Operating System- CS604 Assignment # 4 solution fall 2010

[Xpert]

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Due Date
Your assignment must be uploaded before or on 28th Jan 2010.
Objective
The objective of this assignment is to familiarize with Memory Management.

Instructions
Ø Avoid Plagiarism. No marks will be given in case of cheating or copying from the internet or from other students.
Ø Submit the assignment through your account on VULMS. No assignment will be accepted through email after the due date.
Ø

Question # 1

The logical address space of the process is 32 pages and the size of each page is 4 Kb. There are 256 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer to the following questions:

• Calculate the size of Logical Address?
• Find the size of the Logical Memory?
• Calculate the size of Physical Address?
• Determine the size of the Physical Memory?

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[Xpert]

Solution can be seen in the images

[imtiazyar]

The logical address space of the process is 8 pages and the size ofeach page is 1 Kb. There are 64 frames in the main memory. The sizeof a frame is equivalent to the size of a page. With the giveninformation, answer to the following questions:

1 Calculate thesize of Logical Address?
2 Find the size ofthe Logical Memory?
3 Calculate the size of Physical Address?
4 Determine the sizeof the Physical Memory?

Solution
Logical address space has 8bits..therefore it requires 3 bits to denote the page [since 23 =8]
Size of each page is 1Kb i.e. 1024 words..therefore it requires 10bits to denote the words . [since 1024 = 210]

1) Therefore size of logical address = 3 bits (for page) + 10bits ( for words)
= 13 bits

2) size of logical memory = 8 pages each of size 1kb
= 8Kb

3) 64 frames in the main memory.
Therefore it requires 6 bits to denote 64 frames[since 26 =64]
it requires 10 bits to denote words in eachframe [ since size of frame = size of page = 1Kb]
Therefore size of physical address = 6 (denotesframes) + 10 (words in page)
= 16 bits

4) size of physical memory = 64 frames of size 1Kb each
=64Kb

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Another Example

The logical address space of the process is 16 pages and the size of each page is 2 Kb. There are 128 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer the following questions

• Calculate the size of Logical Address?
• Find the size of the Logical Memory?
• Calculate the size of Physical Address?
• Find the size of the Physical Memory?

Solution:
Given data is

The number of frames in the main memory =128
The size of the frame = 2kb
a)
Size of the logical address = page number + page offset
= Size of logical address= |p| + |d|=15 bits
b)
The size of the logical memory = number of pages * page size
= 16*2kb
=32kb
c)
Size of the physical address = frame number (f) + offset (d)
= 7(128)bits +11bits
= 7 bits +11bits
= 18bits
d)
Size of the physical memory = number of frames * frame size
=128 *2kb
=256kb

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