Lets say a factory machine runs as follows:

  1. Day 1: $3$ widgets per minute $\times 1000$ minutes $\times 1$ kg per widget $= 3000$ kg.
  2. Day 2: $5$ widgets per minute $\times 800$ minutes $\times 1.5$ kg per widget $= 6000$ kg.
  3. Day 3: $8$ widgets per minute $\times 300$ minutes $\times 0.5$ kg per widget $= 1200$ kg.

So, over the $3$ days the factory machine has run for $2100$ minutes. I am assuming that if I multiply the weighted average number of widgets with the weighted average kg per widget and multiply this by the number of minutes over the three days $(2100)$ this should equal $10,200$ kg $(3000+6000+1200)$.

My weighting calculation is below for widgets per minute and kg per widget.

$$(3\cdot1000\cdot1)+(5\cdot800\cdot1.5)+(8\cdot300\cdot0.5)/(1000\cdot1)+(800\cdot1.5)+(300\cdot0.5) = 4.833$$ widgets per minute.

$$(3\cdot1000\cdot1)+(5\cdot800\cdot1.5)+(8\cdot300\cdot0.5)/(3\cdot1000\cdot1)+(5\cdot800\cdot1.5) = 1.085$$ kg per widget. However $4.833$ widgets $\times 1.085$ kg per widget $\times 2100$ mins $= 11,012$ kg over the three days. Not the correct $10,200$kg.

Why is this the case? I have tried weighting by total for the day and this still does not produce the correct answer. I have been using Excel and have noticed that if I keep either "widgets per minute" or "kg per widget" the same across all three days and only change the other variable I get a correct answer. However when both variable differ across the three days the total becomes distorted as you can see from the example above.

Is it not possible to find a weighted average using more than one weighting factor? I can't get my head around this.

Thank you in advance for any answers.


Your weighting calculation is incorrect, otherwise double weighting should work. The correct calculation is

  1. Widgets per minute, weighting calculation: $$\frac{3\cdot1000+5\cdot800+8\cdot300}{1000+800+300}=\frac{94}{21}$$ widgets per minute.
  2. Kg per widget, weighting calculation: $$\frac{1\cdot3\cdot1000+1.5\cdot5\cdot800+0.5\cdot8\cdot300}{3\cdot1000+5\cdot800+8\cdot300}=\frac{102}{94}$$

Hence, total number of kg $$\text{widgets per min}\times\text{kg per widget}\times\text{mins}=\frac{94}{21}\times\frac{102}{94}\times2100=102\cdot100=10200\text{kg}$$

  • $\begingroup$ Perfect. Thankyou! $\endgroup$
    – user325899
    Mar 26 '16 at 10:06
  • $\begingroup$ @user325899 You are welcome. If this answers your question you can accept the answer to indicate it. $\endgroup$
    – Jimmy R.
    Mar 26 '16 at 10:50

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