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In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time ?
in Operating System by (5 points) | 1.7k views

2 Answers

+2 votes
0.3 * 8 + 0.7 * 20 = 2.4 + 14 = 16.4 ms.

(but page table in register??)
by Veteran (420k points)
0
Sir questions says page table in register .. but practically page table stored in main memory only. Thats why for this particular question page table access time is negligible.
+1
I suppose it is Page Table Base Register. Page Table cannot be put in register I suppose. And in this question, we are asked for Page fault service time, so this part is immaterial (we have to consider the part only after page fault occurs).
0

@Arjun sir.

Page table can also reside in registers. In some older systems, the page table is implemented as a set of dedicated registers.

The DEC PDP-11 is an example. 

The hardware implementation of the page table can be done in several ways. In the simplest case, the page table is implemented as a set of dedicated registers. These registers should be built with very high-speed logic to make the paging-address translation efficient. Every access to memory must go through the paging map, so efficiency is a major consideration. The CPU dispatcher reloads these registers, just as it reloads the other registers. Instructions to load or modify the page-table registers are, of course, privileged, so that only the operating system can change the memory map. The DEC PDP-11 is an example of such an architecture. The address consists of 16 bits, and the page size is 8 KB . The page table thus consists of eight entries that are kept in fast registers. 

- From Galvin 9th edition page no. 372

+1 vote
Question lyk.. If i ve to go from A to B,
1. Probability of Taking bus is 70% with cost 20 unit.
2. Probability of taking taxi is 30% with cost 8 unit.

Average cost is 70% times 20 + 30% time 8. = 0.70*20 + 0.30*8 = 14 + 2.4 = 16.4 ms
by Veteran (60.2k points)

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