Return on investment and marginal return

roi() divides each channel's total contribution by its total spend. mroi() asks the different and more useful question of what the next currency unit would return at current spend.

Usage

roi(contributions, spend)

mroi(spend_level, coefficient, type = "hill", delta = 0.01, ...)

Arguments

contributions: A data frame from contributions(), or a named numeric vector of total contribution per channel.
spend: A data frame of raw (untransformed) spend with one column per channel, or a named numeric vector of total spend per channel.
spend_level: Current spend level for the channel, in the same units the saturation curve was fitted on.
coefficient: The channel's fitted coefficient.
type: Saturation curve type.
delta: Increment used for the numerical derivative, as a fraction of spend_level.
...: Passed to saturate(), for example half_max and shape.

Value

roi() returns a data frame with columns channel, contribution, spend and roi, ordered by descending ROI, with roi set to NA where spend is zero. Channels appearing in only one of the two inputs are omitted, and reported when that happens. mroi() returns a single number.

Details

Average ROI and marginal ROI answer different questions and are routinely confused. Average ROI is a scorecard: what did this channel return over the period as a whole. Marginal ROI is the decision variable: what would the next unit return. Only the second one should drive a reallocation, and the two can point in opposite directions.

Where they sit relative to each other depends on the shape of the curve, and it is worth being precise because the usual summary of this is wrong.

For a concave response – saturate_hill() with shape = 1, Michaelis–Menten, negative exponential, saturate_power() – marginal ROI is below average ROI everywhere, and the gap widens the further up the curve a channel sits. Reallocating on average ROI then systematically over-funds channels that are already saturated, which are exactly the channels that look best on an average-ROI table.

For an S-shaped response – saturate_hill() with shape > 1, the usual way to represent a threshold effect – the curve is convex below its inflection point, and there marginal ROI is above average ROI. A channel in that region is under-funded: each extra pound works harder than the pounds already spent, because the channel has not yet reached the pressure at which it starts to pay. Ruling that out by assuming marginal is always lower is how a threshold channel stays starved.

mroi() differentiates the fitted response curve numerically, so it works for any curve saturate() supports, and it is finite and correct at spend_level = 0.

Marginal return when the regressor was adstocked

mroi() is the slope of the saturation curve with respect to the quantity the coefficient multiplies – the transformed media – at a single level. It is a property of the curve, not yet a return on spend, and two things separate the two.

Carryover spreads a unit of spend over many periods. Under a normalised kernel the weights sum to one, so the total extra response to one more unit of spend is roughly the curve's slope, arriving over the kernel's length. Multiplying by the kernel's first weight (1 - decay) gives only the response inside the period of spend; comparing that with an average ROI that counts every period's carryover compares a part with a whole, and understates slow channels several-fold.

And the slope at the mean adstocked level is not the mean of the slope across periods, which matters for flighted media and S-shaped curves.

marginal_roi() handles both by re-running the whole transform on a slightly larger budget. Use it for anything that feeds a budget decision, and use mroi() to read the shape of a curve.

Examples

data(mm_weekly)
north <- mm_weekly[mm_weekly$geo == "north", ]
channels <- c("tv", "video", "search", "social", "display")
truth <- attr(mm_weekly, "truth")

transformed <- as.data.frame(Map(
  function(x, d, h, sh) media_transform(
    x, adstock = list(decay = d),
    saturation = list(half_max = h, shape = sh)),
  north[channels], truth$decay[channels], truth$half_max[channels],
  truth$shape[channels]
))

model_data <- transformed
model_data$week <- seq_len(nrow(model_data))
model_data$price <- north$price
model_data$seasonality <- north$seasonality
model_data$holiday <- north$holiday

fit <- stats::lm(north$revenue ~ ., data = model_data)
contrib <- contributions(transformed, fit, index = north$date)

roi(contrib, north[channels])
#>   channel contribution  spend      roi
#> 1  social    132708.89  36716 3.614470
#> 2  search    208306.77  74560 2.793814
#> 3   video    102461.17  53248 1.924226
#> 4 display     40852.88  26937 1.516608
#> 5      tv    349628.86 266978 1.309579

# Average and marginal return are different questions. marginal_roi()
# re-runs the whole transform on a 1% larger budget, so carryover and
# flighting are both accounted for. Display's average pound returns about
# 1.5, but its next pound returns less than it costs.
avg <- roi(contrib, north[channels])
for (ch in channels) {
  m <- marginal_roi(north[[ch]], coefficient = stats::coef(fit)[[ch]],
                    adstock = list(decay = truth$decay[[ch]]),
                    saturation = list(half_max = truth$half_max[[ch]],
                                      shape = truth$shape[[ch]]))
  cat(sprintf("%-8s average %.2f  marginal %.2f\n",
              ch, avg$roi[avg$channel == ch], m$mroi))
}
#> tv       average 1.31  marginal 1.25
#> video    average 1.92  marginal 1.20
#> search   average 2.79  marginal 1.35
#> social   average 3.61  marginal 2.09
#> display  average 1.52  marginal 0.93