7. Given the following for a hybrid allocation scheme: - size of index block = 256 bytes - size of data block = 8 Kbytes - size of disk block pointer = 8 bytes (to either index or data block) - An i-node consists of 2 direct block pointers, 1 single indirect pointer, 1 double indirect pointer, and 1 triple indirect pointer. Solution: (a) What is the maximum size (in bytes) of a file that can be stored in this file system? Number of single indirect or double indirect pointers in an index block 256 bytes / 8 bytes = 32 Number of direct data blocks = 2 Max file size in blocks = number of direct data blocks + number of data blocks with one level of indirection + number of data blocks with two levels of indirection + number of data blocks with triple levels of indirection 2 + 32 + 1,024 + 32,768 = 33,826 data blocks Max file size = max file size in blocks * size of data block 270,608 bytes = 33,826 * 8 (b) How many data blocks are needed for storing a data file of 1 Gbytes? Number of data blocks needed = size of file / size of data block = 1 Gigabyte = 1,073,741,824 Bytes / 33,826 data blocks =31,743