Complexity lies in this question is.... how much size of small holes can you created?
with n processes ==> you can create minimum 1 hole ===> you can identify The maximum Process can be denied
P1 |
P2 |
P3 |
hole |
212 |
114 |
100 |
474 |
Therefore if a process comes with size 475 KB you can't allocate.
with n processes ==> you can create maximum n+1 holes ===> you can identify The minimum Process can be denied
hole |
P1 |
hole |
P2 |
hole |
P3 |
hole |
|
212 |
|
114 |
|
100 |
|
hole size = $\frac{Total \; size\; available}{no.of holes}$ = $\frac{474}{4}$ = 118.5
Therefore two holes get 118 KB and two holes can get 119 KB
Therefore if a process comes with size 120 KB you can't allocate.
( if you have doubt that why holes must be equal size, then if you decrease size of one hole then one of the remaining holes size must be increased then you can allocate 120 KB process )