Multiply sets - Revision history

Connor at 15:02, 2 August 2012

2012-08-02T15:02:12Z

Connor at 10:38, 2 August 2012

2012-08-02T10:38:50Z

Connor at 10:38, 2 August 2012

2012-08-02T10:38:07Z

Connor at 10:37, 2 August 2012

2012-08-02T10:37:18Z

Connor at 11:59, 30 July 2012

2012-07-30T11:59:16Z

Connor at 11:24, 30 July 2012

2012-07-30T11:24:54Z

Connor at 11:19, 30 July 2012

2012-07-30T11:19:01Z

Connor: Created page with "Multiply across a result set.

schema:scott
DROP TABLE interest ..."

2012-07-30T11:18:27Z

Created page with "Multiply across a result set. <div class='ht'> <div class=params>schema:scott</div> <source lang=sql class='tidy'>DROP TABLE interest</source> <source lang=sql class='setup'> ..."

New page

Multiply across a result set.
<div class='ht'>
<div class=params>schema:scott</div>
<source lang=sql class='tidy'>DROP TABLE interest</source>
<source lang=sql class='setup'> CREATE TABLE interest(
yr INTEGER,
rate DECIMAL );
INSERT INTO interest VALUES (2002,0.05);
INSERT INTO interest VALUES (2003,0.04);
INSERT INTO interest VALUES (2004,0.05);
INSERT INTO interest VALUES (2005,0.03);
</source>
<div>
In this example we are attempting to see how much money has been obtained over 4
years due to interest.
As the added up rate is not accurate 5 + 4 + 5 + 3 = 17%
We need to instead find the logarithm of the compound interest and then we need to sum that.
SELECT SUM(LN(1+rate/100)) FROM interest
Then we inverse or take the exponent of the logarithm with
SELECT EXP(SUM(LN(1+rate/100))) FROM interest
and then finally to get the amount after 4 years we times this amount by 100 ($100).
</div>
<source lang='sql' class='def'>SELECT EXP(SUM(LN(1+rate/100)))*100
FROM interest</source>
<div class="ecomm e-mysql" style="display: none"></div>
</div>
{{Hacks Ref}}

@@ Line 20: / Line 20: @@
 <p>and then finally to get the amount after 4 years we times this amount by 100 (£100).</p>
 </div>
-<source lang='sql' class='def'>SELECT EXP(SUM(LN((rate/100)+1)))*100
+<source lang='sql' class='def'>
-FROM interest</source>
+SELECT EXP(SUM(LN((rate/100)+1)))*100
 <div class="ecomm e-sqlserver" style="display: none"></div>
 </div>
 {{Hacks Ref}}

@@ Line 1: / Line 1: @@
-Multiplying across a result set allows for interest rates to calculated correctly.
+<p>Multiplying across a result set allows for interest rates to calculated correctly.</p>
-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
+<p>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.
+to get 17% (£117) isn't correct.</p> <p>To get the correct results you have to follow the steps given here.</p>
 <div class='ht'>
 <div class=params>schema:scott</div>

@@ Line 1: / Line 1: @@
 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 instead the correct result is obtained differently.
+to get 17% (£117) isn't correct. To get the correct results you have to follow the steps given here.
 <div class='ht'>
 <div class=params>schema:scott</div>

@@ Line 1: / Line 1: @@
-Multiply across a result set.
+Multiplying across a result set allows for interest rates to calculated correctly.
 <div class='ht'>
 <div class=params>schema:scott</div>
@@ Line 12: / Line 14: @@
 </source>
 <div>
 <p>We need to instead find the logarithm of the compound interest and then we need to sum that.</p>
 <p>SELECT SUM(LN((rate/100)+1)) FROM interest</p>
 <p>Then we inverse or take the exponent of the logarithm with</p>
 <p>SELECT EXP(SUM(LN((rate/100)+1))) FROM interest</p>
-<p>and then finally to get the amount after 4 years we times this amount by 100 ($100).</p>
+<p>and then finally to get the amount after 4 years we times this amount by 100 (£100).</p>
 </div>
 <source lang='sql' class='def'>SELECT EXP(SUM(LN((rate/100)+1)))*100
 FROM interest</source>
-<div class="ecomm e-mysql" style="display: none"></div>
+<div class="ecomm e-sqlserver" style="display: none"></div>
 </div>
 {{Hacks Ref}}

@@ Line 12: / Line 12: @@
 </source>
 <div>
-In this example we are attempting to see how much money has been obtained over 4
+<p>In this example we are attempting to see how much money has been obtained over 4
-years due to interest.
+years due to interest.</p>
-As the added up rate is not accurate 5 + 4 + 5 + 3 = 17%
+<p>As the added up rate is not accurate 5 + 4 + 5 + 3 = 17%</p>
-We need to instead find the logarithm of the compound interest and then we need to sum that.
+<p>We need to instead find the logarithm of the compound interest and then we need to sum that.</p>
-SELECT SUM(LN((rate/100)+1)) FROM interest
+<p>SELECT SUM(LN((rate/100)+1)) FROM interest</p>
-Then we inverse or take the exponent of the logarithm with
+<p>Then we inverse or take the exponent of the logarithm with</p>
-SELECT EXP(SUM(LN((rate/100)+1))) FROM interest
+<p>SELECT EXP(SUM(LN((rate/100)+1))) FROM interest</p>
-and then finally to get the amount after 4 years we times this amount by 100 ($100).
+<p>and then finally to get the amount after 4 years we times this amount by 100 ($100).</p>
 </div>
 <source lang='sql' class='def'>SELECT EXP(SUM(LN((rate/100)+1)))*100

@@ Line 6: / Line 6: @@
    yr INTEGER,
    rate INTEGER );
-INSERT INTO interest VALUES (2002,0.05);
+INSERT INTO interest VALUES (2002,5);
-INSERT INTO interest VALUES (2003,0.04);
+INSERT INTO interest VALUES (2003,4);
-INSERT INTO interest VALUES (2004,0.05);
+INSERT INTO interest VALUES (2004,5);
-INSERT INTO interest VALUES (2005,0.03);
+INSERT INTO interest VALUES (2005,3);
 </source>
 <div>
@@ Line 16: / Line 16: @@
 As the added up rate is not accurate 5 + 4 + 5 + 3 = 17%
 We need to instead find the logarithm of the compound interest and then we need to sum that.
-SELECT SUM(LN(1+rate/100)) FROM interest
+SELECT SUM(LN((rate/100)+1)) FROM interest
 Then we inverse or take the exponent of the logarithm with
-SELECT EXP(SUM(LN(1+rate/100))) FROM interest
+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).
 </div>
-<source lang='sql' class='def'>SELECT EXP(SUM(LN(1+rate/100)))*100
+<source lang='sql' class='def'>SELECT EXP(SUM(LN((rate/100)+1)))*100
 FROM interest</source>
 <div class="ecomm e-mysql" style="display: none"></div>
 </div>
 {{Hacks Ref}}

Multiply sets - Revision history

Connor at 15:02, 2 August 2012

Connor at 10:38, 2 August 2012

Connor at 10:38, 2 August 2012

Connor at 10:37, 2 August 2012

Connor at 11:59, 30 July 2012

Connor at 11:24, 30 July 2012

Connor at 11:19, 30 July 2012

Connor: Created page with "Multiply across a result set. schema:scott DROP TABLE interest ..."

Connor: Created page with "Multiply across a result set.

schema:scott
DROP TABLE interest ..."