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To derive Eq 3-72 in Chap 3-5-14, 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
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n! 1 Pb (q) = q!(n q)! m
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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! 6-8
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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 3-19) If q records
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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 6-9
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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
6-10
The successive terms of the Poisson distribution are related to each other by the iteration mean 6-11 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)
6-12
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)
6-13
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)
6-14
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 6-15
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) 6-16 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 4-2-4
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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 6-18 6-17
and the number of records in the primary le is then m P 1u, and in the over ow le is 6-19 o = n m P 1u The over ow area contains the tails of all the Poisson-distributed record sequences Using Eq 6-5 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 6-20 = 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 key-to-address transformation is not doing its job
Example 6-7
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