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! Exercise 15.4.4: In Example 15.6 we discussed the join of two relations R and S, with 1000 and 500 blocks, respectively, an
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a)

We effectively have to perform two nested-loop joins of 500 and 250blocks, respectively, using 101 blocks of memory. Such a join takes 250 + 500*250/100 = 1500 disk I/O's, so two of them takes 3000. Tothis number, we must add the cost of sorting the two relations, which takes four disk I/O's per block of the relations, or another 6000. The total disk I/O cost is thus 9000.

b)

1000+1000*500/100=6000*4=24000 i/o

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