Consider the file of Example 1 with r = 30,000 fixed-length records of size R = 100 bytes stored on a disk with block size B = 1024 bytes. The file has b = 3000 blocks, as calculated in Example 1. To do a linear search on the file, we would require b/2 = 3000/2 = 1500 block accesses on the average. Suppose that we construct a secondary index on a nonordering key field of the file that is V = 9 bytes long. As in Example 1, a block pointer is P = 6 bytes long, so each index entry is Ri = (9 + 6) = 15 bytes, and the blocking factor for the index is bfri = (B/Ri) = (1024/15) = 68 entries per block. In a dense secondary index such as this, the total number of index entries ri is equal to the number of records in the data file, which is 30,000. The number of blocks needed for the index is hence bi = (ri/bfri) = (30,000/68) = 442 blocks.