Price Floors And Ceilings

Why Price Floors and Ceilings Matter

An unconstrained pricing model identifies the price that best serves a chosen objective when any price in the model’s domain is available. In practice, the permissible price may be restricted. A hypothetical insurer might require a minimum premium for a policy, or impose a maximum price for a product. These requirements create a price floor and a price ceiling. The best permissible price can therefore differ from the unconstrained optimum.

The key is to distinguish the unconstrained answer from the feasible answer. A floor does not automatically mean “charge the floor,” and a ceiling does not automatically mean “charge the ceiling.” If the unconstrained optimum lies between the bounds, it remains optimal. If it lies outside the permitted interval, the relevant boundary may determine the best feasible price.

Definition: Let $p$ be the price decision and $J(p)$ the objective being maximized. A price floor $L$ and price ceiling $U$ impose the constraints $L \leq p \leq U$, where $L \leq U$. The feasible price interval is $[L,U]$. The constrained optimum is a price in that interval that gives the largest value of $J(p)$.

The objective matters: $J(p)$ could represent expected revenue or expected underwriting profit, for example. The bounds restrict the choices; they do not change what the objective measures. The earlier material on pricing objectives and demand curves provides the ingredients for building $J(p)$. Here the focus is on how to optimize it subject to lower and upper price bounds.

When a Bound Replaces the Unconstrained Optimum

Suppose $J(p)$ is differentiable and strictly concave on the relevant price range, with an unconstrained maximum at $p_0$. Strict concavity means that the objective’s slope decreases as price increases; it rules out a second, separated peak. If $p_0$ is feasible, it remains the constrained optimum. If it is below the floor, the objective is still decreasing over the feasible interval, so the floor is best. If it is above the ceiling, the objective is still increasing over the feasible interval, so the ceiling is best.

Under those conditions, the constrained answer is the unconstrained optimum limited to the feasible interval:

p^*=\min\!\bigl(U,\max(L,p_0)\bigr).

This expression is sometimes described as clipping $p_0$ to $[L,U]$. Its use depends on the stated conditions: there must be a single unconstrained maximum, and the objective must be concave across the relevant interval. Without those conditions, simply clipping an unconstrained local optimum need not find the best feasible price; it may be necessary to compare feasible candidates using the objective itself.

Result: For a differentiable, strictly concave objective with unique unconstrained maximum $p_0$, the constrained optimum is $L$ when $p_0<L$, $p_0$ when $L\leq p_0\leq U$, and $U$ when $p_0>U$. At equality, the unconstrained optimum is itself a boundary price.

Why the Boundary Rule Holds

The boundary rule follows from the direction of the objective’s slope. At an interior maximum $p_0$, the derivative is zero. Strict concavity makes the derivative decrease with price. Consequently, prices below $p_0$ have a positive derivative, and prices above $p_0$ have a negative derivative.

If $p_0<L$, every feasible price satisfies $p\geq L>p_0$. Thus $J'(p)<0$ throughout the feasible interval: raising price from the floor lowers the objective, so $L$ is best. If $p_0>U$, every feasible price satisfies $p\leq U<p_0$, and therefore $J'(p)>0$: raising price toward the ceiling improves the objective, so $U$ is best. If $L\leq p_0\leq U$, the unconstrained maximizer is already feasible and no feasible price can have a larger objective value. This establishes the rule.

Write the feasible interval.
Record the floor $L$ and ceiling $U$, and check that $L\leq U$.

Find the unconstrained optimum.
Use the objective function and its assumptions to calculate $p_0$.

Locate $p_0$ relative to the bounds.
If it is inside the interval, keep it; if it is outside, examine the corresponding boundary.

Verify the feasible choice.
Check the derivative direction or compare objective values, and state the resulting price and objective value.

Worked Applications

The following examples use hypothetical demand functions. In each case, the model’s demand is nonnegative over the prices considered, and the objective is expected value per period. The calculated price is a model result, not a statement about an actual insurer or product.

Worked Example: A Minimum Price Binds

A hypothetical policy has demand $q(p)=1200-10p$ policies per period when the price is $p$ dollars. Expected cost is a constant $c=\$40$ per policy. The insurer maximizes expected underwriting profit, $J(p)=(p-c)q(p)$, subject to a floor of $\$85$ and a ceiling of $\$110$. Find the constrained optimum and its profit.

Method. First derive the unconstrained profit maximum, then apply the feasible interval. This avoids the common mistake of reporting the unconstrained price without checking whether it is permitted.

Substitute the demand and cost:

$$ J(p)=(p-40)(1200-10p) =-10p^2+1600p-48000. $$

Differentiate and set the derivative equal to zero:

J'(p)=-20p+1600=0 \quad\Longrightarrow\quad p_0=80.

The second derivative is $J''(p)=-20<0$, so this is a strictly concave profit function with a unique unconstrained maximum. The permitted interval is $[85,110]$, and $p_0=80<85$. The floor replaces the unconstrained optimum, giving $p^*=\$85$.

At the selected price, demand is $q(85)=1200-10(85)=350$ policies. Profit per policy is $85-40=\$45$, so total expected profit is $350\times \$45=\$15{,}750$. As a check on the boundary choice, at the ceiling demand is $q(110)=100$, and profit is $100\times \$70=\$7{,}000$. Since the profit function is decreasing for every feasible price above $80$, the floor is the maximum over the entire permitted interval. The final price is $\$85.00$, and expected profit is $\$15{,}750.00$.

Worked Example: A Maximum Price Binds

For a hypothetical product, demand is $q(p)=720-6p$ policies per period and expected cost is $c=\$40$ per policy. The objective is expected underwriting profit. Prices must lie between $\$45$ and $\$75$. Find the constrained price and verify that the ceiling is binding.

Method. Use the derivative to determine the unconstrained optimum, then check the objective’s direction throughout the feasible interval. The trap is to treat a ceiling as the answer without first establishing that higher feasible prices improve the objective.

The profit function and its derivative are:

J(p)=(p-40)(720-6p)=-6p^2+960p-28800, \qquad J'(p)=960-12p.

Setting $J'(p)=0$ gives $960-12p=0$, so $p_0=80$. Also, $J''(p)=-12<0$, confirming a unique maximum. Since $80>75$, the unconstrained optimum is above the feasible interval. For every $p$ in $[45,75]$, $J'(p)=960-12p\geq 960-12(75)=60>0$. Profit is increasing throughout that interval, so the maximum occurs at the ceiling: $p^*=\$75$.

At $p=75$, demand is $q(75)=720-6(75)=270$, and profit per policy is $75-40=\$35$. Expected profit is $270\times \$35=\$9{,}450$. At the floor, demand is $q(45)=450$, the margin is $\$5$, and profit is $450\times \$5=\$2{,}250$, which is lower. The final price is $\$75.00$, and expected profit is $\$9{,}450.00$.

Worked Example: The Unconstrained Optimum Remains Feasible

A hypothetical policy has demand $q(p)=500-5p$, with constant expected cost $c=\$30$ per policy. The insurer maximizes expected underwriting profit, subject to a floor of $\$60$ and a ceiling of $\$70$. Determine the optimal price and show why neither bound replaces it.

Method. Calculate the unconstrained maximum, check that it falls inside the permitted interval, and evaluate the objective there. Comparing both endpoints also provides a numerical check.

The profit function is:

$$ J(p)=(p-30)(500-5p)=-5p^2+650p-15000. $$

Its derivative is $J'(p)=-10p+650$, so the stationary point satisfies $-10p+650=0$, giving $p_0=65$. Since $J''(p)=-10<0$, this is the unique unconstrained maximum. The value $\$65$ lies in $[60,70]$, so it is also the constrained optimum.

At $\$65$, demand is $500-5(65)=175$, and the margin is $\$65-\$30=\$35$. Thus profit is $175\times \$35=\$6{,}125$. At the floor, demand is $500-5(60)=200$, the margin is $\$30$, and profit is $\$6{,}000$. At the ceiling, demand is $500-5(70)=150$, the margin is $\$40$, and profit is also $\$6{,}000$. Both endpoint values are below the value at $\$65$, confirming the result. The final price is $\$65.00$, and expected profit is $\$6{,}125.00$.

Common Errors and Practical Checks

A frequent error is to calculate an unconstrained optimum correctly and stop there. The result is incomplete until it has been checked against both bounds. Another is to assume that the presence of a floor means the floor must bind, even when a feasible interior optimum produces a higher objective.

For example, in the first worked example, reporting $\$80$ as the price would be wrong because it violates the $\$85$ floor. The correction is not to substitute $\$85$ without checking the objective: verify that $p_0=\$80$ lies below the interval, and that profit decreases for feasible prices above it. Those checks establish that $\$85$ is best. Similarly, in the ceiling example, the ceiling is best because the objective is increasing throughout the feasible interval—not merely because a ceiling exists.

Check feasibility. A price outside $[L,U]$ cannot be the constrained answer, however attractive its unconstrained objective value.
Check the model’s shape. The clipping rule requires the stated concavity and unique-maximum conditions. For a nonconcave objective, compare feasible candidates rather than assuming the rule applies.
Check the objective value. A price is not an objective value. Report both the selected price and the resulting revenue or profit, with units.
Check the interval itself. If $L>U$, there is no feasible price satisfying both bounds; the model’s requirements must be reconciled before optimization.

Floors and ceilings are simple constraints, but they can change the chosen price and the objective achieved. A disciplined solution identifies the objective, calculates the unconstrained optimum under its assumptions, checks feasibility, and verifies any binding boundary. This approach is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each hypothetical pricing problem, identify the unconstrained optimum, apply the bounds, and compute the requested objective value.

A product has demand $q(p)=600-4p$, and the objective is revenue $J(p)=p q(p)$. The feasible interval is $[80,120]$. What price maximizes revenue, and what is the revenue at that price? (Answer: $p^*=\$80$; revenue $=\$22{,}400.00$.)
A policy has demand $q(p)=1000-10p$, constant cost $c=\$30$ per policy, and expected-profit objective $J(p)=(p-c)q(p)$. The feasible interval is $[55,85]$. Find the optimal price and expected profit. (Answer: $p^*=\$65$; expected profit $=\$12{,}250.00$.)
A product has demand $q(p)=720-6p$, cost $c=\$40$ per policy, and objective $J(p)=(p-c)q(p)$. The floor is $\$45$ and the ceiling is $\$75$. Which bound, if either, determines the optimum, and what is the profit there? (Answer: the ceiling binds; $p^*=\$75$; expected profit $=\$9{,}450.00$.)
An objective is $J(p)=9000-3(p-48)^2$, with $40\leq p\leq55$. Find the unconstrained maximum and the constrained maximum, including the objective value. (Answer: $p_0=\$48$ is feasible, so $p^*=\$48$ and $J(p^*)=9000$.)
A product has demand $q(p)=900-9p$, and the objective is revenue $J(p)=p q(p)$. Prices must satisfy $40\leq p\leq60$. Find the constrained optimum and revenue, showing whether a bound binds. (Answer: the unconstrained optimum is feasible, so $p^*=\$50$; revenue $=\$22{,}500.00$.)

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