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I've seen the integer version $$m_n = m_{n-1} + \frac{a_n-m_{n-1}}{n}.$$ But how do I calculate it where there are amounts, which is $$A=[3, 5, 4]$$ and the corresponding quantities $$Q=[1.5, 2.5, 1]$$ Below version doesn't work, I'm doing it like this $$m_{1.5}=0+\frac{3-0}{1.5}=2$$ $$m_{4}=2+\frac{5-2}{4}=2.75$$ Which is wrong $m_4$ should have been $$\frac{3+5}{1.5+2.5}=2$$

In this case, do I calculate the average incrementally in correct way?

Example for clarification First I bought 1.5kg apple for 3 dollars, then 2.5kg for 5 dollars, then 1kg for 4 dollars. I need to calculate the average price per kg of apple incrementally

Please read the comments below

Dulguun Otgon
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  • What is $m_{1.5}$ supposed to mean? $n$ have to be an integer in $m_n$. I think you are a confusing the number of the element with the value of the element. – Winther Dec 28 '14 at 13:10
  • Oh the problem I have to figure out is that let's say I bought apples several times. First 1.5kg for 3 dollars, second 2.5kg for 5 dollars, 1kg for 4 dollars. I need to calculate the average price per kg of apple incrementally. – Dulguun Otgon Dec 28 '14 at 13:17
  • Then you first have to calculate the price per kilo first: $P = [3/1.5, 5/2.5, 4/1.0] = [2,2,4]$ then do incremental averaging on this array. – Winther Dec 28 '14 at 13:26
  • Tried that way the answer comes wrong. It should be for A=[3] Q=[1.5] m=2,A=[3,5] Q=[1.5,2.5] m=2, A=[3,5,4] P=[1.5,2.5,1] m=2.4 – Dulguun Otgon Dec 28 '14 at 13:29
  • This is what I get: $m_1 = a_1 = 3/1.5 = 2.0$, $m_2 = m_1 + \frac{a_2-m_1}{2} = 2.0 + \frac{5/2.5-2.0}{2.0} = 2.0$, $m_3 = m_2 + \frac{a_3-m_2}{2} = 2.0 + \frac{4/1.0 - 2.0}{3.0} = 2.666$. What is wrong with this? Why do you think $2.4$ should be the correct result? – Winther Dec 28 '14 at 13:34
  • Now I think I know what you mean. You want $m = \frac{3+5+4}{1.5+2.5+1.0}$ right? This cannot be done with incremental averaging since the result on step $n$ depends on all the previous prices and there is no simple linear relationship between $m_n$ and $m_{n-1}$. – Winther Dec 28 '14 at 13:39
  • Ok thanks, but how accurate is that approach? – Dulguun Otgon Dec 28 '14 at 14:12
  • You need to store another variable, which is the total quantity seen so far, say $t_n$. Then the update formulas are $t_n = t_{n-1} + q_n$ and $m_n = (t_{n-1}m_{n-1} + a_nq_n)/t_n$. You didn't need to keep track of $t_n$ before because it was always equal to $n$. –  Jan 09 '15 at 21:12

1 Answers1

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Well, it's quite easy: let $a_i$ be the prices and $q_i$ the respective quantities. Let $A_n = a_1 +a_2 + \cdots a_n$ , $Q_n = q_1 +q_2 + \cdots q_n$

Then the average price is

$$M_n = \frac{A_n}{Q_n}=\frac{A_{n-1}+a_n}{Q_{n-1}+q_n}=\frac{Q_{n-1} M_{n-1} +a_n }{Q_{n-1}+q_n}$$

You cannot obtain the recursion in terms of $M_{n-1}$ (and $n$) alone, you need to store $Q_{n-1}$ (instead of $n$; indeed, you can think of it as a "effective" $n$).

In the particular case in which $q_n=k$ (constant), then $Q_n= n k$ and

$$M_n= \frac{ (n-1) k M_{n-1} +a_n}{k n}=M_{k-1} + \frac{a_n/k-M_{n-1}}{ n}$$

leonbloy
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  • Ok it doesn't get any more accurate than this $$M_n = \frac{A_n}{Q_n}=\frac{A_{n-1}+a_n}{Q_{n-1}+q_n}=\frac{Q_{n-1} M_{n-1} +a_n }{Q_{n-1}+q_n}$$ – Dulguun Otgon Jan 14 '15 at 08:17