Operating Systems代写 | os代做 | 操作系统代写 – CSI3131 Operating Systems

CSI3131 Operating Systems

Operating Systems代写 | os代做 | 操作系统代写 – 这道题目是OS的代写任务,涵盖了Operating Systems等方面的内容

scheme代写 代做scheme

Tutorial 7
Memory Management Solution
1. Describe two differences between logical and physical addresses.
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 in to a physical address by the
memory management unit (MMU) (when address binding occurs at execution time). Therefore, physical
addresses are generated by the MMU.
2. Consider a system in which a program can be separated into two parts: code and data. The
CPU knows whether it wants an instruction (instruction fetch) or data(data fetch or store).
Therefore, two baselimit register pairs are provided: one for instructions and one for data.
The instruction baselimit register pair is automatically read-only, so programs can be shared
among different users. Discuss the advantages and disadvantages of this scheme.
The major advantage of this  scheme is that it is an effective mechanism for code and data sharing. For
example, only one copy of an editor or a compiler code needs to be kept in memory, and this code can be
shared by all processes needing access to the editor or compiler code. Another advantage is protection of
code against erroneous modification. The only disadvantage is that the code and data must be separated,
which is usually adhered to in a compiler-generated code.
3. Why are page/frame sizes always powers of 2?
Recall that paging is implemented by breaking up an address into a page and offset number. It is most
efficient to break the address into X page bits and Y offset bits, the offset represents the position of an
octet from the start of the page. Since both the page size and frame size are the same, translating a logical
address to a physical address only requires the replacement of the page number by the frame number (a
concatenation operation). 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.
4. Consider a logical address space of eight pages of 1024 bytes each, mapped onto a physical
memory of 32 frames.
How many bits are there in the logical address? Identify in the logical address the bits
used as an offset and the bits used as the page number.
Logical address: 13 bits (8 pages X 1024 bytes/page = 8196 bytes  requires 2^13 addresses)
Bits for offset: Bits 0 to 9 (10 bits for offset into 1024, 2^10 byte page), 3 bits identify page number (2^3 =
8 pages)
How many bits are there in the physical address? Identify in the physical address the bits
used as an offset and the bits used as the frame number.
Physical address: 15 bits (32 frames X 1024 byes/frame = 32768 bytes, requires 2^15 addresses).
Bits for offset: Bits 0 to 9 (10 bits for offset into 1024, 2^10 byte frame), 5 bits identify frame number
(2^5 = 32 frames)
5. Consider a computer using dynamic contiguous allocation, where each allocated block is a
multiple of 2kb. Consider the following situation:
Consider the following sequence of allocating/deallocating
  • Allocate process P7, size 7kb
  • Deallocate process P
  • Deallocate process P
  • Allocate process P8, size 12 kb
  • Allocate process P9, size 11 kb
  • Allocate process P10, size 5k
a) Draw pictures describing the final allocation using first-fit and best-fit allocation

P 1 ( 3 K)

0 K 8 K 16 K 24 K 32 K 40 K 48 K 56 K 64 K

P 8 ( 12 K) P 4 ( 4 K) P 9 ( 11 K) P 6 ( 5 K) P 7 ( 7 K) P 6 ( 6 K) P 10 ( 5 K)

First fit

P 1 ( 3 K) P 10 ( 5 K)

0 K 8 K 16 K 24 K 32 K 40 K 48 K 56 K 64 K

P 9 ( 11 K) P 4 ( 4 K) P 8 ( 12 K) P 6 ( 5 K) P 7 ( 7 K) P 6 ( 6 K)

Best fit
b) What is the internal fragmentation at the end? What about external fragmentation?
Internal Fragmentation: 5K (5 processes have 1 k unused)
External: 6K  Three holes of 2k unused by processes
6. Consider a memory containing 16 blocks, with the allocation described by the following
Inverted Page Table (An entry contains Process number and block number within the process:

####### 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

####### P1,4 P2,2 P2,4 P1,1 P1,0 P2,3 P3,1 P1,3 P2,1 P1,5 P3,0 P2,0 P1,

From the above Inverted Page Table, complete the Page tables for Processes P1 and P

####### P1 P

Frame # Valid
Frame # Valid
4 1 13 1
3 1 8 1
14 1 1 1
7 1 5 1
0 1 2 1
10 1 - 0
Note that P2 does not use as many pages as P1. In fact, if you examine process P3, you will see that
only 2 frames have been allocated to this process.
7. Consider a computer with 24 bit logical address space, 64MB of physical memory, 1kb
pages, with each page table entry taking 4 bytes.
Describe the format of logical address for this computer, by answering the following
a) Draw a bar, indicating how many bits define the offset and how many bits define the
logical page #.

14 bits 10 bits

Page # Offset

b) Do you need hierarchical page tables? In not, explain why. If yes, separate the bits for the
logical page # into the corresponding parts (in the answer for a)). How many levels of
hierarchy do you need?
No, hierarchical pages are not necessary. A page table will take 2^14 * 4 = 64 Kbytes which represents 0.
% of physical memory space or 0.4 % of the process logical space
c) Assume you have decided to use Inverted Page Tables (IPT) instead. How big should the
IPT be? (assume that 2 bytes are sufficient to store PID, you have to answer for yourself
how many bytes you need to store the logical page #).
For a 64 Mbytes address and a 1kb page size, 10 bits are used for offset and 16 bits (2 bytes) are used for
the frame number, giving 2^16 = 64 K (64X1024) entries in the IPT.
With a PID of two bytes, and allocating 2 bytes for the logical page number (14 bits, + 2 control bits, such
as the dirty bit), i.e. 4 bytes per entry, the size of the IPT would be 64 K X 4 bytes = 256 Kbytes
Alternative: extrapolating from the 4 byte entry in the Page table, two bytes are used for the frame
number, and two bytes for control bits (dirty bits, permission bits, etc). Given that the control bits are
carried over to the IPT, then an entry in the IPT would contain 2 bytes for the PID, 2 bytes for the logical
page number (only 14 bits used), and 2 bytes for control bits, giving 6 bytes for an entry in the IPT. In this
case, the size of the IPT would be 64K entries X 6 bytes/entry = 384 Kbytes.
8. Consider the following simple
memory management system
combining segmentation and paging
(illustrated in the figure to the right):
A process consists of at most 4
segments, where each segment itself is
paged, with pages/frames of size 2kb.
The page table of each segment
contains 8 entries, each storing
reference to the corresponding
physical frame. The segment table
entries refer to the page tables of the
corresponding segments.
Because of relatively small size, both
segment table and segment page
tables are stored directly in the
processs PCB (process control
block), without occupying another
physical frames.
Answer the following questions:
a. What is the maximum size of each

16kb (8x 2kb)

b. What is the largest logical address
for the process?
65535 = 2^16 - 1 (16kb x 4 segments,
starting with 0)
c. Give the format of the logical
xx yyy zzzzzzzzzzz

xx segment # yyy page # within the segment zzzzzzzzzzz(11 bits) offset within a page

d. Explain how the logical address is translated to the physical address.
  1. take xx and use it as an index into the segment table to get the pointer to the page table of that segment
  2. take yyy and use it as an index in this page table, get the frame #fffff
  3. concatenate the offset to the frame number to get the physical address fffffzzzzzzzzzzz
Segment Table
Page Tables
Physical Page