Two marks Questions with Answers

Two Marks Questions with Answers

Q.1 State the function of memory manager.

Ans. Functions of memory manager

• Allocates primary memory to processes.

• Maps process address space to primary memory.

• Minimizes access time using cost-effective memory configuration.

• May use static or dynamic techniques

Q.2 Name two differences between logical and physical addresses.

Ans. A logical address does not refer to an actual existing address; rather, it refers to an abstract address in an abstract address space. Contrast this with a physical address that refers to an actual physical address in memory. A logical address is generated by the CPU and is translated into a physical address by the memory management unit (MMU). Therefore, physical addresses are generated by the

Q.3 What is address binding?

Ans. The process of associating program instructions and data to physical memory addresses is called address binding, or relocation.

Q.4 What is memory?

Ans. Memory is a device used to store the data and instructions required for any operation.

Q.5 What is called pages?

Ans. The address space is usually broken into fixed-size blocks, called pages. Each page resides either in main memory or on disk.

Q.6 Explain the consequences of swapping.

Ans.

a. Context switching time is fairly high.

b. Only idle process must swap.

c. Transfer time is directly proportional to the amount of memory swapped.

d. It increase the O.S. overheads.

Q.7 What is an overlay? What is the use of it?

Ans. It allows a process to execute despite to the system having insufficient physical memory. The idea of overlay is to keep in memory only those instructions and data that are needed at any given time. When other instructions are needed, they are loaded into space that was occupied previously by instructions that are no longer needed. Overlay do not require any special support from the operating system.

Q.8 What is internal fragmentation?

Ans. Internal fragmentation exists when the smallest available block is larger than the requested memory. It is the memory which is internal to a partition, but is not being used.

Q.9 Distinguish between internal and external fragmentation.

Ans. Internal fragmentation is the area in a region or a page that is not used by the process it is allocated to. The space is wasted until the process terminates. External fragmentation occurs when there is enough free space to satisfy a request for memory, but none of the free holes between processes in memory is large enough to satisfy the request.e

Q.10 Bring out the disadvantages of fixed partition multiprogramming.

Ans. Disadvantages of fixed partition multiprogramming

a. Limited degree of multiprogramming.

b. Internal fragmentation.

c. Placement policies.

Q.11 What are memory pages and segments?

Ans. Logical memory is also broken into blocks of the same size called pages. It divides a program into a smaller block called segments, each of which is allocated to memory independently.

Q.12 What are the consequences of multiprogramming with fixed partitioning and variable partitioning?

Ans.

1. Fixed partitioning

• Advantages

i) Simple to implement.

ii) Overhead is less.

• Disadvantages

i) Suffer from internal fragmentation.

ii) Inefficient use of memory.

2. Variable partitioning

• Advantages

i) Multiprogramming.

ii) Better utilization of processor and I/O devices.

iii) Easy to implement.

• Disadvantages

i) Fragmentation.

ii) Require contiguous memory location.

iii) Some memory may never be utilized.

Q.13 Explain how memory can be dynamically allocated using first fit, best fit and worst fit strategies.

Ans.First fit: Take the first available hole in memory that is of adequate size. Fast but leads to fragmentation.

Best fit: Take the hole that most closely matches (atleast as big as) the size of the program. Usage efficient but slow still leads to fragmentation.

Worst fit : Take the biggest possible hole. Aims to reduce the number of little and wasted memory holes. Still leads to fragmentation.

Q.14 Describe non-contiguous memory allocation.

Ans. For N memory blocks, the loss of 0.5 N blocks is possible due to external fragmentation. This means that one-third of memory is not usable. But compaction is too costly to perform regularly. External fragmentation arises because we are trying to allocate memory contiguously. We can deal with external fragmentation if we can allow process memory to be non-contiguous.

Q.15 What are the disadvantages of single contiguous memory allocation ?

Ans.

Disadvantages:

1. Memory is not fully utilized. 2. Poor utilization of processors.

Q.16 Define external fragmentation ?

Ans.

Total memory space exists to satisfy a request, but it is not contiguous; storage is fragmented into a large number of small holes.

Q.17 What is a page ?

Ans. Divide logical memory into blocks of same size called pages.

Q.18 What is frame table ?

Ans. Allocation and availability of the frame information is kept in a data structure called a frame table. Frame table has one entry for each physical page frame, indicating whether the latter is free or allocated and if it is allocated, to which page of which process or processes.

Q.19 Define zero level paging.

Ans. There is no paging is used. The TLB is large enough to hold all the entries which is required for processing.

Q.20 What is the purpose of paging the page tables ?bbe

Ans. In certain situations the page tables could become large enough that by paging the page tables, one could simplify the memory allocation problem (by ensuring that everything is allocated as fixed size pages as opposed to variable sized chunks) and also enable the swapping of portions of page table that are not currently used.

Q.21 Explain the advantages of inverted page tables.

Ans.

Advantages

1. It decreases the amount of memory needed to store each page table.

2. It limit the search using TLB.

3. Separate hardware is not required.

Q.22 What is a translation look aside buffer used for?

Ans.

Problem with paging is that, extra memory references to access translation tables can slow programs down by a factor of two or three. Too many entries in translation tables to keep them all loaded in fast processor memory. To solve this problem TLB is used. TLB is used to store a few of the translation table entries.

Q.23 Why should paging be used by operating systems ?

Ans. Paging permits the physical address space of process to be non-contiguous. Discuss the salient features and merits of multilevel paging and inverted page tables.

Q.24 Features and merits of multilevel paging

Ans.

1. It solves the problem of large logical address space.

2. Operating system can leave partitions unused until a process needs them.

3. It reduces main memory use.

Features and merits of inverted page tables

1. It decreases the amount of memory needed to store each page table.

2. Associative memory access is very fast which is used in inverted page table.

Q.25 Explain what the use of a page table is and how it is used.

Ans. Use of page table,

1. It shows the frame location for each page of the process.

2. Processor uses page table to produce a physical address.

Q.26 What is page frame?

Ans. Physical memory unit is called page frame.

Q.27 Consider a logical address space of eight pages of 1024 words each, mapped onto a physical memory of 32 frames. How many bits are there in the logical address and in the physical address.

Ans. Addressing within a 1024-word page requires 10 bits because 1024 = 2¹⁰. Since the logical address space consists of 8 = 2³ pages, the logical addresses must be 10+ 3 = 13 bits. Similary, since there are 32 = 2⁵ physical pages, physical addresses are 5+10 15 bits long.

Q.28 Why are pages sizes always powers of 2 ?

Ans. Paging is implemented by breaking up an address into page and offset number. It is most efficient to break the address into X page bits and Y offset bits, rather than perform arithmetic on the address to calculate the page number and offset. Because each bit position represents a power of 2, splitting an address between bits results in a page size that is a power of 2.

Q.29 Define swap space.

Ans. Secondary memory holds those pages that are not present in main memory. The secondary memory is usually a high speed disk. It is known as the swap device and the section of disk used for this purpose is known as swap space.

Q.30 What do you mean by page fault?

Ans. A page fault is a trap to the software raised by the hardware when a program accesses a page that is mapped in the virtual address space, but not loaded in

Q.31 Define TLB.

Ans.: TLB is the memory cache of the most recently used memory management unit.

Q.32 Name the dynamic memory allocation techniques based on size.

Ans: Segmentation.

Q.33 What is segmentation ?

Ans. Segmentation divides a program into a number of smaller blocks called segments.

Q.34 Explain the type of fragmentation present in paging and segmentation systems.

Ans. Paging: Internal fragmentation.

Q.35 Why are segmentation and paging sometimes combined into one scheme?

Ans. Segmentation and paging are often combined in order to improve upon each other. Segmented paging is helpful when the page table becomes very large. A large contiguous section of the page table that is unused can be collapsed into a single segment table entry with a page table address of zero. Pages segmentation handles the case of having very long segments that require a lot of time for allocation. By paging the segments, we reduce wasted memory due to external fragmentation as well as simplify the allocation.

Q.36 Why should we use virtual memory ?

Ans. Virtual memory allows execution of partially loaded processes.

Q.37 What is virtual memory? Mention its advantages.

Ans. Virtual memory is a technique that allows the execution of processes that are not completely in memory. Advantage of virtual memory is that programs can be larger than physical memory.

Q.38 What happens if the process tries to use a page that was not brought into memory?

Ans. Access to a page marked invalid causes a page-fault trap. Invalid bit indicates that the page either is not valid (not in logical address space) or is valid but is currently on the disk.

Q.39 Consider a memory system with a cache access time of 10ns and a memory access time of 110ns assume the memory access time includes the time to check the cache. If the effective access time is 10% greater than the cache access time, what is the hit ratio H?  AU CSE: May-17

Ans. Effective Access Time = H × T_cache + (1-H) × T_memory

1.1 × T_cache = H × T_cache + (1-H) × T_memory

1.1 × 10 = H × 10 + (1 - H) 110

11 = H × 10 + 110 - 110 × H

- 99 = -100 H

H = 99/100

H = 0.99

Q.40 What is a difference between instruction? Which of the following instructions should be privileged and only allowed to execute in kernel mode ?

a user-level instruction and a privileged

(a) User level instruction

(b) privileged instruction

Ans. A privileged instruction is an instruction that can only be executed in kernel mode. User level instruction is executed in user area.

Q.41 What is demand paging ?

Ans. : Determines when a page should be brough into memory. Demand paging only brings pages into main memory when a reference is made to a location on the page.

Q.42 What is meant by Belady'sanomaly ?

Ans.

For some page replacement algorithms, the page fault rate may increase as the number of allocated frames increases.

Q.43 What is optimal page replacement ?

Ans. The optimal policy selects for replacement the page that will not be used for longest period of time.

Q.44 Differentiate between global and local page replacement algorithms.

Ans. Global page replacement considers all unblocked pages in main memory as candidates for replacement regardless for which process owns a particular page. Local page replacement chooses only among the resident pages of the process that generated the page fault in selecting a page to replace.

Q.45 What is the key distinction between FIFO and optimal algorithms ?

Ans. The key distinction between FIFO and optimal algorithms is that FIFO uses the time when a page was brought into memory; the optimal uses the time when a page is to be used (future).

Q.46 Define a cache hit.

Ans. When the CPU refers to memory and finds a required word in cache it is termed as cache hit.

Q.47 Define hit ratio.

Ans. The ratio of the number of hits divided by the total CPU references to memory is the hit ratio.

Q.48 Define a miss.

Ans. When the CPU refers to memory and if the required word is not found in cache it is termed as miss.

Q.49 Write a formula for average memory access time.

Ans. Average memory access time = Hit time + Miss rate x Miss penalty

Q.50 Mention the significance of LDT and GDT in segmentation.

Ans. A descriptor table is simply a memory array of 8-byte entries that contain descriptors. The Local Descriptor Table (LDT) is a memory table used in protected mode and containing memory segment descriptors. The Global Descriptor Table is a data structure used in 80286. A: anA

Q.51 Will optimal page replacement algorithm suffer from Belady's anomaly? Justify your answer.            AU: CSE: May-17

Ans. Optimal replacement never suffers from Belady's anomaly. It was always believed that an increase in the number of page frames would always result in the same number or fewer page faults.

Q.52 Consider the following page-reference string 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. How many page faults ratio would occur for the FIFO page replacement algorithm? Assuming three is four frames.

page faults = 12         page faults ratio = 100%

Q.53 What are the counting based page replacement algorithms ?

Ans. Counting-based page replacement algorithms are least frequently used (LFU) and most frequently used (MFU).

Q.54 How do you limit the effects of thrashing?

Ans. To limit the effect of thrashing we can use local replacement algorithm. With local replacement algorithm, if the process starts thrashing, it cannot steal frames from another process and cause the latter to thrash as well. The problem is not entirely solved.

Q.55 What is temporal locality ?

Ans: Recently referenced items are likely to be referenced again in the near future. This is often caused by special program constructs such as iterative loops, process stacks, temporary variables or subroutines.

Q.56 Define spatial locality.

Ans: This refers to the tendency for a process to access items whose addresses are near one another.

Q.57 What do you mean by 'Thrashing' ?

Ans: It is a situation in which a process is spending more time paging than executing.

Q.58 How does the system detect thrashing ?

Ans: The system can detect thrashing by evaluating the level of CPU utilization as compared to the level of multiprogramming.

Q.59 Consider the following segmentation table.

What are the physical addresses for the logical addresses 3400 and 0110? AU CSE/IT: May-19

Ans. :

Logical address 3,400 :

Segment = 3, offset = 400, length = 580

Physical address = offset + base = 400 + 1327

Physical address = 1727

Logical address 0, 110:

Segment = 0, offset = 110, length = 600

Physical address = offset + base = 110 + 219

Physical addrres = 329

Q.60 What is thrashing and how to resolve this problem? AU CSE/IT: May-19

Ans. 'Thrashing'

 It is a situation in which a process is spending more time paging than executing.

Ans. To limit the effect of thrashing we can use local replacement algorithm. With local replacement algorithm, if the process starts thrashing, it cannot steal frames from another process and cause the latter to thrash as well. The problem is not entirely solved.