Answered Essay: 1-If a program has 471 bytes and will be loaded into page frames of 126 bytes each, assuming the job begins loading at the first page (page 0) in memor

Please help with Paged Memory Allocation:

1-If a program has 471 bytes and will be loaded into page frames of 126 bytes each, assuming the job begins loading at the first page (page 0) in memory 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 2e sy apogmm l be laded to page ames of 126 bytes each, assuming the job begins loading at the first page (page 0) in memory and the instruction to be used is at byte 577, answer the followin,g questions: c. How many pages are needed to store the entire job? d. Compute the page number and the exact displacement for each of the byte addresses where the data is stored

1-If a program has 471 bytes and will be loaded into page frames of 126 bytes each, assuming the job begins loading at the first page (page 0) in memory 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 2-Let’s say a program has 1010 bytes and will be loaded into page frames of 126 bytes each, assuming the job begins loading at the first page (page 0) in memory and the instruction to be used is at byte 577, answer the following questions: c. How many pages are needed to store the entire job? d. Compute the page number and the exact displacement for each of the byte addresses where the data is stored

Expert Answer

 

a) Given program size is 471 bytes

page frame can hold 126 bytes

total no. of pages required for storing entire job is = 471/126 =4pages

b) to compute the page number and the exact displacement for each of the byte address where the data is stored

Given tha the job begins loading at the first page(page 0) in memory and the instruction to be used is at

132/126 = 1 = page 1

132-(126*1) = 6 = 5th byte in the page1

c). Given program size is 1010 bytes

page frame can hold 126 bytes

total no. of pages required for storing entire job is = 1010/126 = 8 pages

d) to compute the page number and the exact displacement for each of the byte address where the data is stored

Given tha the job begins loading at the first page(page 0) in memory and the instruction to be used is at

(577/126) = 4 =(page is from 0) so, 4th page

577 – (4 * 126) = 577-504 = 73 th byte (504 bytes are in pages 4)

so, 577 byte is at page 4 and 73 th byte

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