Step Functions for Tiered Fees

Why Tiered Fees Are Step Functions

A tiered fee stays constant while an input remains within a specified range, then changes when the input crosses a threshold. For example, a parking facility might charge one amount for up to one hour and a higher amount for more than one hour but no more than two hours.

The fee does not increase continuously with every small increase in time. Instead, it takes a fixed value on each interval. A function with this kind of behavior is called a step function.

Definition: A step function is a function that has a constant output on each of a collection of input intervals. The constant output may change from one interval to another. For a fee model, the input is often a nonnegative quantity such as time or weight, and the output is the charge.

To build a useful model, identify the input, its contextual domain, the thresholds where the fee changes, and the charge on each resulting interval. The endpoints matter: a threshold value must be assigned to exactly one interval.

From Fee Schedule to Piecewise Rule

Suppose the input intervals are determined by thresholds \(b_0<b_1<\cdots<b_n\), and the charge on each interval is a constant. A fee schedule that says “up to \(b_1\),” then “more than \(b_1\) and up to \(b_2\),” can be represented using right-closed intervals:

[b_0,b_1],\quad (b_1,b_2],\quad \ldots,\quad (b_{n-1},b_n].

The notation \([b_0,b_1]\) includes both endpoints; \((b_1,b_2]\) excludes \(b_1\) and includes \(b_2\). This matches the wording “up to” at each upper threshold. If the schedule instead says “at least \(b_1\) but less than \(b_2\),” the interval should include its lower endpoint and exclude its upper endpoint. The written conditions, not a universal convention, determine which endpoint belongs where.

Once the intervals are settled, put the appropriate constant charge beside each one. The result is a piecewise rule: each input is checked against the interval conditions, and the function returns that interval’s fee.

Worked Example: Parking by the Started Hour

A parking lot charges 4 dollars for the first hour or any part of it, then 3 dollars for each additional hour or part of an hour. Parking for zero hours costs nothing. Let \(t\) be the parked time in hours, with \(t\geq 0\), and let \(P(t)\) be the fee in dollars.

At exactly zero hours, the fee is \(0\). For a positive time up to and including one hour, the fee is \(4\). For a time greater than one hour and up to two hours, the fee is \(4+3=7\). For a time greater than two hours and up to three hours, it is \(4+3+3=10\). Thus the rule begins:

P(t)= \begin{cases} 0, & t=0,\\ 4, & 0<t\leq 1,\\ 7, & 1<t\leq 2,\\ 10, & 2<t\leq 3. \end{cases}

For example, at \(t=1.5\), the condition \(1<t\leq2\) applies, so \(P(1.5)=7\) dollars. At \(t=2\), the same interval applies and the fee is still \(7\) dollars; at \(t=2.1\), the next interval applies and the fee is \(10\) dollars. These values agree with the rule that any part of the next hour incurs the next charge.

More generally, for each positive integer \(n\), a parked time satisfying \(n-1<t\leq n\) occupies the \(n\)th started hour. The fee is \(4+3(n-1)\) dollars. At \(t=3.4\), for instance, \(n=4\), so \(P(3.4)=4+3(4-1)=4+9=13\) dollars. This calculation checks against the schedule: the first hour costs 4 dollars, and the next three started hours cost 3 dollars each.

Why the Interval Conditions Work

Ordered thresholds divide the input range into consecutive pieces. If the intervals exclude the previous upper threshold and include the next one, then no input at a shared threshold is counted twice or left out. On each piece the fee is fixed, so assigning that fee produces a function.

Justification: Let \(b_0<b_1<\cdots<b_n\) be thresholds, and consider the intervals \([b_0,b_1],(b_1,b_2],\ldots,(b_{n-1},b_n]\). Every input in \([b_0,b_n]\) belongs to one of these intervals: it lies between some consecutive thresholds, and a threshold \(b_i\) belongs to the interval ending at \(b_i\). The intervals do not overlap, because each interval after the first excludes its left endpoint, which is the right endpoint of the preceding interval. Therefore, assigning one constant fee to each interval gives exactly one output for every input in the domain.

This justification also explains why endpoint choices cannot be treated as decoration. The schedule must specify which tier applies exactly at a threshold. Once that choice is made consistently, each input has an unambiguous fee.

Worked Examples with Different Tier Schedules

Worked Example: Shipping by Weight Band

A carrier charges 6 dollars for a package weighing more than 0 and at most 2 kilograms, 9 dollars for a package weighing more than 2 and at most 5 kilograms, and 14 dollars for a package weighing more than 5 and at most 10 kilograms. Let \(w\) be the package weight in kilograms and \(S(w)\) its charge in dollars. The schedule covers the contextual domain \(0<w\leq10\).

Translating each band directly gives:

S(w)= \begin{cases} 6, & 0<w\leq2,\\ 9, & 2<w\leq5,\\ 14, & 5<w\leq10. \end{cases}

For a 2-kilogram package, \(w=2\) satisfies \(0<w\leq2\), so \(S(2)=6\) dollars. For a 3-kilogram package, \(2<3\leq5\), so \(S(3)=9\) dollars. For a 5.4-kilogram package, \(5<5.4\leq10\), so \(S(5.4)=14\) dollars.

Each result can be checked against the verbal bands: 2 kilograms is included in “at most 2,” 3 kilograms is above 2 and at most 5, and 5.4 kilograms is above 5 and at most 10. A weight of 10 kilograms is covered by the last band; a weight above 10 kilograms is not covered by this schedule, so this rule does not specify its charge.

Worked Example: Laundry by Machine Time

A machine charges 1.25 dollars for a positive cycle lasting at most 20 minutes, 2 dollars for a cycle lasting more than 20 and at most 40 minutes, and 2.75 dollars for a cycle lasting more than 40 and at most 60 minutes. Let \(m\) be the cycle time in minutes, and let \(L(m)\) be the fee in dollars. The described domain is \(0<m\leq60\).

L(m)= \begin{cases} 1.25, & 0<m\leq20,\\ 2, & 20<m\leq40,\\ 2.75, & 40<m\leq60. \end{cases}

For \(m=20\), the first condition applies, so \(L(20)=1.25\) dollars. For \(m=20.5\), the first condition fails because \(20.5\not\leq20\), while \(20<20.5\leq40\), so \(L(20.5)=2\) dollars. For \(m=40\), the second condition applies and \(L(40)=2\) dollars. For \(m=55\), the last condition applies and \(L(55)=2.75\) dollars.

These evaluations check the boundary behavior as well as the interior of each tier: exactly 20 minutes remains in the first tier, exactly 40 minutes remains in the second, and a time just over 40 minutes enters the third.

A Common Boundary Mistake

A frequent error is to include a shared threshold in both neighboring intervals. For the parking schedule, a mistaken start might be:

P(t)= \begin{cases} 4, & 0<t\leq1,\\ 7, & 1\leq t\leq2. \end{cases}

At \(t=1\), both conditions hold: \(0<1\leq1\) and \(1\leq1\leq2\). The rule therefore assigns both 4 dollars and 7 dollars to the same input, which is not a function. The correction is to exclude the shared endpoint from one interval:

P(t)= \begin{cases} 4, & 0<t\leq1,\\ 7, & 1<t\leq2. \end{cases}

Now \(t=1\) belongs only to the first interval, while \(t=1.01\) belongs only to the second. The corrected rule matches “first hour or any part of it; more than one hour and up to two hours.”

Practical check: Test every threshold itself, and test an input just to either side when those inputs are in the domain. Confirm that the fee agrees with the exact wording of the schedule and that every covered input receives only one fee.

What a Step Function Says About the Charge

Within one tier, changing the input does not change the fee. For example, every shipping weight in \(2<w\leq5\) kilograms has the same charge of 9 dollars. The input varies across that interval, but the output remains constant. At the next threshold, the output changes to the next listed amount.

This makes a step function a suitable model for charges based on a category or a started unit: a parking hour, a shipping-weight band, or a machine-time bracket. It is not suitable for a rule where the charge increases proportionally at every moment or for every additional fraction of a kilogram. The fee schedule must support the constant-on-an-interval assumption.

Before using a tiered-fee rule, check three things:

Domain: Which inputs does the schedule cover? A weight schedule ending at 10 kilograms does not define a charge above 10 kilograms.
Endpoints: Does the wording say “up to,” “more than,” “at least,” or “less than”? Use those words to decide which interval includes each threshold.
Output: Is each value a fixed charge for its entire interval, and are its units clear? The input might be measured in hours while the output is measured in dollars.

A careful interval model does more than list prices. It makes the schedule usable for arbitrary inputs, including inputs exactly at thresholds, and makes clear what the model does and does not specify.

Check Your Understanding

For each question, use the interval conditions and the wording of the schedule to justify your answer.

A garage charges 5 dollars for up to 1 hour and 8 dollars for more than 1 hour and up to 2 hours. Write a piecewise rule for the fee when zero hours costs nothing.
Using the garage schedule in question 1, what fee applies at exactly 1 hour? What fee applies at 1.2 hours?
A delivery costs 7 dollars for \(0<w\leq3\) kilograms and 11 dollars for \(3<w\leq6\) kilograms. What is the charge for a 3-kilogram parcel, and why?
Why is the rule with intervals \(0<t\leq1\) and \(1\leq t\leq2\) ambiguous at \(t=1\)?
A fee schedule lists tiers only through 8 kilograms. What can you conclude about the fee for a 9-kilogram parcel from that schedule alone?

What You'll Learn

Why Tiered Fees Are Step Functions

From Fee Schedule to Piecewise Rule

Worked Example: Parking by the Started Hour

Why the Interval Conditions Work