MADE EASY 2019

Question

Answer 1

No of block address which can be contained = 8*2^10*8/2^5 = 2^11B

Hence max file supported = 1 * 2^11*2^11*2^11*8*2^10B = 2^46B

Since, 2^46 > 13423956. So 1 disk is sufficient.

Sorry for the formatting, but the answer is correct.

Comments

But according to me 1 should be the answer.

---

they have said that there will be 2 disk accesses ; 1 for indirect block and 2nd for block containing the data;

but i node is already in the memory(as they have mentioned in the question), how it can be involved in disk access;

so I also think that 1 should be the answer;

lets wait if some one says that it is 2

---

ok