# (Homework Solution): Question 1 (5 marks) Consider two relations called Invoice and Customers.

Question 1 (5 marks) Consider two relations called Invoice and Customers. Imagine that the Customers relation has 10 p SELECT* FROM Customers INNER JOIN Invoice ON Customers.Cust_ID Invoice.Cust_ID; We wish to evaluate an equijoin between Customers and Invoice, with an equality condition Customers.Cust ID Invoice.Cust_ID. There are 502 buffer pages available in memory for this operation. Both relations are stored as (unsorted) heap files. Neither relation has any indexes built on it. Consider the alternative join strategies described below and calculate the cost of each alternative. Evaluate the algorithms using the number of disk I/O’s (i.e. pages) as the cost. For each strategy, provide the formulae you use to calculate your cost estimates. a) Page-oriented nested loops join. Consider Customers as the outer relation. (1 mark) b) Block-oriented nested loops join. Consider Customers as the outer relation.(1 mark) c) Sort-Merge join. Assume that Sort-Merge Join can be done in 2 passes. (1 mark) d) Hash Join (1 mark) e) What would be the lowest possible cost to perform this query, assuming that no indexes are built on any of the two relations, and assuming that sufficient buffer space is available? What would be the minimum buffer size required to achieve this cost? Explain briefly. (1 mark)

Assume I/O to be 100tuples/page – C’r and 80tuples/page – I’s then

Let C be the 1,000 customers and I be the 5,000 invoices pages then Cost can be calculated using following formula:

Cost = C+C*I = 1,000+1,000*5,000=5,001,000 (considering Customers as outer relation)

b.) Block-oriented nested loops join –

Cost=C+(C/(B-2))*N

B pages of Memory buffer = 502

then Cost will be = 1,000+(1,000/(502-2))*5,000 = 11,000

c.) Sort-Merge join –

Cost = (Cost to sort Customer+ Cost to sort Invoice) = ((No.of passes for each write and for each read page*C)+(No.of passes for each write and for each read page*I) + C + I)

= 4*1,000+ 4*5,000 + 1,000 + 5,000 = 30,000

d.) Hash Join –

Cost to partition : 2C (read and write) = 2*1,000 = 2,000

Cost to partition : 2I (read and write ) = 2*5,000 = 10,000

Cost to join partition : C+I = 6,000

Total : 3*(C+I) = 3*(1,000+5,000)=18,000

e.) Minimum cost to perform the query if no indexes are built will be

C+C*C’r*(Indexes given+1) = 1,000+1,000*100(0+1) = 101,000

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