a. b. u i ho pre e put circle waiting 1024 (10%) A multiprogramming operating sy
ID: 3874430 • Letter: A
Question
a. b. u i ho pre e put circle waiting 1024 (10%) A multiprogramming operating system uses paging. The available memory is 4096KB, and each frame is 4KB. The program A needs 359KB. The program B needs 111KB. The program C needs 987KB. The program D needs 234KB. The program E needs 577KB. User 1 loads program A, B and executes them. User 2 loads program B, C and executes. User 3 loads program B, D, E and executes. How many frames are unused? How much memory is wasted in internal fragmentation? If the system has only 1024KB, at least how large the swap space is needed in hard disk to run all the programs? 44 If a data locate in physical memory address Ox035738, what is the frame number of this address? (hint: the size of frame/page is 4KB, the physical memory address Ox000000 is the starting address of frame Ox000) One program contains two pages. After loading to the memory, the page Ox000 is mapped to frame Ox005, the page Ox001 mapped to frame Ox022. What is the physical memory address of the 4735th instruction (count from the beginning) in this program?Explanation / Answer
Answer a,b,c)
Available memory is = 4096 KB.
Each frame is 4 KB.
Total frames = 4096/4= 1024
A = 359 KB = 90 frames
B = 111 KB = 28 frames
C = 987 KB = 247 frames
D = 234 KB = 59 frames
E = 577 KB = 145 frames
Unused frames = 455.
Memory wasted in internal fragmentation = .25 + .25 + .25 + .50 + .75 = 2 KB
Avialable memory is changes to 1024 KB.
So that is 1024/4 = 256 frames.
Swap space is a space on the hard disk that is used to store information from and to the RAM when the RAM is full.
Ideally a system with 1024 KB of memory should have the double space ie. 2048 KB as swap space.
Having this 2048 KB as swap space makes sure that all the programs run smooth.
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