TMC 1234: Operating Systems
Session 2013/2014, Semester 2
TUTORIAL 3
1. Explain the differences between a page memory allocation scheme and a segment memory
allocation scheme.
A paged memory allocation scheme jobs are divided into fixed-size partition called pages (physically
divided). A segmented memory allocation scheme jobs are divided into a number of distinct logical
units called segments (logically divided).
2. List the memory allocation scheme that required the entire job to be stored in main memory.
Memory allocation scheme that required the entire job to be stored in main memory:
i.
Single User Contiguous Scheme
ii.
Fixed partition
iii.
Dynamic partition
iv.
Paged memory allocation
3. List down the step on how to calculate the exact address of the instruction of byte 214 in main
memory when using paged memory allocation. Assume that each page size is 100 and use the
below Page Map Table (PMT) for Job 1.
Job Page No
0
1
2
3
Page Map table for Job 1
Page Frame No
8
10
5
11
For example, to find the exact location of byte 214 (where the system will find the value to load to
Register 1), the operating system will do the following:i.
Compute the page number and displacement of the requested byte
2
100 214
200
14
Page number = 2 and Displacement =14
ii.
iii.
iv.
Go to PMT (Page Map Table) and retrieve the appropriate page frame number.
Compute the starting address of the page frame by multiplying the page frame number
times the page frame size.
ADDR_PAGE_FRAME = PAGE_FRAME_NUMBER * PAGE_SIZE
= 5*100
Finally, calculate the exact address of the instruction in memory by adding the
displacement to the starting address.
INSTR_ADDR_IN_MEM = ADDR_PAGE_FRAME + DISPL
INSTR_ADDR_IN_MEM = 500 +14
4. Describe how the function of the Page Map Table (PMT) differs in paged vs. segmented/demand
paging memory allocation.
In both memory allocation schemes, the PMT tracks page assignments to page frames, but in the
paged scheme the PMT is referenced directly from the Job Table, and in segmented/demand paged,
it is referenced from the Segment Map Table (SMT), which is referenced from the Job Table.
5. Demand paging allows the user to run jobs with less memory than the size of the job. How is this
possible?
Demand paging allows user to run jobs with less main memory than the size of the job by
manipulating a high-speed direct access storage device. Pages will be passed quickly from secondary
storage to main memory and back again. Therefore, it gives the appearance as if the jobs required
always in main memory when, in reality they are swapped back when other jobs need the place in
main memory and will be stored back when required. This is also known as virtual memory
implementation.
6. What is the cause of thrashing?
Too many page fault causes thrashing. Page fault caused by inefficient page replacement policy, or
the size of working set is too small. CPU may spend the majority of its time taking care of page
swaps, leaving little time for useful work.
7. What is the working set?
A job’s working set is the set of pages residing in memory that can be accessed directly without
incurring a page fault.
8. Given that main memory is composed of three page frames for public use and that a seven-page
program (with pages a, b, c, d, e, f, g) requests pages in the following order:
a, b, a, c, d, a, e, f, g, c, b, g
a. Using the FIFO page removal algorithm, do a page trace analysis indicating page faults with
asterisks (*). Then compute the failure and success ratios.
(The first-in page, the oldest swappable page, is shown in bold.)
Page request
Page fault
Page Frame 1
Page Frame 2
Page Frame 3
a
b
a
c
d
a
e
f
g
c
b
g
* *
* * *
a a a a d d
b b b b a
c c c
*
d
a
e
*
f
a
e
*
f
g
e
*
f
g
c
*
b b
g g
c c
Failure ratio = 10 / 12 = 83%
Success ratio = 2 / 12 = 17%
b. Increase the size of memory so it contains four page frames for public use. Using the same page
requests as above and FIFO, do another page trace analysis and compute the failure and success
ratios.
Page request
Page fault
Page Frame 1
Page Frame 2
Page Frame 3
Page Frame 4
a
b a
c
d a
e
f
g
c
b g
* *
* *
a a a a a a
b b b b b
c c c
d d
*
e
b
c
d
*
e
f
c
d
*
e
f
g
d
*
e
f
g
c
*
b
f
g
c
b
f
g
c
Failure ratio = 9 / 12 = 75%
Success ratio = 3 / 12 = 25%
c. Did the result correspond with your intuition? Explain.
The extra page frame increases performance only fractionally (8%), even though one might think it
should be much better. The improvement would be much greater if the program had a small
working set, which might minimize the number of page faults (especially if the working set consisted
of four pages).
9. Given that main memory is composed of three page frames for public use and that a program
requests pages in the following order:
a, d, b, a, f, b, e, c, g, f, b, g
a. Using the FIFO page removal algorithm, perform a page trace analysis indicating page faults with
asterisks (*). Then compute the failure and success ratios.
(The first-in page, the oldest swappable page, is shown in bold.)
Page request
Page fault
Page Frame 1
Page Frame 2
Page Frame 3
a
*
a
d b a f b e c
* *
*
* *
a a a f f f f
d d d d d e e
b b b b b c
Failure ratio = 9 / 12 = 75%
Success ratio = 3 / 12 = 25%
g
*
g
e
c
f
*
g
f
c
b g
*
g g
f f
b b
b. Using the LRU page removal algorithm, perform a page trace analysis and compute the failure
and success ratios.
(The least recently used page is shown in bold.)
Page request
Page fault
Page Frame 1
Page Frame 2
Page Frame 3
a
*
a
d b a f b e c g f
* *
*
* * * *
a a a a a e e e f
d d d f f f c c c
b b b b b b g g
Failure ratio = 9 / 12 = 75%
Success ratio = 3 / 12 = 25%
b g
*
f f
b b
g g
c. Which is better? Why do you think it is better? Can you make general statements from this
example? Why or why not?
It is curious here that the two algorithms result in such miserable success ratios, even though we
would expect LRU to be more efficient. Clearly, with seven pages competing for three page frames,
and with no obvious working set, there will be many page faults. Look for an original analysis for this
answer.
10. Given that main memory is composed of four page frames for public use, use the following table to
answer all parts of this problem:
Page Frame
0
1
2
3
Time When
Loaded
126
230
120
160
Time When Last
Referenced
279
280
282
290
Referenced bit
Modified Bit
0
1
1
1
0
0
1
1
a. The contents of which page frame would be swapped out by FIFO?
Page Frame 2.
b. The contents of which page frame would be swapped out by LRU?
Page Frame 0
11. Given three subroutines of 700, 200, and 500 words each, if segmentation is used then the total
memory needed is the sum of the three sizes (if all three routines are loaded). However, if paging is
used then some storage space is lost because subroutines rarely fill the last page completely, and
that results in internal fragmentation. Determine the total amount of wasted memory due to
internal fragmentation when the three subroutines are loaded into memory using each of the
following page sizes:
a. 100 words
b. 600 words
c. 700 words
d. 900 words
a. Subroutine
Needs
Internal Fragmentation.
S1 with 700 words
7 pages of 100 words
0 words
S2 with 200 words
2 page of 100 words
0 words
S3 with 500 words
5 pages of 100 words
0 words
Total
0 words
b. Subroutine
Needs
Internal Fragmentation.
S1 with 700 words
2 pages of 600 words
500 words
S2 with 200 words
1 page of 600 words
400 words
S3 with 500 words
1 page of 600 words
100 words
Total
c. Subroutine
Needs
1000 words
Internal Fragmentation
S1 with 700 words
1 page of 700 words
0 words
S2 with 200 words
1 page of 700 words
500 words
S3 with 500 words
1 page of 700 words
200 words
Total
d. Subroutine
Needs
700 words
Internal Fragmentation.
S1 with 700 words
1 page of 900 words
200 words
S2 with 200 words
1 page of 900 words
700 words
S3 with 500 words
1 page of 900 words
400 words
Total
12. Given the following Segment Map Tables for two jobs:
SMT for Job 1
Segment
Number
0
1
2
3
4
Memory
Location
4096
6144
9216
2048
7168
1300 words
SMT for Job 2
Segment
Number
0
1
2
Memory
Location
2048
6144
9216
a. Which segments, if any, are shared between the two jobs?
Three segments are shared by both jobs, those starting at locations: 2048, 6144 and 9216.
b. If the segment now located at 7168 is swapped out and later reloaded at 8192, and the segment
now at 2048 is swapped out and reloaded at 1024, what would the new segment tables look like?
SMT for Job 1
Segment
Memory
Number
Location
0
4096
1
6144
2
9216
3
1024
4
8192
SMT for Job 2
Segment
Memory
Number
Location
0
1024
1
6144
2
9216