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ean 128 barcode excel Analysis Techniques in Software
Analysis Techniques Make Quick Response Code In None Using Barcode encoder for Software Control to generate, create QR image in Software applications. Quick Response Code Decoder In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. To derive Eq 372 in Chap 3514, use was made of a relationship which gave the probability of occurrence of collisions in direct access slots given n entries into m slots This result will now be derived using the tools developed Each record has a probability of 1/m of being assigned to any one slot The probability for one slot to receive q records is given by the binomial distribution QR Code 2d Barcode Generator In C#.NET Using Barcode drawer for .NET framework Control to generate, create QR Code image in .NET applications. Encoding QRCode In .NET Framework Using Barcode creator for ASP.NET Control to generate, create QR Code image in ASP.NET applications. Collision Probability
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Data Matrix 2d Barcode Drawer In None Using Barcode creator for Software Control to generate, create DataMatrix image in Software applications. Encode UPCA Supplement 5 In None Using Barcode creator for Software Control to generate, create UPC Code image in Software applications. 1 1 m Code128 Drawer In None Using Barcode generator for Software Control to generate, create Code128 image in Software applications. Print UPC  13 In None Using Barcode printer for Software Control to generate, create EAN13 image in Software applications. For the case that n 1, m 1, q n, we can approximate this expression by the Poisson distribution with the mean density n/m: Pb (q) = e (n/m) (n/m)q q! 68 Barcode Creation In None Using Barcode creation for Software Control to generate, create barcode image in Software applications. Paint ANSI/AIM Code 39 In None Using Barcode maker for Software Control to generate, create ANSI/AIM Code 39 image in Software applications. Pb (0) is the probability of the slot remaining empty, and Pb (n) is the (very small) probability of this slot receiving all n records (see again Fig 319) If q records Making British Royal Mail 4State Customer Barcode In None Using Barcode creator for Software Control to generate, create Royal Mail Barcode image in Software applications. Print Barcode In .NET Using Barcode encoder for Reporting Service Control to generate, create barcode image in Reporting Service applications. Sec 61 UPC A Printer In Visual Studio .NET Using Barcode encoder for ASP.NET Control to generate, create UCC  12 image in ASP.NET applications. Painting UPC Symbol In Java Using Barcode generator for Java Control to generate, create UPCA Supplement 2 image in Java applications. Statistical Methods
UCC  12 Generator In None Using Barcode creator for Online Control to generate, create Universal Product Code version A image in Online applications. Barcode Printer In Java Using Barcode creation for Android Control to generate, create barcode image in Android applications. appear in one slot, causing q 1 over ows, then the number of accesses to fetch all these records, which form one sequential chain, will be 0 + 1 + 2 + + q 1 = 1 q(q 1) 2 69 Barcode Printer In VB.NET Using Barcode printer for .NET framework Control to generate, create bar code image in Visual Studio .NET applications. UCC.EAN  128 Encoder In Visual Studio .NET Using Barcode creation for Reporting Service Control to generate, create EAN 128 image in Reporting Service applications. Taking the sum of all the Pb (j), j = 1 n, multiplied by their cost in terms of disk accesses, gives for any one slot the expected access load AL ALslot =
j(j 1) Pb (j) 2 610 The successive terms of the Poisson distribution are related to each other by the iteration mean 611 P (j) = P (j 1) j The mean is here n/m, so that using the Poisson distribution for Pb ALslot = Since at j = 0 : 1n 2m
(j 1) P (j 1) 612 P (j 1) = 0 (j 1)P (j 1) = and at j = 1 : j 1=0 (j 1)P (j 1) = j=2 k=1
k Pb (k) 613 where k = j 1 The expected record load RL per slot is the mean, which in turn is equal to the mean of the cumulative frequency distribution, speci cally the sum of the products of the number of records per slot i and the probability of the occurrence of the event P (i) RLslot n = = m i P (i) 614 and when m n the series terms are equal, so that the expected increase of accesses for over ows due to collisions per record p is p= 1n AL = RL 2m 615 If multiple records (Bfr = B/R) can be kept in a bucket, the probability of another record entering a slot can be predicted using Pb (q) as derived above The number of buckets will be m/Bfr and the number of collisions per bucket will be higher There will be no over ows, however, until Bfr records have entered the bucket, so that the cost of accessing over ows AL becomes less For this case Johnson61 derives similarly, n 1 n p= (k Bfr)(k Bfr 1) P (k) 616 2 m/Bfr k=B fr 1
Of course, the processor s computing e ort, cix, to search within one bucket increases now An evaluation of this e ect in indexed les is given in Chap 424 Analysis Techniques
Collision Expectations for Direct Files The results obtained above can be further
developed to give the expected over ow space requirement or the average over ow chain length, Lc, due to collisions These can then be compared with observations from the operational le The probability of a slot being empty was approximated by the zero term of the Poisson distribution so that Pb (0) = e n/m The probability of a slot of the primary le being used is then P 1u = 1 e n/m 618 617 and the number of records in the primary le is then m P 1u, and in the over ow le is 619 o = n m P 1u The over ow area contains the tails of all the Poissondistributed record sequences Using Eq 65 for the of a sequence and the fact that 2 of a Poisson distribution is equal to the mean n/m, we expect a standard deviation of o to be n m 620 = m and this applies of course to the entire le If an observed standard deviation is less, we can deduce that the keys are more uniform than a Poisson distributionbased model expects and that a smaller prime or over ow area may su ce If the observations are worse, the keytoaddress transformation is not doing its job Example 67

