Maximum file size given a particular inode structure?
This explains it for you:
http://www.cis.temple.edu/~ingargio/cis307/readings/stable.html
"The maximum size of a file will be 8KB*(10 + 2**10 + 2**20 + 2**30), that is more than 8TB."Swap 8KB for your 2KB, and adjust the entries for the smaller block size.
2KB*(10 + 2**8 + 2**16 + 2**24)It's not clear to me from your question if the 13 entries include the singles, doubles and triples, or if they are separate, but that should be easy to adjust -- just change the 10 in the expression to a 13.
I think I've adjusted all the math correctly... double check it =| Hope this isn't homework I did for you ;)
How many pointers in 1 block?
each block is 2kb = 2^11 1 disk address is 8b = 2^3 So, in 1 block there are 2^11/2^3 = 2^8 pointers"
How many pointers in the file system?
for 13 direct entries = (2^8)*13 = 3328 for single = (2^8)^2 = 2^16 for double = (2^8)^3 = 2^24 for triple = (2^8)^4 = 2^32 total pointer is :3328 + 2^16 + 2^24 + 2^32"
Therefore the size of the file system is:
size of the disk is : total of pointer*size of the pointer , which is around 34 GB "