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Program No. Base Length 0 1220 768 1 2300 1536 2 90 384 3 480 512 4 2000 256 3. (24 points=6+2+6+6+1+1+2) We are using

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Program No. Base Length 0 1220 768 1 2300 1536 2 90 384 3 480 512 4 2000 256 3. (24 points=6+2+6+6+1+1+2) We are using a demand paging model in this question. Our operating system uses frames of size 512 bytes. bytes. Give the number of frames needed a. The programs must be split into pages of for: Prog 0 Progl Prog2 Prog 3 Prog 4 b. What is the minimum number of frames that you need in order to have the smallest number of page faults irrespective of the order in which your system will process the interrupts. c. Now assume that at a certain instant you have five frames and they are loaded with the following pages: Second Page Third Page First Page First Page First Page of Progo Of Progl Of Prog4 Of Prog3 Of Prog2 Give the indicated address following logical address: 1500 in Progl in pto form in f+0 form form absolute physical address d.Under the same circumstances as in Part c, give the logical addresses corresponding to the following physical addresses 2000 in fto form p+o form in e. What is the lowest invalid logical address in Prog2 f. Is there any external fragmentation? g. Is there any internal fragmentation. If so give an example and state how many bytes

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