1) a)
Dis Adv: More bits per page means more overhead in updating them. Also,
relatively more space is needed to accommodate bitmap.

Adv: The earlier scheme (2 bit scheme mentioned in 11.7) guarantees 70%
utilization only. The space utilization is improved with the increased
number of bits per page.

Marks: 1(disadv)+2(adv)

1) b) Search for the free block from where we stopped search earlier,
instead of searching from the beginning. Assuming that not many blocks
are released, we would avoid searching some part of the list which we
have searched in the last search (to be precise, in last few searches).

Marks: 2 marks

2)
Every transaction requires 2 IOs (one read, one write). So, 1200 transactions
require 2400 IOs. Assuming only one disk is present, it takes
2400 * 10ms = 24sec. The requirement is to finish 1200 transactions in 1 sec.

Consider RAID0(no redundancy):
If we have only one disk, it takes 24 secs. So, to finish it 1200 transactions
in 1 sec, we need 24 disks.

Consider RAID1:
Every write operation is done on two disk (redundancy). So, now every transaction
requires 3 IO operations(one read, two write operations). Now we need need
36 disks to finish 1200 transactions in a second.

Marking scheme: 5 marks (RAID0) and RAID1/5: upto 3 marks (and upto maximum for the assignment)

3) Define the importance_of_page as (rate_of_reference * cost)/(size).
The algorithm is as follows: 
	 Find importance_of_page for all pages.
	 Sort the list on importance_of_page.
	 Evict the least important pages in the order till we get enough
	 free space to serve the current request.

This is the greedy approach to serve the request. Finding the optimal
combination of the pages which should be evicted is a hard problem.

Marking scheme: 10 marks for above method/formula (even exponential algo is also ok as far as this homework is concerned).