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Lecture 28: Size/Cost Estimation, Recovery

Lecture 28: Size/Cost Estimation, Recovery. Monday, December 6, 2004. Outline. Cost estimation: 16.4 Recovery using undo logging 17.2. Size Estimation. The problem: Given an expression E, compute T(E) and V(E, A) This is hard without computing E Will ‘estimate’ them instead.

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Lecture 28: Size/Cost Estimation, Recovery

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  1. Lecture 28:Size/Cost Estimation, Recovery Monday, December 6, 2004

  2. Outline • Cost estimation: 16.4 • Recovery using undo logging 17.2

  3. Size Estimation The problem: Given an expression E, compute T(E) and V(E, A) • This is hard without computing E • Will ‘estimate’ them instead

  4. Size Estimation Estimating the size of a projection • Easy: T(PL(R)) = T(R) • This is because a projection doesn’t eliminate duplicates

  5. Size Estimation Estimating the size of a selection • S = sA=c(R) • T(S) san be anything from 0 to T(R) – V(R,A) + 1 • Estimate: T(S) = T(R)/V(R,A) • When V(R,A) is not available, estimate T(S) = T(R)/10 • S = sA<c(R) • T(S) can be anything from 0 to T(R) • Estimate: T(S) = (c - Low(R, A))/(High(R,A) - Low(R,A)) • When Low, High unavailable, estimate T(S) = T(R)/3

  6. Size Estimation Estimating the size of a natural join, R ||A S • When the set of A values are disjoint, then T(R ||A S) = 0 • When A is a key in S and a foreign key in R, then T(R ||A S) = T(R) • When A has a unique value, the same in R and S, then T(R ||A S) = T(R) T(S)

  7. Size Estimation Assumptions: • Containment of values: if V(R,A) <= V(S,A), then the set of A values of R is included in the set of A values of S • Note: this indeed holds when A is a foreign key in R, and a key in S • Preservation of values: for any other attribute B, V(R ||A S, B) = V(R, B) (or V(S, B))

  8. Size Estimation Assume V(R,A) <= V(S,A) • Then each tuple t in R joins some tuple(s) in S • How many ? • On average T(S)/V(S,A) • t will contribute T(S)/V(S,A) tuples in R ||A S • Hence T(R ||A S) = T(R) T(S) / V(S,A) In general: T(R ||A S) = T(R) T(S) / max(V(R,A),V(S,A))

  9. Size Estimation Example: • T(R) = 10000, T(S) = 20000 • V(R,A) = 100, V(S,A) = 200 • How large is R ||A S ? Answer: T(R ||A S) = 10000 20000/200 = 1M

  10. Size Estimation Joins on more than one attribute: • T(R ||A,B S) = T(R) T(S)/(max(V(R,A),V(S,A))*max(V(R,B),V(S,B)))

  11. Histograms • Statistics on data maintained by the RDBMS • Makes size estimation much more accurate (hence, cost estimations are more accurate)

  12. Histograms Employee(ssn, name, salary, phone) • Maintain a histogram on salary: • T(Employee) = 25000, but now we know the distribution

  13. Histograms Ranks(rankName, salary) • Estimate the size of Employee ||Salary Ranks

  14. Histograms • Eqwidth • Eqdepth

  15. Histograms • Assume: • V(Employee, Salary) = 200 • V(Ranks, Salary) = 250 • Main property: Employee ⋈Salary Ranks=Employee1⋈Salary Ranks1’  ...  Emplyee6⋈Salary Ranks6’ • A tuple t in Employee1 joins with how many tuples in Ranks1’ ? • Answer: with T(Rank1)/T(Rank) * T(Rank)/250 = T1 ‘/ 250 • Then T(Employee ⋈Salary Ranks) = = Si=1,6 Ti Ti’ / 250 = (200x8 + 800x20 + 5000x40 + 12000x80 + 6500x100 + 500x2)/250 = ….

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