Mathematical Question to Isolate Gross Profit Drivers

[nanceediane]

In the Setup below, I am trying to isolate the impact of change in Mix from the impact of change in individual product GP's to explain the Combined/Overall -2.3% drop in gross profit. First, I hold Mix unchanged to isolate product GP impact and then I hold product GP unchanged to isolate Mix. However, these two components do not add up to exactly -2.3% and I cannot understand why they would not mathematically.

Does anyone understand what is incorrect in this scenario?

Nancy

SETUP	% Total Sales (Mix)		Gross Profit
	Yr 1	Yr 2	Change		Yr 1	Yr 2	Change
Product X	40.0%	25.0%	-15.0%		30.0%	60.0%	30.0%
Product Y	60.0%	75.0%	15.0%		40.0%	25.0%	-15.0%
Combined	100.0%	100.0%	0.0%		36.0%	33.8%	-2.3%
HOLD MIX	% Total Sales (Mix)		Gross Profit
	Yr 1	Yr 2	Change		Yr 1	Yr 2	Change
Product X	40.0%	40.0%	0.0%		30.0%	60.0%	30.0%
Product Y	60.0%	60.0%	0.0%		40.0%	25.0%	-15.0%
Combined	100.0%	100.0%	0.0%		36.0%	39.0%	3.0%	(A) Product GP Change Impact
HOLD PROD GP	% Total Sales (Mix)		Gross Profit
	Yr 1	Yr 10	Change		Yr 1	Yr 10	Change
Product X	40.0%	25.0%	-15.0%		30.0%	30.0%	0.0%
Product Y	60.0%	75.0%	15.0%		40.0%	40.0%	0.0%
Combined	100.0%	100.0%	0.0%		36.0%	37.5%	1.5%	(B) Sales Mix Change Impact
							4.5%	Total GP Change = A + B
							Why doesn't this equal -2.3%?

 

[JeffM]

Your intuition is, sadly, wrong.

For example, consider the relationship
revenue = number of units sold * price per unit.

\(\displaystyle Let\ r = revenue;\)

\(\displaystyle u = units\ sold;\ and\)

\(\displaystyle p = price\ per\ unit \implies\)

\(\displaystyle r = p * u.\)

Now suppose both price per unit and number of units sold change.

\(\displaystyle p + \Delta p = new\ price\ per\ unit.\)

So \(\displaystyle \dfrac{\Delta p}{p} = relative\ change\ in\ price.\)

With me so far?

\(\displaystyle u + \Delta u = new\ number\ of\ units\ sold.\)

So \(\displaystyle \dfrac{\Delta u}{u} = relative\ change\ in\ number\ units\ sold.\)

Still with me?

So \(\displaystyle new\ revenue = (p + \Delta p)(u + \Delta u) = pu + p\Delta u + u\Delta p + \Delta p \Delta u =\)

\(\displaystyle r + p\Delta u + u\Delta p + \Delta p \Delta u = r + \Delta r.\)

So the dollar change in revenue is \(\displaystyle \Delta r = p \Delta u + u \Delta p + \Delta p \Delta u.\)

I hope you are still OK. If not, ask a question.

So the relative change in revenue is

\(\displaystyle \dfrac{\Delta r}{r} = \dfrac{p \Delta u + u \Delta p + \Delta p \Delta u}{pu} = \dfrac{\Delta p}{p} + \dfrac{\Delta u}{u} + \dfrac{\Delta p \Delta u}{pu}.\)

So to get an exact answer, you need to add three terms, one showing the effect of the change in price assuming no change in units sold, one showing the effect of the change in units sold assuming no change in price, and one accounting for the interaction of both changes. However, if the relative changes are both individually small, the interaction between them is negligible, which means the sum of the two relative changes will give a good approximation.

\(\displaystyle \dfrac{\Delta r}{r} \approx \dfrac{\Delta p }{p} + \dfrac{\Delta u}{u}.\)

 

[nanceediane]

Jeff,


Thank you for taking all the time to walk me through this process. Your explanation was very clear. I appreciate it.

Nancy