Answer:
Rank of the algorithms in terms of how efficiently they use memory is:
- Best-fit: this create the least amount of extra/waste memory. (3+3+4) = 10MB
- First-fit: this create (35+3+12) = 50MB worth of memory extra/wastage
- Worst-fit: this create (35+40+4) = 79MB worth of memory extra/wastage.
Step-by-step explanation:
First-fit algorithm will assigned the first available space that can fit the process to it without considering the amount of space to be wasted.
Best-fit algorithm will assigned the best available space that will fit the process to it while minimizing wastage of space.
Worst-fit algorithm will look for the largest space and assigned it to the process.
Given ten memory partitions of
85MB = M1
45MB = M2
57MB = M3
49MB = M4
82MB = M5
65MB = M6
17MB = M7
53MB = M8
74MB = M9
40MB = M10
Processes of size 50 MB, 42 MB, and 70MB
50MB = P1
42MB = P2
70MB = P3
Using First-fit
P1 will be assigned to M1 as it is the first available memory to contain P1. Then M1 = (85 - 50) = 35MB
P2 will be assigned to M2 as it is the first available memory to contain P2. Then M2 = (45 - 42) = 3MB
P3 will be assigned to M5 as it is the first available memory to contain P3. Then M5 = (82 - 70) = 12MB
Using Best-fit
P1 will be assigned to M8 as it is the best available memory to contain P1. Then M8 = (53 - 50) = 3MB
P2 will be assigned to M2 as it is the best available memory to contain P2. Then M2 = (45 - 42) = 3MB
P3 will be assigned to M9 as it is the best available memory to contain P3. Then M9 = (74 - 70) = 4MB
Using Worst-fit
P1 will be assigned to M1 as it is the largest available memory to contain P1. Then M1 = (85 - 50) = 35MB
P2 will be assigned to M5 as it is the largest available memory to contain P2. Then M5 = (82 - 42) = 40MB
P3 will be assigned to M9 as it is the largest available memory to contain P3. Then M9 = (74 - 70) = 4MB
Best-fit efficiently manage the memory partition, followed by first-fit and then worst-fit.