If a program has 471 bytes and will be loaded into page frames of 100 bytes each

Question

If a program has 471 bytes and will be loaded into page frames of 100 bytes each, and the instruction to be used is at byte 132, answer the following questions: a. How many pages are needed to store the entire job? B Compute the page number and the exact displacement for each of the byte addresses where the data is stored. (Remember that page numbering starts at zero) If a program has 471 bytes and will be loaded into page frames of 100 bytes each, and the instruction to be used is at byte 132, answer the following questions: a. How many pages are needed to store the entire job? B Compute the page number and the exact displacement for each of the byte addresses where the data is stored. (Remember that page numbering starts at zero)

Explanation / Answer

Given,

471 bytes

page frame 100 bytes each

the intructions to be used is 132 byte

a) How many pages are needed to store the entire job

By Dividing the length of the program by the page size, and round up to the nearest integer.

471/100 + 1 = 5 pages

b) Compute the page number and the exact displacement for each of the byte addresses where the data is stored. (Remember that page numbering starts at zero)

So here the first page begins at address 0.

The page size is 256, so page 1 the second page,

offset displacement 0 corresponds to address 256.

Address 266 would be offset 10 on page 1.

Address 514 would be page 2,

offset 2.

Then,

address = 256 * (page) + displacement

To store page number = 3 bits
To store offset = 7 bits

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