Positive Part¶

Definition¶

For a real number $u$ , its positive part is

u^+=\max\{u,0\}=\begin{cases}u,&u\ge0,\\0,&u<0.\end{cases}

In the newsvendor model, $(y-D)^+$ is leftover stock and $(D-y)^+$ is unmet demand.

Interpretation¶

The positive part keeps a number when it is positive and replaces it by $0$ otherwise. In the newsvendor model it splits one difference $y-D$ into two counts that cannot be negative: leftover stock and unmet demand.

The graph of the positive part of u, equal to 0 for negative u and to u for positive u, with the line u itself dashed.

The graph is flat at height $0$ for $u<0$ and follows the line $u^+=u$ for $u\ge0$ $0$ ; the dashed piece of the line $u$ below zero is what the positive part cuts off. Like the absolute value, it has a corner at $u=0$ $0$ , and in fact $u^+=\tfrac{1}{2}(u+|u|)$ $∣ u ∣)$ .

A common slip is to compute $(y-D)^+$ as $y^+-D^+$ $D^{+}$ . For $y=10$ and $D=12$ the leftover is $(-2)^+=0$ $0$ , not $10-12=-2$ $- 2$ .

Key Results¶

Both leftover and shortage quantities are nonnegative.
For a fixed realized demand, at least one of them is zero.
For discrete demand, expectations are computed by probability-weighting the positive parts for every demand outcome.

Examples¶

Defining the negative part by $u^-=\max\{-u,0\}$ gives the identity $u=u^+-u^-$ $u^{-}$ .

If $y=10$ and $D=7$ $7$ , leftovers are 3 and unmet demand is zero.
If $y=10$ and $D=12$ $12$ , leftovers are zero and unmet demand is 2.

Prerequisites¶

Real numbers

Sources¶

OR Modelling · Week 6

Aliases: ramp function, plus function, positive part function

Last updated: 20 October 2026 03:02