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.
