# Forecasting From a Single Performance of a Marketing Machine Having an Inverse Relationship between Input and Output Ted Mitchell.

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Forecasting From a Single Performance of a Marketing Machine Having an Inverse Relationship between Input and Output Ted Mitchell

Three Weaknesses of Presenting a Marketing Performance as the Single Performance of a Two-Factor Machine 1) Overly Simplifies a Complex Process by assuming other marketing inputs are held constant or not important in the analysis 2) Can only deal with a single input being converted into a single output 3) Need to know the in advance if the relationship between input and output is positive or inverse

The Inverse Relationship In a single performance of a Two-Factor Marketing Machine is 1) difficult to explain and conceptualize 2) difficult to use for a simple forecast

The Two-Factor Model Is easily seen as equation in which the conversion rate, r, reflects a positive relationship Increase the product’s quality, the information about the product, the convenience of the transaction’s location and you will increase the quantity demanded

The Inverse Relationship Between the size of the price tag, P, and number of items sold, Q Is the best example of an inverse relationship between Marketing Inputs and Outputs when the size of the price tag is decreased then the quantity demanded will increase

The Inverse Relationship is Presented as slope-origin version of the demand machine Demand, Q = (conversion rate, r) x (inverse of price tag, 1/P) The conversion rate, r = Q/(1/P) awkward to explain how the increase in the size of the price tag is reducing the quantity demanded

A slope-origin presentation of the inverse relationship between price and quantity sold from a single performance 0, 0 Inverse of the price tag, 1/P Amount of Product Sold, Q Slope-Origin equation Q = r x 1/P Slope-Origin equation Q = r x 1/P Conversion rate is the slope, r = Q/(1/P)

A Single Performance of a Pricing Machine has been recorded The price tag was P = \$3.50 on each item sold The number of items sold was Q = 2,000 cups of coffee You want to forecast the number of cups that will be sold if the price tag is set a P = \$4.00 on each cup You know there is an inverse relationship between the size of the price tag and the quantity sold An increase in price is expected to reduce the number of cups sold and forecasted quantity is expected to be less than 2,000 cups What is the forecasted number of cups sold, Q*, when the price tag on a cup is \$4.00?

Example of a Simple Forecast Given an Inverse Relationship Recorded Performance Calibrated performance Proposed Price and calibrated conversion rate Forecasted Demand, Q* InputPrice Tag size P = \$3.50 1/P = 1/\$3.50 1/P = 0.2857 1/P = 1/\$4.00 1/P = 0.25 Conversion rate, r r = Q/(1/P) r = 2,000/0.2857 r = 7,000 r = 7,000 OutputQuantity sold Q = 2000 cups Q = 2,000 cupsForecasted Demand, Q* Q = r x !/P Q* = 1,750 cups will be sold

Example of a Simple Forecast Given an Inverse Relationship Recorded Performance Calibrated Performance Proposed Price and calibrated conversion rate Forecasted Demand, Q* InputPrice Tag size P = \$3.50 1/P = 1/\$3.50 1/P = 0.2857 1/P = 1/\$4.00 1/P = 0.25 Conversion rate, r r = Q/(1/P) r = 2,000/0.2857 r = 7,000 r = 7,000 OutputQuantity sold Q = 2000 cups Q = 2,000 cupsForecasted Demand, Q* Q = r x !/P Q* = 1,750 cups will be sold

Example of a Simple Forecast Given an Inverse Relationship Recorded Performance Calibrated Performance Proposed Price and Calibrated conversion rate Forecasted Demand, Q* InputPrice Tag size P = \$3.50 1/P = 1/\$3.50 1/P = 0.2857 1/P = 1/\$4.00 1/P = 0.25 Conversion rate, r r = Q/(1/P) r = 2,000/0.2857 r = 7,000 r = 7,000 OutputQuantity sold Q = 2000 cups Q = 2,000 cupsForecasted Demand, Q* Q = r x !/P Q* = 1,750 cups will be sold

Example of a Simple Forecast Given an Inverse Relationship Recorded Performance Calibrated Performance Proposed Price and calibrated conversion rate Forecasted Demand, Q* InputPrice Tag size P = \$3.50 1/P = 1/\$3.50 1/P = 0.2857 1/P = 1/\$4.00 1/P = 0.25 Conversion rate, r r = Q/(1/P) r = 2,000/0.2857 r = 7,000 r = 7,000 OutputQuantity sold Q = 2000 cups Q = 2,000 cupsForecasted Demand, Q* Q* = r x !/P Q* = 1,750 cups will be sold

Using a calibrated single performance to forecast a new quantity sold at a proposed price of \$4.00 per cup 0, 0 Inverse of the Price tag, 1/P Amount of Product Sold, Q Slope-Origin equation Q* = 7,000 x 1/\$4.00 Q* = 1,750 cups Slope-Origin equation Q* = 7,000 x 1/\$4.00 Q* = 1,750 cups Calibrated Conversion rate is the slope, r = Q2,000/(1/\$3.50) r = 7,000 1/\$3.50 = 0.2857 1/\$4.00 = 0.25 Q = 2,000 cups Q* = 1,750 cups

The Forecast of Q* = 1,750 cups sold is 1) relatively easy to forecast with an inverted input 2) hard to interpret the meaning of an inverted price (a basic definition of customer value) 3) hard to interpret the conversion rate as rate (7,000 fold increase?) 4) likely to be very inaccurate given it is based on a single observed performance

The great blessing Is that forecasting the demand using a slope- origin model calibrated from a single performance is not used very much!! It is used as a naïve or benchmark forecast Today we normally have several observed performances to work from A calibrated performance based on the average of many observations is likely to be used as a benchmark or as a forecast for a small business

Any Questions About forecasting from a single calibrated performance of an Inverse relationship?

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