the main memory. Here pages are stored in the main memory on demand i.e.only when a page is required for execution of a process it is allocated a frame into main memory.
Segmentation supports the user view of memory because it divides a process into segments in such a way that the meaning of the code does not change after dividing it.
Dynamic partitioning helps in eliminating the internal fragmentation by providing the partitions dynamically but using this method of partitioning external fragmentation can’t be avoided. So compaction is a method used to overcome the external fragmentation problem in dynamic partitioning.
Using fixed partitioning we can store a process in fixed partitions of main memory. By doing this only the pages required for the successful execution of a process are bring into memory and this way instead of storing only a single process in main memory, multiple processes can be stored now. Hence degree of multiprogramming increases.
 Question 77
Function of memory management unit is:
 A Address translation B Memory allocation C Cache management D All of the above
Operating-Systems       Memory-Management       UGC NET CS 2015 Dec - paper-3
Question 77 Explanation:
Memory management unit is used for converting the logical address into the physical address. It does not allocate memory or it is not responsible for the cache management. It’s task is to compare the logical address with the limit if it is less than the limit, it adds the logical address to the base value and provides the physical address. There are 77 questions to complete.