OS6

Question 1

State the goal of memory protection.

Answer 1

To protect memory access to prevent malicious or buggy user programs corrupting other programs.

Question 2

What is a logical address space defined by?

Answer 2

Base: the smallest legal address. Limit: the size of the range.

Question 3

State the problem of address binding.

Answer 3

When programs are brought into memory to create processes, where to put them in memory given the program will refer to memory locations.

Question 4

Describe the 3 address binding schemes.

Answer 4

Compile time: if memory location is known at compile time, absolute code can be generated. Load time: otherwise, generate relocatable code. Execution time: if the process can be moved during execution, binding is delayed until run time.

Question 5

State the differences between logical and physical address.

Answer 5

Logical address is generated by the CPU. Physical address is the address seen by the memory unit.

Question 6

Compare logical and physical addresses in different address binding schemes.

Answer 6

Identical in compile-time and load-time binding. Differ in execution-time binding.

Question 7

Define logical/physical address space.

Answer 7

The logical/physical address space is the set of all logical/physical addresses generated by a program.

Question 8

Why is hardware support needed for address binding?

Answer 8

To perform mapping from logical to physical addresses at real time.

Question 9

Define the memory management unit.

Answer 9

It is the piece of hardware that maps logical to physical addresses at run time.

Question 10

Describe how the memory management unit works.

Answer 10

The base register is replaced with the relocation register. Add the value in the relocation register for every address generated by the user process when it is sent to memory.

Question 11

What address binding scheme does the memory management use?

Answer 11

Execution-time binding.

Question 12

State how using a memory management unit makes user programs simpler.

Answer 12

User programs only deal with logical addresses, never seeing physical addresses.

Question 13

What is linking?

Answer 13

Linking combines different object code modules to create a program.

Question 14

Describe static linking.

Answer 14

All libraries and program code are combined into the binary program image.

Question 15

Describe dynamic linking.

Answer 15

Linking is postponed to execution time.

Question 16

What is dynamic linking commonly used for?

Answer 16

Linking system and shared libraries.

Question 17

Describe how dynamic loading works.

Answer 17

Calls are replaced with a stub. When called, the stub replaces itself with the address of the routine, and executes the routine.

Question 18

State the advantages of dynamic loading.

Answer 18

Better memory usage as unused routines are never loaded. Useful when large amounts of code are needed infrequently.

Question 19

State the requirement for a library to be dynamically loaded.

Answer 19

Requires it to be compiled with relocatable addresses.

Question 20

Describe contiguous allocation.

Answer 20

Each process is put in one chunk of memory.

Question 21

What is variable partitioning?

Answer 21

Contiguous memory allocation where partitions are created at runtime.

Question 22

How is the main memory partitioned?

Answer 22

Low memory for kernel and interrupt vectors. High memory for user processes.

Question 23

What do the relocation registers protect?

Answer 23

User processes from each other. OS code and data from being modified.

Question 24

What is swapping?

Answer 24

When memory requested exceeds physical memory in machine, temporarily move processes from main memory to storage.

Question 25

State and explain the performance impact of swapping.

Answer 25

Time to transfer to/from storage is proportional to the amount of memory swapped. Context switches therefore become very expensive.

Question 26

State the challenges in address binding when swapping.

Answer 26

Depending on address binding method, processes swapped out may need to be swapped into the same physical address.

Question 27

What are holes?

Answer 27

Blocks of available memory of various sizes scattered throughout memory.

Question 28

State the dynamic allocation problem.

Answer 28

How to satisfy a request of a given size from a list of free holes.

Question 29

State the 3 strategies for dynamic allocation.

Answer 29

First fit: allocate the first hole that is big enough. Best fit: allocate the smallest hole that is big enough. Worst fit: allocate the largest hole.

Question 30

What are the better strategies for dynamic allocation?

Answer 30

First fit and best fit.

Question 31

What is fragmentation?

Answer 31

Fragmentation results in memory being unused and unusable.

Question 32

Explain external fragmentation.

Answer 32

Occurs when free memory exists to satisfy a request but is not contiguous.

Question 33

Explain internal fragmentation.

Answer 33

Occurs when allocated memory is larger than requested memory - the memory internal to a partition is unused.

Question 34

State the storage utilisation of first-fit.

Answer 34

For N blocks allocated, 0.5N blocks lost to fragmentation.

Question 35

Describe compaction.

Answer 35

Reduce external fragmentation by combining all free memory into one large block.

Question 36

State the requirements for compaction.

Answer 36

Relocation is dynamic and done at execution time.

Question 37

Describe segmentation.

Answer 37

A program is viewed as a collection of segments.

Question 38

How can a process access memory with segmentation?

Answer 38

The user program specifies the segment number and offset within the segment.

Question 39

State the format of logical address under segmentation.

Answer 39

The pair

Question 40

Describe the segment table.

Answer 40

Segment table maps logical to physical addresses using a table of base and limit entries.

Question 41

What does the segment table base register do?

Answer 41

Points to the segment table's location in memory.

Question 42

What does the segment table length register do?

Answer 42

Indicates the number of segments used by a program.

Question 43

How does segmentation offer memory protection?

Answer 43

Each entry in the segment table has: a validation bit indicating legal/illegal segment, and read/write/execute privileges.

Question 44

State the problem when executing a jump statement from a shared segment.

Answer 44

The jump will be to the segment with the same segment number as the shared segment, so all programs sharing the segment must reference it with the same segment number.

Question 45

State the problem with forcing all programs sharing a segment to refer to it with the same segment number.

Answer 45

It is difficult to find a common shared segment number as the number of programs sharing the segment increases.

Question 46

Describe the two ways to handle sharing segments.

Answer 46

Information of the shared segment is stored in each process segment table. Each segment is assigned a unique system segment number, the process segment table then maps from a process segment number to SSN.

Question 47

State the advantage of using a system segment number.

Answer 47

Duplicates of common information are not stored in each process segment table. No chance the segment information can get out of sync between processes.