Given a disk block has the size of 4096M formatted to FAT. The size of each block is 64K. Calculate the size of the FAT table.
My solution:
Number of blocks = disk size / block size = (4096 * 2^20) / (64 * 2^10) = 2^16 blocks.
Assume using FAT16, since we have 2^16 blocks -> have 2^16 entries, each entry needs to store 16 bits.
=> Size of FAT table = 2^16 * 16 = 2^20 bits = 128KB.
I'm preparing for the final exam and the funny thing is that my teacher told me to self-study virtual memory so I'm not sure that my solution and explanation are correct or not. Please help me point out if I'm doing wrong. Thanks for reading.
I've already had the answer to this question and found out I was correct. Thank you and peace out!