Cgs3763 Final Exam Deck 2 (Exam Questions)

Front 1

Using page fault frequency scheme for controlling thrashing, suppose we have an upper limit of 20 page faults per 100 references and a lower limit of 4 page faults per 100 references. A process is currently experiencing 22 page faults per 100 references. To prevent the process from thrashing we should:

Back 1

Increase its page frame allocation.

Generally, increasing page frames allocated to a process will reduce the page fault frequency. 22/100 > 20/100, so we want to reduce it.

Front 2

Assume a main memory of 1024KB is partitioned into 32 fixed size partitions each 32KB in size. Suppose that every process that will run on this system is between 1KB and 20KB. What is the minimum and maximum size of internal fragmentation that can occur?

Back 2

Min = 12KB
Max = 31KB

Internal fragmentation is wasted space within the partition. A 1KB process leaves 31KB wasted; 20KB leaves 12KB wasted

Front 3

Address Binding Can be done at:

Back 3

Compile/Assembly, Load, and Execution time.

Not during program development time.

Front 4

The primary advantage of a virtual memory system is:

Back 4

User Programs can be larger than the physical memory of the system.

Front 5

Assume that a process whose size is 500 bytes [addresses 0-499] has generated a logical address of 60 and the MMU currently has a value of 1400 in the relocation register for that process. When the MMU calculates the physical address that currently corresponds to the generated logical address, it will calculate:

Back 5

1460

Front 6

In a pure demand-paged virtual memory system, the max number of page faults that a single process can incur is equal to the number of different pages that it will reference during its execution. (T/F)

Back 6

False

Front 7

A relocation register allows the system loader to place relocatable programs at any location in the memory it so desires. The MMU then resets the relocation register value after loading the program. (T/F)

Back 7

True

Front 8

The backing store is:

Back 8

a high speed disk used to temporarily hold memory images of processes in execution that have been context-switched out to make room for other processes.

Front 9

A race condition

Back 9

results when several threads try to access and modify the same data concurrently.

Front 10

If address binding is to occur at load time, then the compiler or assembler must generate

Back 10

relative code

Front 11

Given a paged memory system where each page contains 16 bytes (2^4), the physical memory contains a total of 2048 bytes (2^11) and the logical memory consists of 256 bytes (2^8). How many bits are required to uniquely represent the set of different pages and displacements?

Back 11

Page Number = 4 bits
Displacement = 4 bits

2^4 bytes per page; so 4 bits are needed.

256 bytes of logical memory, divided into 16 byte pages = 16 pages (2^4); which means 4 bits to uniquely represent each page.

Front 12

The working set model for page frame allocation in a virtual memory environment was designed specifically to

Back 12

Eliminate thrashing and Maintain an optimal level of multiprogramming

Front 13

For any of the algorithms in the class of page replacement algorithms known as "stack algorithms", the number of page faults for a given reference string will be the same as the number of page faults for the reverse of the reference string. (T/F)

Back 13

True

Front 14

Given the logical address 0xAEF9 (Hex) with a page size of 256 bytes, what is the page number?

Back 14

0xAE

256 = 2^8; this means 8 bits are needed for the displacement. The total string is 16 bits long (4 hex digits). Therefore the displacement is 'F9' and the page number is
AE'

Front 15

Programs that employ dynamically linked addressing are loaded using dynamic run time loading. (T/F)

Back 15

False

Front 16

____ can be used to prevent busy waiting when implementing a semaphore.

Back 16

Waiting Queues

Front 17

____ is the dynamic storage allocation algorithm which results in the smallest leftover hole in memory

Back 17

Best Fit

Front 18

In the enhanced second chance algorithm, which of the following ordered pairs represents a page that would be the best choice for replacement?

Back 18

(0,0)

Reference bit, Modify bit

0,0 = Neither recently used nor modified

Front 19

The ____ is an approximation of a program's locality.

Back 19

Working Set

Front 20

Optimal page replacement ____.

Back 20

is used mostly for comparison with other page replacement schemes.

(It is not physically implementable because it requires looking into the future.)

Front 21

Assume the value of the base and limit registers are 1200 and 350, respectively. What is the legal address space?

Back 21

Between 1200 and 1550, inclusive.

Front 22

Assume a system has a TLB hit ratio of 90%. It requires 15 ns to access the TLB, and 85 ns to access main memory. What is the effective memory access time in nanoseconds for this system?

Back 22

108.5 ns

90% of the time it takes 100ns, 10% of the time it takes 185ns (TLB+memaccess*2).
(100*0.9) + (185*0.1) = 108.5

Front 23

What is true of the direct access method?

Back 23

It allows programs to read and write records in no particular order.

Front 24

A mount point is ____.

Back 24

the location within the file structure where the file system is to be attached.

Front 25

The difference between simple paging and simple segmentation is:

Back 25

Paging has no external fragmentation and segmentation has no internal fragmentation

Front 26

The primary goal of an operating system is:

Back 26

to make the underlying hardware more convenient to use.

Front 27

The architecture of a modern computer system consists of a common bus which connects the CPU and IO devices to the memory subsystem. We can describe this system as:

Back 27

interrupt driven in which the CPU and IO device controllers compete for memory cycles.

Front 28

The long-term scheduler is responsible for placing processes into the job pool based upon some selection protocol. (T/F)

Back 28

True

Front 29

Which of the following is not a "state" in which a given process can currently be in:

Back 29

temporary

(can be waiting, running, ready, suspended and so on...)

Front 30

A clustered system ____.

Back 30

gathers together multiple CPU's to accomplish computational work.

Front 31

The ____ refers to the number of processes in memory.

Back 31

degree of multiprogramming

Front 32

Which is least likely to be included in the PCB?
a - contents of registers belonging to process
b - number of interrupts generated thus far
c - state of the process
d - time at which process entered system

Back 32

b - the number of interrupts generated by the process thus far.

Front 33

Historically, the design of OS's has focused on:

Back 33

efficient use of system resources.

Front 34

An IO bound process is

Back 34

a process that requires more time in IO service than CPU services

Front 35

Which of the following is not a class of interrupts?'
a - Hardware
b - User
c - Timer
d - Program
e - All above

Back 35

b - User

Front 36

Round Robin scheduling degenerates to first come first serve scheduling is the time quantum is too long. (T/F)

Back 36

True

Front 37

The short term scheduler is responsible for removing processes from the ready list and placing them in the correct IO waiting queue. (T/F)

Back 37

False

Front 38

As you move up the memory hierarchy in a modern computer, the following typically holds true:

Back 38

Memory becomes faster
Amount of memory decreases
Memory becomes more expensive

Front 39

Consider a system in which threads are supported only at the user level. Suppose we have two processes, P1: 1 thread and P2: 50 threads. Assume each process receives the same length time slice, how much slow is a thread in P2 executed than in P1?

Back 39

50 times slower

Front 40

The rate of a periodic task in a hard real-time system is ____, where p is a period and t is the processing time.

Back 40

1/p

Front 41

One of the main advantages to the layered approach in OS design is:

Back 41

the system can be constructed in a modular fashion.

Front 42

Which of the following is not a goal of uniprocessor scheduling?
a - increase processor efficiency
b - improve throughput
c - improve response time
d - increase memory utilization

Back 42

d - increase memory utilization

Front 43

____ is the number of processes that are completed per time unit.

Back 43

Throughput

Front 44

A(n) ____ refers to where a process is accessing/updating shared data.

Back 44

critical section

Front 45

A significant problem with priority scheduling algorithms is ____.

Back 45

starvation

Front 46

The Shortest Process Next scheduling protocol prioritizes processes based upon the length of their expected service time. Since this means you must predict the future, such a protocol cannot actually be implemented and is only useful as a theoretical tool. (T/F)

Back 46

False

Front 47

General priority scheduling can lead to starvation. To alleviate this problem, implement:

Back 47

aging the process.

Front 48

Interrupts may be triggered by hardware or software. (T/F)

Back 48

True

Front 49

Processor scheduling techniques that use a feedback mechanism are basically penalizing processes that have been executing for long periods of time, thus tending to prefer processes which have not been executing for long periods of time. (T/F)

Back 49

True

Front 50

Convert 1024 (decimal) to binary.

Back 50

100 0000 0000