I have not read Buddy Sytem till the date but thanks buddy I googled it and found an easiest video to understand basics of 'Buddy System':
https://m.youtube.com/watch?v=ZI3ybYeHOdI
Now, come to the sum:
16kB
Process P1(7 kB): 16 divides in two parts: 8 kB(Part-A) and 8 kB(Part-B) and P1 fits in one of the blocks.
Process P2(4 kB): Second part (Part-B) of 8k divides in two parts(Part-B.1 and Part-B.2): 4 kB and 4 kB and P2 fits in one of the blocks.
P1 releases so 8 kB is free again.
Process P3(6 kB): Goes directly into 8 kB part which is recently got free from P1(Part-A)
Process P4(3 kB): We have Part-B.2 still free with us which is a size of 4 kB.
so after executing each steps we have 1 kB internal fragmentation in Part-B.2 and 2 kB internal fragmentation in Part-A(where process-P3 is running)
So coming to the total internal fragmentation I'm bit confused about Process-P1 as it is released already then should we have to count its 1 kB internal fragmentation or not?(plz tell me)
Total= 1 + 2 + (1 in doubt) = 3 or 4 kB