Question

**How much internal fragmentation would you have - in
paging Cassume a page size of n bytes) in**

**a) the best case**

**b) worst case**

**c) average case**

2) Assuming a page size of 100 bytes in paging, it a program

references logical address 235, tell me:

a) what the page number and the offset

(displacement) will be

b)describe how the page number and the offset will be used to

come up with the actual physical address and how many

memory accesses will required to do that

Answer #1

**Que:-** How much internal fragmentation would you
have - in paging (**assume a page size of n bytes**)
in

**a) the best case**

**b) worst case**

**c) average case**

**Ans:-** It is given that size of a page in paging
memory management technique in OS is **n bytes**

**a)** In **best case** there will
**0 (zero) byte** **internal
fragmentation**, It can be the case when the process size is
the multiple of page size. (For example if process of size 400
bytes and the page size is of 100 bytes then there will be exactly
4 pages required to accomodate the process.

b) In **worst case** there will (**n- 1)
bytes** **internal fragmentation**, It can be
the case when the process required **M whole pages plus one
byte,** in such case total number of pages required
**(M+1).** In M+1^{th} page only 1 byte will
be used and remaining n-1 bytes will be internal fragmentation.

c) In **average case** there will **be n/2
bytes internal fragementation.** In such case last page of
the process will be half filled and remaining half page will be
internal fragmentation.

**Que
2)** Assuming a page size of 100
bytes in paging, if a program references logical
address 235, tell me: ****

**a)
**what
the page number and the offset (displacement)
will be ?

**b)** Describe how the page number
and the offset will be used to come up with
the actual physical address and how many memory
accesses will required to do that?

**Ans-2:- **GIven that

**page size =100**

**Logical address (referenced by program) A=
235**

**a) The logical address will be at 3rd page
(**because as page size is 100, logical addresses 0-99 will
be at PAGE0, 100-199 will be PAGE1, 200-299 will be at PAGE2.

So **page number will be 2**, because in paging
page numbers starts from 0 (Zero).

Offset = A mod Page_size

= 235 mod 100

= 35

**b)** To obtain the physical address first we see
the frame number in the page table. multiply (Frame number-1) by
page size then add the offset in that. By that way way we can
calculate the physical address and access the particular logical
address on a particular page in the main memory.

**For accessing a logical address**

**Step-1:** first we find the page number on the
basis of page size. (Page# = logical address/Page_size)

**Step-2:** Search the Page map table in the memory
and check whether it is in main memory or not on the basis of
bit.

**Step-3:** If it is there then directly access the
logical address else there will be a page fault and first that page
will be swap into the main memory from the secondary storage and
make the entry in PMT and access the address at the page in that
particular frame in the main memory.

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B) larger page table overhead
C) efficient disk I/O
D) All of the above

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