JOB 1 is divided into three pages: P0,P1,P2 AND 1K EACH( remember that 1k = 1024

Question

JOB 1 is divided into three pages: P0,P1,P2 AND 1K EACH( remember that 1k = 1024)
^^ this is what top cut off portion says
OPERATING SYSTEMS

a) Job 1 is divided into three pagés! PU, PI, ll 1K 1024.) b). Main memory is divided into eight frames of IK each, with the first two (FO and F1) reserved for the operating system. c) Job 1's pages are mapped into the following frames: frames POF2 P2 =-=-=>F4 d) The instruction ADD AX,[1238] starts at byte 637 (of Job 1, not in main memory). Answer the following questions: 6) On what page of Job 1 does the instruction reside? 6.1 On PO 6.2 On P1. 6.3 On P2 6.4 Non of above. 7) What is the offset into the page of the instruction? 7.1 637 bytes. 7.2 2685 bytes 7.3 1238 bytes 7.4 2K 7.5 1K 8) At what real memory location does the instruction start? 8.1 637 bytes. 8.2 2685 bytes 8.3 1238 bytes 8.4 2K 8.5 1K 9) The location in the program 1238 is part of which page and which frame? 9.1 PO and F6. 9.2 PO and F2. 9.3 P1 and Fl 9.4 P2 and F4 9.5 P1 and F6

Explanation / Answer

Ans (6)

Since the instruction starts at 637 bytes of Job1 it must reside in Page P0. Because 637 is smaller than 1024 for it to go on next page. Hence the option is 1

Ans(7)

since Page P0 is the first page of Job1 and 637 is lesser than 1024 the offset will be 637 itself. Hence option 1 is answer

Ans(8)

Since the instruction is in Page P0 and at offset 637. Memory location of P0 is F2. Thus we have F1 and F0 which are 1024 bytes each before F2. So the F0 + F1 will be 2048 and plus the offset we need to add as we are considering memory segment F2 completely. Thus 2048 + 637 = 2685 Hence option 2 is answer

Ans(9)

1238 is greater than 1024 that is greater page size. Thus it will be in page P1. But we need to subtract the bytes equivalent of size of page P0. Thus we have 1238 - 1024 = 14 bytest offset in page P1. as page P1 resides at memory location F6 the correct option is option 5

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