1 / 15

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.

lhyde
Download Presentation

Lecture 28: Size/Cost Estimation, Recovery

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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 = ….

More Related