3. Multiprogramming with Dynamic Partitions: memory is treated as one large cont

Question

3. Multiprogramming with Dynamic Partitions: memory is treated as one large contiguous space. Main memory consists of 64 units. Process requests arrive in the following order: P, requests 10 memory units memory units requests 16 P, requests 29 memory units P4 requests 8 memory units Show what memory contiguous memory allocation methods. Determine the first and last addresses of programs in each partition and calculate the total fragmentation. looks like when using the following First Address belonging to P Last Address belonging to P First Address belonging to P2 Last Address belonging to P2 First Address belonging to P3 Last Address belonging to P3 First Address belonging to P Last Address belonging to P4 Total Fragmentation Observations? See Algorithms in Appendix A pp. 533-535 (edition 7 pp. 521 523) hat does memory look like after the following has Problem: W happened? P2 completes Ps requests 8 memory units P, completes P, completes

Explanation / Answer

Question 1

I explains this with following example.

-> Total main memory is 64 units.

-> Assume that addresses start with 0. So last address of main memory is 63.

-> The following is initial scenario. The starting address is below.

-> After 1st process request OS allocates 10 units to memory

-> After 2nd process request OS allocates 16 units to memory

-> After 3rd process request OS allocates 29 units to memory

-> After 4th process request OS allocates 8 units to memory

-> 1 Unit of memory remains free.

First address belonging to P1 = 0

Last address belonging to P1 = 9

First address belonging to P2 = 10

Last address belonging to p2 = 25

First address belonging to P3 = 26

Last address belonging to P3 = 54

First address belonging to P4 = 55

Last address belonging to P4 = 62

Total Fragmentation = 1 Unit

Question 2

After p2 completes, p5 will come with 8 unit memory request. OS will allocate memory to p5 in free space where p2 has vacated.

-> So starting address of P5 is 10 and ending address of P5 is 17.

-> Then p4 and p3 completed. So OS deallocates space in momory.

-> So following is final scenario after the given process sequences.

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