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The Question is: I am attempting to figure the amount of data that is stored on DLTtape III tapes. The tapes are 10/20 gb. My calculations dont seem to add up. I have a total of 54,691,570 blocks of data. I multiply the total blocks by 512 and then divide by 1000000000. This gives me 1.3 gb of data. Hmmmm that should fit on one tape. mb should be (blocks * 512)/1000000 gb should be (blocks * 512)/1000000000 Also what is the compaction ratio using a TZ87 table top unit. Your help would really assist me. Edward A. Lucas The Answer is : Please see the OpenVMS FAQ for some details of the problems involved with predicting the capacity of a magtape. Also please note that 54,691,570 halfkilobyte units is equivalent to 27,345,785 kilobytesized units, or 27,345,785,000 bytes, or roughly 27 gigabytes, uncompressed. (The FAQ also has information on units, both as used by OpenVMS and as used by the storage device ratings.) For reference purposes, here are the uncompressed rated capacities of various DLT and DLTlike media and drives: TK50 95 MB CompacTape TK70 195 MB CompacTape II TZ85 2.5 GB CompacTape III (DLT III) TZ86 5 GB CompacTape III (DLT III) TZ87 10 GB CompaqTape III (DLT III) TZ88 20 GB DLT IV, CompaqTape III, DLT IIIXT TZ89 35 GB DLT IV, CompaqTape III, DLT IIIXT SDLT 110/220 110 GB 110/220GB Super DLT SDLT 160/320 160 GB 160/320GB Super DLT For drives with compaction support and assuming 2:1 compaction ratio (compression ratio), double the above values. Applications such as BACKUP can add redundancy groupings and can inject recovery data into the data stream, and can thus reduce the aggregate capacity of the media. Accordingly, you may or may not choose to use this capability of BACKUP and/or the capabilities of the particular tape drive.
