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GATE | Gate IT 2007 | Question 12

  • Last Updated : 28 Jun, 2021

The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 bytes per page is
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.

Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.
How many page faults will occur ?
(A) 0
(B) 4
(C) 7
(D) 8


Answer: (C)

Explanation:


Address            Page faults    last byte in memory
0100         page fault,      199 
0200         page fault,      299
0430        page fault,      529 
0499         no page fault
0510         no page fault 
0530         page fault,     629
0560            no page fault 
0120         page fault,     219
0220         page fault,     319
0240            no page fault 
0260            no page fault 
0320        page fault,     419
0410            no page fault
So, 7 is the answer- (C)


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