Assume a program has 510 bytes and will be loaded into frame

Assume a program has 510 bytes and will be loaded into frames of 256 bytes each, and the instruction to be used is at byte 495, answer the following questions: How many pages are needed to store the entire job? What is the page number and the displacement for the instruction in question?

Solution

1 Frame = 256 bytes

So 510 bytes of program needs = 2 Frames

a.

Size of page is equal to size of frame . So pages needed to store the entire job = 2 Pages

b.

For instruction at byte 495,

Page number = 2

displacement = 495 - 256 = 239

page 1(256 bytes)

page 2(256 bytes)