Timeline for Incremental averaging (different case)
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| S Jan 14, 2015 at 8:13 | history | bounty ended | Dulguun Otgon | ||
| S Jan 14, 2015 at 8:13 | history | notice removed | Dulguun Otgon | ||
| Jan 14, 2015 at 8:13 | vote | accept | Dulguun Otgon | ||
| Jan 9, 2015 at 22:52 | answer | added | leonbloy | timeline score: 0 | |
| Jan 9, 2015 at 21:12 | comment | added | user856 | 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 $n$. | |
| S Jan 8, 2015 at 6:59 | history | bounty started | Dulguun Otgon | ||
| S Jan 8, 2015 at 6:59 | history | notice added | Dulguun Otgon | Canonical answer required | |
| Jan 8, 2015 at 6:58 | history | edited | Dulguun Otgon | CC BY-SA 3.0 |
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| Dec 28, 2014 at 14:12 | comment | added | Dulguun Otgon | Ok thanks, but how accurate is that approach? | |
| Dec 28, 2014 at 13:39 | comment | added | Winther | 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 $m_{n-1}$. | |
| Dec 28, 2014 at 13:34 | comment | added | Winther | 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? | |
| Dec 28, 2014 at 13:29 | comment | added | Dulguun Otgon | 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 | |
| Dec 28, 2014 at 13:26 | comment | added | Winther | 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. | |
| Dec 28, 2014 at 13:20 | history | edited | Dulguun Otgon | CC BY-SA 3.0 |
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| Dec 28, 2014 at 13:17 | comment | added | Dulguun Otgon | 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. | |
| Dec 28, 2014 at 13:10 | comment | added | Winther | 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. | |
| Dec 28, 2014 at 12:50 | history | asked | Dulguun Otgon | CC BY-SA 3.0 |