Multiply sets

Multiplying across a result set allows for interest rates to calculated correctly.

In this example we get the interest after 4 years imagine over the 4 years we have rates 5%, 4%, 5% and 3% adding these rates to get 17% (£117) isn't correct.

To get the correct results you have to follow the steps given here.

schema:scott

DROP TABLE interest

 CREATE TABLE interest(
  yr INTEGER,
  rate INTEGER );
INSERT INTO interest VALUES (2002,5);
INSERT INTO interest VALUES (2003,4);
INSERT INTO interest VALUES (2004,5);
INSERT INTO interest VALUES (2005,3);

We need to instead find the logarithm of the compound interest and then we need to sum that.

SELECT SUM(LN((rate/100)+1)) FROM interest

Then we inverse or take the exponent of the logarithm with

SELECT EXP(SUM(LN((rate/100)+1))) FROM interest

and then finally to get the amount after 4 years we times this amount by 100 (£100).

SELECT EXP(SUM(LN((rate/100)+1)))*100
  FROM interest

SQL HACKS

Hack 10 Converting subqueries into joins

Hack 11 Converting aggregate subqueries into joins

Hack 14 Generate Combinations

Hack 16 Search for a String across columns

Hack 24 Multiply Across a Result Set

Hack 25 Keep a running total

Hack 25.5 Splitting and combining columns

Hack 26 Include the rows your JOIN forgot

Hack 30 Calculate the maximum/minimum of two fields

Hack 33 Get values and subtotals in one shot

Hack 50 Combine tables containing different data

Hack 51/52 Display rows as columns

Hack 55 Import Someone Else's Data

Hack 62 Issue Queries Without Using a Table

Hack 63 Generate rows without tables

Hack 70 Combine your queries

Hack 72 Extract a subset of the results

Hack 78 Break it down by Range

Hack 88 Test two values from a subquery

Hack 98 Find and stop long running queries

Hack 100 Run SQL from a Web Page