Planning for More Than One Possible Outcome

Botie receives boxes of sensors. Some arrive undamaged, while others need replacements. Ordering for just the most common outcome can leave him unprepared for the less common ones.

We need two pieces of information: what could happen and how likely each possibility is. A distribution keeps these together. We can then calculate an average requirement or compare the consequences of different stocking decisions.

Expected damage = (0 × 0.5) + (1 × 0.3) + (2 × 0.2) = 0.7
Most likely individual count: 0 (probability 0.5)

No single outcome describes every delivery. Change the probability shares, keeping their total at one, and compare the weighted count with the individual outcomes.

0.7

Let's make a table with one position for each possible count of damaged sensors. Beside that count, put its probability. Every probability must be nonnegative, and together they must add to one, because our listed outcomes exhaust the possibilities.

A count takes discrete values. Temperature is different: a continuous model describes it with a density. To find the probability of a temperature range, we take the area under that density over the range. The density's height at a single temperature is not a probability.

A distribution is a description of uncertainty under a model, not a guarantee about the next delivery.

Imagine receiving many deliveries under the same conditions. Outcomes with larger probabilities should contribute more to the average. Multiply each count by its probability and add the contributions.

The result need not be an outcome that one delivery can actually have. We cannot receive a fraction of a broken sensor, but we can have a fractional average across deliveries.

E[damage] = 0 × 0.5 + 1 × 0.3 + 2 × 0.2
E[damage] = 0.7

Expectation answers an average question. It does not replace the whole distribution.

Suppose each missing spare causes a delay. For a proposed stock level, calculate the shortage in each outcome, keeping only positive shortages. Weight those shortages by their probabilities.

We can repeat this calculation for several stock levels. If holding spares also costs money, include that cost before comparing decisions. The cheapest choice depends on both the distribution and the consequences.

Stock one spare: shortages = [0, 0, 1]
Expected shortage = 0 × 0.5 + 0 × 0.3 + 1 × 0.2 = 0.2

The most likely outcome does not necessarily produce the lowest expected cost.

import numpy as np
counts = np.array([0., 1., 2.])
probabilities = np.array([0.25, 0.5, 0.25])
contributions = counts * probabilities
print(contributions.tolist())
print(float(contributions.sum()))

[0.0, 0.5, 0.5]
1.0

The probabilities might have come from historical frequencies or a fitted model. A new supplier or a change in packaging can make those values inappropriate. Compare predictions with later deliveries and retain rare outcomes when they matter.

Expected cost is one decision rule. A service requirement might instead limit the chance of any shortage. State which objective is being used before choosing a stock level.

Botie’s workshop computes an expectation. The choice of stocking objective remains a separate decision.

A shipment can contain zero, one or two damaged sensors. Its expected damage count can lie between those integers because it averages across possible shipments, weighted by their probabilities. It is not a claim that a particular shipment contains a fraction of a sensor.

If Botie must order whole replacement units, rounding the expectation is one possible rule. Whether it is useful depends on the consequences of too many or too few replacements. A prediction summary and an action are different objects.

Suppose an unused replacement costs one unit, while a shortage costs five per missing replacement. For each stock choice, calculate the cost under every damage outcome and then take its probability-weighted average.

This is the same expectation operation, now applied to cost instead of damage count. It connects a probability distribution to a decision without pretending the most likely outcome is the only one that matters.

import numpy as np
damage = np.array([0,1,2])
probability = np.array([.5,.3,.2])
for stock in [0,1,2]:
    unused = np.maximum(stock-damage, 0)
    shortage = np.maximum(damage-stock, 0)
    print(stock, round(float(probability@(unused+5*shortage)), 2))

0 3.5
1 1.5
2 1.3

The distribution may itself be estimated from a small or changing collection of shipments. A precisely computed expectation can therefore depend on uncertain inputs.

Repeat the decision calculation under plausible alternative damage rates and shortage costs. If the preferred stock changes easily, the decision needs more evidence or an explicit tolerance for risk. Arithmetic precision does not remove uncertainty about the model.