Posted in December 2, 2009 ¬ 3:10 amh.adminNo Comments »
Inch Magnetic Programming storage/records for a magnetic tape?
Okay I'm stuck on this execise in my Understanding Operating Systems book. Chapter 7:device management.
Given the following characteristics for a magnetic tape:
density=1600 bpi
speed= 200 inches/second
size=2,400 feet
start/stop time= 3 ms
number of records to be stored= 200,000 records
size of each record= 160 bytes
block size= 10 logical records
IBG= 0.5 inch
Find the following:
a. number of blocks needed
b. size of the block in bytes.
c. Time required to read one block.
d. Time required to write all of the blocks.
e. amount of tape used for data only,in inches
f. total amount of tape used(data+ IBGs), in inches
Sheesh....that tape format has been obsolete for almost three decades! And the speed and start/stop time are properties of the tape drive, not the tape itself or the recorded data file. Different drives would read and write the same tape file at different speeds.
Anyway, the computation is easy.
a. 200,000 logical records / 10 logical records/block = 20,000 blocks
b. 160 bytes/record * 10 records/block = 1,600 bytes/block
c. (1,600 bytes/block / 1,600 byte/in) = 1 in/block; 1 in/block / 200 in/sec = 0.005 sec/block
d. (solve last--see below)
e. 20,000 blocks * 1 in/block = 20,000 in
f. 20,000 gaps * 1.5 in/(block+gap) = 30,000 in
(Note: This is 2500 ft. so the file won't fit on a single reel. Strange problem.)
d. This is a puzzler. You can't write the blocks without gaps. Nobody would have had any use for the length or time to read or write data only, but that seems to be what's asked for in the light of questions e and f. If so, calculate the time as:
20,000 in (data length from part e) / 200 in/sec = 100 sec. (or 1 min : 40 sec)
If the gaps are to be included, make that 50% bigger (1.5 in instead of 1 in per block), but then the real problem would have to include rewind time for the first reel (speed not specified in the problem) and human time to dismount the first reel and mount another.
But wait, there's more!
If a bonehead programmer didn't fully buffer the output, the tape would have to write a block plus a gap, then stop while the program was filling a new block, then start up again to write the next block and gap. If this is the case you are supposed to compute, then add 20,000 block * (3 ms to stop + 3 ms to start) = 120 sec.
See what *fun* computers used to be? :^)
See why I said "bonehead"? Failing to fully buffer the output increased the run time of the program to over double the optimum run time.
PS: In 1979, most data centers were upgrading to 6250 BPI tape drives and media. 1600 was being used only for compatibilty with older hardware and/or to use with small data files during the lifetime of the 1600 BPI tape on hand.
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Old 9-track tape drive and 10,5 inch magnetic tape.
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