Chapter 8 Homework Solutions
8.1, 8.3, 8.4, 8.6, 8.10, 8.15, 8.17
Problem 8.1
Virtual Page
Number
Valid
Bit
Reference
Bit
Modify
Bit
Page Frame
Number
0
1
1
0
4
1
1
1
1
7
2
0
0
0
-
3
1
0
0
2
4
0
0
0
-
5
1
0
1
0
• Page table with page size 1024
a. Describe how, in general, a virtual address generated by the CPU is
translated into a physical main memory address.
• Split the binary address into a virtual page number (VPN) and an offset
• use VPN as an index into the page table
• extract the page frame number from the page table entry
• concatenate the page frame number with the offset to get physical memory
address
Problem 8.1
Virtual Page
Number
Valid
Bit
Reference
Bit
Modify
Bit
Page Frame
Number
0
1
1
0
4
1
1
1
1
7
2
0
0
0
-
3
1
0
0
2
4
0
0
0
-
5
1
0
1
0
b. What physical address, if any, correspond to the virtual address:
(i) 1052
1052 = 1024+ 28 = 00..01 0000011100 so VPN = 1 and offset = 28
VPN 1  Frame number 7, so the physical address = 7*1024+28 = 7196
(ii) 2221
2221 = 2 × 1024 + 173, so the VPN = 2, page fault
(iii) 5499
5499 = 5 × 1024 + 379
VPN 5  PFN 0, so physical address = 0 × 1024+379 = 379)
Problem 8.3
a. How much memory space is needed for the user page table of Fig. 8.4?
4 Mbytes
Extra credit question: Who is buried in Grant’s tomb?
Problem 8.3
b. Hashed inverted table for above, 20-bit PN  6-bit hash number.
Entry = page number, frame number, chain pointer; 3 overflow entries per
table entry. What is the inverted page table size?
• Number of rows: 26+3=256 entries.
• Each entry consist of:
20 (page number) + 20 (frame number) + 8 bits (chain index) = 48 bits = 6 bytes.
• Total: 256 × 6= 1536 bytes
• Alternate possibility: 3 overflow includes initial entry  128 × 6 = 768
Problem 8.4
Virtual Page
Number
Page
Frame
Time
Loaded
Time
Referenced
R Bit
M Bit
2
0
60
161
0
1
1
1
130
160
1
0
0
2
26
162
1
0
3
3
20
163
1
1
• Time is number of ticks since process load time.
• Page fault has occurred at time 164. Which page is replaced and why:
a. FIFO
Page Frame Number (PFN) 3 since it was loaded longest ago at time 20
b. LRU
PFN 1 since it was referenced longest ago at time 160
c. Clock
Clear R in PFN 3 (oldest loaded), clear R in PFN 2 (next oldest loaded),
victim PFN is 0 since R=0
Problem 8.4
Virtual Page
Number
Page
Frame
Time
Loaded
Time
Referenced
R Bit
M Bit
2
0
60
161
0
1
1
1
130
160
1
0
0
2
26
162
1
0
3
3
20
163
1
1
• Time is number of ticks since process load time.
• Page fault has occurred at time 164. Which page is replaced and why:
d. Optimal, assuming page reference string 4,0,0,0,2,4,2,1,0,3,2
Replace the page in PFN 3 since VPN 3 (in PFN 3) is used furthest in the
future
Problem 8.4
Virtual Page
Number
Page
Frame
Time
Loaded
Time
Referenced
R Bit
M Bit
2
0
60
161
0
1
1
1
130
160
1
0
0
2
26
162
1
0
3
3
20
163
1
1
e. Given above (just before page fault), assume page reference string
4,0,0,0,2,4,2,1,0,3,2. How many page faults if working set policy with LRU
were used with a window size 4 and a fixed allocation?
There are 6 faults, indicated by *
Problem 8.6
Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace:
1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3
a. Show the successive pages residing in the four frames using the LRU
replacement policy. Compute the hit ratio in main memory. Assume the
frames are initially empty.
Hit ratio = 16/33
PF1
PF2
PF3
PF4
1
1
F
0
1
0
F
2
1
0
2
F
2
1
0
2
-
1
1
0
2
-
7
1
0
2
7
F
6
1
6
2
7
F
7
1
6
2
7
0
1
6
0
7
F
1
1
6
0
7
2
1
2
0
7
F
0
1
2
0
7
3
1
2
0
3
F
0
1
2
0
3
4
4
2
0
3
F
5
4
5
0
3
F
1
4
5
0
1
F
5
4
5
0
1
2
4
5
2
1
F
4
4
5
2
1
5
4
5
2
1
6
4
5
2
6
F
7
4
5
7
6
F
6
4
5
7
6
7
4
5
7
6
2
2
5
7
6
F
4
2
4
7
6
F
2
2
4
7
6
7
2
4
7
6
3
2
4
7
3
F
3
2
4
7
3
2
2
4
7
3
3
2
4
7
3
Problem 8.6
Process with 8 pages of VM, fixed allocation of 4 page frames. Page trace:
1,0,2,2,1,7,6,7,0,1,2,0,3,0,4,5,1,5,2,4,5,6,7,6,72,4,2,7,3,3,2,3
a. Show the successive pages residing in the four frames using the FIFO
replacement policy. Compute the hit ratio in main memory. Assume the
frames are initially empty.
Hit ratio = 16/33
PF1
PF2
PF3
PF4
1
1
F
0
1
0
F
2
1
0
2
F
2
1
0
2
-
1
1
0
2
-
7
1
0
2
7
F
6
6
0
2
7
F
7
6
0
2
7
0
6
0
2
7
1
6
1
2
7
F
2
6
1
2
7
0
6
1
0
7
F
3
6
1
0
3
F
0
6
1
0
3
4
4
1
0
3
F
5
4
5
0
3
F
1
4
5
1
3
F
5
4
5
1
3
2
4
5
1
2
F
4
4
5
1
2
5
4
5
1
2
6
6
5
1
2
F
7
6
7
1
2
F
6
6
7
1
2
7
6
7
1
2
2
6
7
1
2
4
6
7
4
2
F
2
6
7
4
2
c. Compare the two policy’s effectiveness with respect to this trace.
The two policies are equally effective for this particular page trace.
7
6
7
4
2
3
6
7
4
3
F
3
6
7
4
3
2
2
7
4
3
3
2
7
4
3
Problem 8.10
•
•
•
•
•
•
•
•
•
•
•
Page size = 4 Kbytes; page table entry = 4 bytes; 64-bit address space;
top level page table fits into a single page.
How many levels of page tables are needed?
A one-page page table would point to
4K/4 = 1024 = 210 pages, addressing a
total of 210 × 212 = 222 bytes.
The address space however is 264 bytes.
Adding a second layer of page tables,
the top page table would point to
210 page tables, addressing a total of
232 bytes.
Continuing this process, we can see that
5 levels do not address the full 64 bit address space, so a 6th level is required.
But only 2 bits of the 6th level are required, not the entire 10 bits.
So instead of requiring your virtual addresses be 72 bits long, you could mask out
and ignore all but the 2 lowest order bits of the 6th level.
This would give you a 64 bit address.
Your top level page table then would have only 4 entries.
Yet another option is to revise the criteria that the top level page table fit into a
single physical page and instead make it fit into 4 pages.
This would save a physical page, which is not much.
Problem 8.15
• Define the mean working set size after the kth reference as
1 k
sk ()   | W (t ,) |
k t 1
and define the missing page probability after the kth reference as
1 k
mk ()   F (t ,)
k t 1
where F(t,) = 1 if a page fault occurs at virtual time t and is 0 otherwise.
• Consider the following page reference sequence:
12345213323454511325
a. Draw a diagram similar to that of Figure 8.19 for the reference sequence
for the values  = 1,2,3,4,5,6.
Problem 8.15
Problem 8.15
• Define the mean working set size after the kth reference as
1 k
sk ()   | W (t ,) |
k t 1
and define the missing page probability after the kth reference as
1 k
mk ()   F (t ,)
k t 1
where F(t,) = 1 if a page fault occurs at virtual time t and is 0 otherwise.
• Consider the following page reference sequence:
12345213323454511325
b. Plot s20() as a function of 
c. Plot w20() as a function of 
s20(Δ) is an increasing function of Δ. m20(Δ) is a nonincreasing function of Δ.
Problem 8.17
• A task is divided into four equal sized segments and the system has built an
eight-entry page descriptor table for each segment. Thus the system has a
combination of paging and segmentation. The page size is 2 Kbytes.
a. What is the maximum size of each segment?
8 × 2K = 16K
b. What is the maximum logical address space for the task?
16K × 4 = 64K
c. What is the maximum physical address space for the system?
232 = 4 GBytes
Assume that an element in physical location 00021ABC is accessed by this
task. What is the format of the logical address that the task generates for
it?
Problem 8.17
• A task is divided into four equal sized segments and the system has built an
eight-entry page descriptor table for each segment. Thus the system has a
combination of paging and segmentation. The page size is 2 Kbytes.
c. Assume that an element in physical location 00021ABC is accessed by this
task. What is the format of the logical address that the task generates for
it?
Segment:
0
0
1
2
3
7
00021ABC
Page descriptor
table
Main memory
(232 bytes)
Problem 8.17
• A task is divided into four equal sized segments and the system has built an
eight-entry page descriptor table for each segment. Thus the system has a
combination of paging and segmentation. The page size is 2 Kbytes.
c. Assume that an element in physical location 00021ABC is accessed by this
task. What is the format of the logical address that the task generates for
it?
(2)
(3)
(11)
Segment
Page
Offset
X
Y
2BC