Question

Dept. of Computer Science and Information Systems-CSIS College of Arts and Sciences AUK CSIS 310 Textbook Assignment ( 10 marks) AMERICAN UNIVERSITY of KUWAIT Q1) Consider a system consisting of four resources of the same type that are shared by seven processes, each of which needs at most two resources. Show that the system is deadlock or deadlock-free. (5 marks) Q2) Given five memory partitions of 120 KB, 380 KB, 410 KB, 232 KB, and 540 KB (in order), how would each of the first-fit, best-fit, and worst-fit algorithms place processes of 215 KB, 425 KB, 112 KB, and 328 KB (in order)? Which algorithm makes the most efficient use of memory? (5 marks)

Answer 1

Q.2)

Variable size portion of a memory block

1. First fit

In this algorithm memory are divided into 5 block i.e 120,380,410,232,540(in order)

Process are 215,425,112,328(in order)

the first process will be allocated to the partition which is first sufficient block from the top of Main Memory.

So 215 kB process is not allocated to 120 KB memory block then check for the next block. 380 KB block is sufficient for it so first 215 KB process is allocated to 380KB block and the remaining memory will be 165 KB which can be used for upcoming processes.

Then 425 KB process is allocated to first sufficient block i.e 540 KB remaining 115 KB.

112 KB process is allocated to first sufficient block i.e 120 KB remaining 8 KB

328 KB process is allocated to first sufficient block i.e 410 KB remaining 82 KB

The remaining block is shaded

E 550 . Variable Memory partition Algorithm. D) First Fit , memars block - 120KB 12380 4410-15232- [112 E 215 221 328 E process , #2, 25, 37 215,425,112,328 525 P 1163

Best fit

Best Fit Allocate the process to the block which is the first smallest sufficient block among the free available block.

Process are 215,425,112,328(in order)

First 215 KB process is allocated to first smallest sufficient block i.e 232 KB remaining 17 KB

425 KB process is allocated to first smallest sufficient block i.e 540 KB remaining 115 KB

112 KB process is allocated to first smallest sufficient block i.e 115 KB remaining 3 KB

328 KB process is allocated to first smallest sufficient block i.e 350 KB remaining 52 KB

2) Best first sio -> k 232- E12072 380k 328 530- 425 112 52 153 proces order = 25,425,112, 328

Worst fit

Worst Fit Allocate the process to the block which is the largest sufficient among the freely available block available in the main memory

Process are 215,425,112,328(in order)

First 215 KB process is allocated to first largest sufficient block i.e 540 KB remaining 325 KB

425 KB process is allocated to first largest sufficient block. But there is no memory block which can hold the 425 KB process. This will generate  So algorithm move to the next process.

112 KB process is allocated to first largest sufficient block i.e 410 KB remaining 298 KB

328 KB process is allocated to first largest sufficient block i.e 380 KB remaining 52 KB

Process onder 25,425, 112, 328. No place for 425 kB prauss 120 380 232 328 S2 298 325

According to the question, the second one i.e BEST FIT algorithm is suited for variable partition memory. So write the Best-Fit algorithm is the most effective algorithm.

NOTE

(But actually, the worst fit algorithm is most efficient for variable size partition memory and the best-fit algorithm is most efficient for fix  size partition memory)