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![]() HP OpenVMS Systemsask the wizard |
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The Question is: A customer of ours has three 2 disk shadowsets which should all be identical. When doing a SHOW DEVICE FULL of each shadowset two of the sets shows the Total blocks to be 71074708 blocks, but the other says the Total blocks is 71076708, an increase of 200 0 blocks on the other two. The customer asked for an explanation, any chance you could explain this to me. The disks are all configured via a fibre channel switch and HSG80 controller. Thanks. The Answer is : So you have three shadowsets of two volumes each... There are (at least) three interpretations to your question... One: that one of the shadowsets has differences among its constituent volumes. Volume shadowing is not particularly aware of the host file structure and is thus ignorant of constructs such as free blocks, so clearly there is an inconsistency in the physical contents of the volumes. Assuming a merge is not running, apply current ECOs, and then contact the Compaq Customer Support Center directly for assistance. Two: that there are three shadowsets, and one of the shadowsets differs from the other two shadowsets, but you believe that the shadowsets should be identical. The three volumes may well be entirely identical in terms of user data contents, but various volume-level structures and allocations strategies can lead to small differences in the volume free block counts. Three: the total numbers of blocks reported by the shadowsets differ. This is not permitted by OpenVMS volume shadowing, and you cannot use OpenVMS volume shadowing to shadow disks with differing numbers of disk blocks together -- while the disks can be different geometries (differing tracks, sectors, cylinders; assuming current shadowing software is loaded and running), the total block count must match. (All modern disks are logically perfect, meaning that bad blocks are transparently replaced by known-good blocks, until no more spare blocks are available for replacement. Regardless, the total number of blocks reported by the disk will remain constant.)
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