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Data Processing Techniques - All about the IBM 1130 Computing ...

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Section Subsections Page90 30 20 04printer. For example, an analysis that indicates a1403 speed of 450 1pm must be modified if <strong>the</strong> printerconsidered is a 1403-6, which cannot exceed 340 1pm.The 750-millisecond time for <strong>the</strong> 1132 is based onstandard FORTRAN, which is not overlapped.The 176 (or 100) millisecond time for <strong>the</strong> 1403consists mainly of <strong>the</strong> conversion from EBCDIC to1403 code - <strong>the</strong> 1403 itself is buffered, and <strong>the</strong> timerequired to fill <strong>the</strong> buffer is quite small. The 176milliseconds drops to 100 on a 2.2 microsecondCPU because of <strong>the</strong> faster CPU. See <strong>the</strong> next subsection.<strong>Computing</strong>The 1131 Central <strong>Processing</strong> Unit is available withone of two basic cycle times: 3.6 microseconds(Models 1 and 2) or 2.2 microseconds (Model 3). Inmore basic terms, <strong>the</strong> Model 3 will compute in .61<strong>the</strong> time of <strong>the</strong> Model 1 or 2.However, in this area it is not quite as easy tocalculate <strong>the</strong> improvement to be expected from <strong>the</strong>faster CPU. The problem is that you often don'tknow how much time you were computing before(with a 3.6-microsecond CPU), in which case youcannot possibly tell what effect <strong>the</strong> 2.2-microsecondCPU will have.Let us review <strong>the</strong> previous example: 1442-6 and1132; ten minutes run time, read 600 cards, print600 lines, skip 100 lines. The times in milliseconds,were:Card read 120,000Print and skip 451,600"Everything else" 28,400Total 600,000 (10 minutes)The only way you determined <strong>the</strong> 28,400 millisecondsof "everything else" was by subtracting one knownvalue (I/O times) from ano<strong>the</strong>r known value (totalrun time).If you know that all 28,400 milliseconds werespent in computing, you can calculate that <strong>the</strong> 2.2-microsecond CPU will do <strong>the</strong> same amount of workin 61% of that time, or 17,300 milliseconds, areduction of 1,100 milliseconds or 1.8 minutes.If those 28,400 milliseconds had included any diskoperations, you could not have made <strong>the</strong> above estimate,since you would have had no way to determine<strong>the</strong> split between disk activity and computing. Asidefrom a good estimate, which would be quite anachievement, <strong>the</strong> only way to evaluate <strong>the</strong> effect ofa new CPU in this case would be to take your programto such an <strong>1130</strong>, run it, and time it.

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