hadoop - Large Block Size in HDFS! How is the unused space accounted for?

Question

We all know that the block size in HDFS is pretty large (64M or 128M) as compared to the block size in traditional file systems. This is done in order to reduce the percentage of seek time compared to the transfer time (Improvements in transfer rate have been on a much larger scale than improvements on the disk seek time therefore, the goal while designing a file system is always to reduce the number of seeks in comparison to the amount of data to be transferred). But this comes with an additional disadvantage of internal fragmentation (which is why traditional file system block sizes are not so high and are only of the order of a few KBs - generally 4K or 8K).

I was going through the book - Hadoop, the Definitive Guide and found this written somewhere that a file smaller than the block size of HDFS does not occupy the full block and does not account for the full block's space but couldn't understand how? Can somebody please throw some light on this.

Answer 1

The block division in HDFS is just logically built over the physical blocks of underlying file system (e.g. ext3/fat). The file system is not physically divided into blocks( say of 64MB or 128MB or whatever may be the block size). It's just an abstraction to store the metadata in the NameNode. Since the NameNode has to load the entire metadata in memory therefore there is a limit to number of metadata entries thus explaining the need for a large block size.

Therefore, three 8MB files stored on HDFS logically occupies 3 blocks (3 metadata entries in NameNode) but physically occupies 8*3=24MB space in the underlying file system.

The large block size is to account for proper usage of storage space while considering the limit on the memory of NameNode.