Minimum Premium And Profit Constraints

Portfolio Targets as Price Constraints

An optimization objective identifies what a pricing decision is intended to improve; a constraint specifies what the decision must satisfy. A portfolio may, for example, be required to produce at least a stated amount of premium or underwriting profit. These are portfolio-level requirements: they depend on both the price offered and the expected number of policies that respond to that price.

Let $p$ be a uniform premium per policy and let $Q(p)$ be the expected number of policies sold at that premium. The expected premium volume is $R(p)=pQ(p)$. It is not generally enough to check whether $p$ exceeds a premium floor: a higher price can reduce expected demand enough to lower total premium volume. Likewise, expected profit depends on the margin per policy as well as on the number of policies sold.

Definition: A minimum premium-volume constraint requires $pQ(p)\ge R_{\min}$, where $R_{\min}$ is a specified portfolio premium target. A minimum-profit constraint requires expected portfolio underwriting profit to be at least a specified target $\Pi_{\min}$. A price is feasible only if it satisfies every applicable constraint and any stated price bounds.

Throughout the examples, demand functions and costs describe hypothetical portfolios. Expected policy counts may be noninteger because they are model expectations, not guaranteed realized counts. Assume premium is collected per policy and that the stated demand function applies throughout its specified price domain. These calculations describe expected outcomes; actual outcomes can differ.

How a Premium-Volume Requirement Restricts Price

For a specified price, expected premium volume is the premium per policy multiplied by expected policies sold. Under a linear demand model $Q(p)=a-bp$, where $a>0$, $b>0$, and $0\le p\le a/b$, this volume is $R(p)=p(a-bp)$. It first increases and then decreases: raising the price increases premium per policy but reduces expected demand.

p(a-bp)\ge R_{\min} \quad\Longleftrightarrow\quad bp^2-ap+R_{\min}\le 0

The expression $p(a-bp)$ is a concave quadratic. Its maximum occurs at $p=a/(2b)$, with maximum value $a^2/(4b)$. Therefore, if $R_{\min}>a^2/(4b)$, no price can satisfy the target under this demand model. If the target is no greater than the maximum, solving the quadratic gives a candidate interval between its two roots. That interval must still be intersected with the model’s valid price domain and any other price restrictions.

p_{\pm}=\frac{a\pm\sqrt{a^2-4bR_{\min}}}{2b}

Worked Example: Finding the Premium-Volume Feasible Range

A hypothetical portfolio has expected demand $Q(p)=240-p$ policies for $0\le p\le240$, with $p$ measured in dollars per policy. The portfolio must produce at least $\$12{,}000$ in expected premium volume. Find the feasible price range.

Method. Write expected premium volume as $R(p)=pQ(p)$, impose the minimum, and solve the resulting quadratic inequality.

First, $R(p)=p(240-p)$. The constraint is $p(240-p)\ge12{,}000$, or $p^2-240p+12{,}000\le0$. The discriminant is $240^2-4(12{,}000)=57{,}600-48{,}000=9{,}600$, and $\sqrt{9{,}600}\approx97.9796$.

p_{\pm}=\frac{240\pm\sqrt{9{,}600}}{2} \approx 71.0102,\;168.9898

Because the quadratic $p^2-240p+12{,}000$ opens upward, it is nonpositive between its roots. Thus, with continuously selectable prices, the feasible range is approximately $\$71.01$ to $\$168.99$, inclusive at the exact roots. The maximum premium volume occurs at $p=120$: $120(240-120)=\$14{,}400$, which exceeds the required $\$12{,}000$. As a check within the feasible range, $p=100$ gives $Q=140$ and premium volume $100(140)=\$14{,}000$.

Common mistake avoided. Reporting $\$140$ as the lower boundary is incorrect: it is not a root of the quadratic. At $p=140$, premium volume is $140(100)=\$14{,}000$, so this price is feasible, but prices between the actual lower root, approximately $\$71.01$, and $\$140$ are also feasible. If prices must be set to whole cents, check the rounded endpoints against the constraint: $\$71.01$ gives $71.01(168.99)=\$11{,}999.9799$, just below the target, so the first feasible cent price is $\$71.02$; by symmetry the last is $\$168.98$. The continuous boundary values above are rounded to two decimals for reporting.

Minimum Expected Underwriting Profit

A minimum-profit constraint must specify what is counted as profit. In the simple model used here, each policy has expected variable cost $c$, including the costs explicitly stated in the problem, and the portfolio has fixed expense $F$. Expected underwriting profit is $\Pi(p)=(p-c)Q(p)-F$. If costs vary with policy or price, the relevant expected cost function must replace this simplified expression. In particular, premium volume is not profit: premium must cover expected costs before it contributes to underwriting profit.

With linear demand $Q(p)=a-bp$, this profit function is also a concave quadratic when $b>0$. For a target $\Pi_{\min}$, solve $(p-c)(a-bp)-F\ge\Pi_{\min}$. If the model’s maximum profit is below the target, the feasible set is empty. Otherwise, the quadratic inequality typically defines the prices between two roots, subject to the demand domain and any requirement that the price cover variable cost.

Worked Example: Constraining Price by Minimum Profit

A hypothetical portfolio has demand $Q(p)=200-p$ for $0\le p\le200$. Its expected variable cost is $\$50$ per policy, with no additional fixed expense. The required expected underwriting profit is at least $\$5{,}000$. Find the feasible prices.

Method. Calculate expected contribution per policy, multiply by expected policies, and solve the target inequality.

The expected profit is $(p-50)(200-p)$. The constraint becomes $(p-50)(200-p)\ge5{,}000$. Expanding gives $-p^2+250p-10{,}000\ge5{,}000$, or $p^2-250p+15{,}000\le0$. The discriminant is $250^2-4(15{,}000)=62{,}500-60{,}000=2{,}500$, whose square root is $50$. The roots are $(250-50)/2=100$ and $(250+50)/2=150$.

p^2-250p+15{,}000\le0 \quad\Longrightarrow\quad 100\le p\le150

The quadratic opens upward, so the inequality holds between the roots. Both roots lie in the demand domain, and both exceed the $\$50$ variable cost. At either endpoint, expected profit is exactly $\$5{,}000$: at $p=100$, $(100-50)(200-100)=\$5{,}000$; at $p=150$, $(150-50)(200-150)=\$5{,}000$. At the midpoint, $p=125$, demand is 75 and profit is $75(75)=\$5{,}625$, confirming that the target is attainable.

Common mistake avoided. Do not multiply the premium $p$ by demand and call the result profit. At $p=100$, premium volume is $\$10{,}000$, but profit is only $\$5{,}000$ after the $\$50$ expected variable cost per policy.

Why the Feasible Prices Fall Between the Roots

The interval result follows from the shape of the quadratic, not from a special pricing rule. For a quadratic $g(p)=A p^2+Bp+C$ with $A>0$ and two real roots $r_1\le r_2$, it factors as $g(p)=A(p-r_1)(p-r_2)$. Between the roots, one factor is nonpositive and the other is nonnegative, so their product is nonpositive. Outside the roots, the factors have the same sign and the product is positive. Therefore, an inequality $g(p)\le0$ holds between the roots. A profit constraint can be rearranged into this form; a premium-volume constraint under linear demand can too.

This reasoning also explains two important checks. First, real roots require a nonnegative discriminant. Second, real roots alone do not guarantee feasibility: the target may exceed the quadratic’s maximum, or the roots may fall outside the permitted price domain. Always check the target against the maximum and then intersect the algebraic interval with all model and business bounds.

Define the portfolio outcome.
For premium volume, use $pQ(p)$. For profit, state the per-policy expected cost and any fixed expenses, then construct expected profit.

Impose the target.
Write the requirement as an inequality, such as $pQ(p)\ge R_{\min}$ or $\Pi(p)\ge\Pi_{\min}$.

Solve and test feasibility.
Find the roots or determine that the target exceeds the attainable maximum. Check the price domain and verify boundary values in the original outcome formula.

Intersect constraints.
The final feasible set contains only prices satisfying every requirement simultaneously, including applicable price floors, ceilings, or retention guardrails.

Combining Portfolio Requirements

A minimum premium-volume requirement and a minimum-profit requirement may restrict prices in different ways. Each constraint creates its own feasible set; the portfolio’s feasible set is their intersection. One constraint can eliminate prices that are acceptable under the other, and a combined set can be empty even when each requirement is individually attainable. This is why each requirement should be solved and checked separately before combining the results.

Worked Example: Intersecting Premium and Profit Constraints

A hypothetical portfolio has demand $Q(p)=300-2p$ for $0\le p\le150$. Expected variable cost is $\$40$ per policy, and fixed expense is $\$1{,}000$. The portfolio must achieve at least $\$2{,}000$ in expected underwriting profit and at least $\$10{,}000$ in premium volume. Find the prices satisfying both requirements.

Method. Solve the profit inequality and the premium-volume inequality separately, then take their intersection.

Profit is $\Pi(p)=(p-40)(300-2p)-1{,}000$. The requirement $\Pi(p)\ge2{,}000$ gives $-2p^2+380p-13{,}000\ge2{,}000$, or $p^2-190p+7{,}500\le0$. The discriminant is $190^2-4(7{,}500)=36{,}100-30{,}000=6{,}100$, and $\sqrt{6{,}100}\approx78.1025$. Thus the profit-feasible interval is $(190-78.1025)/2\le p\le(190+78.1025)/2$, or approximately $[55.9488,134.0512]$.

Premium volume is $p(300-2p)$. Requiring it to be at least $10{,}000$ gives $-2p^2+300p-10{,}000\ge0$, equivalently $p^2-150p+5{,}000\le0$. The roots are $(150\pm\sqrt{22{,}500-20{,}000})/2=(150\pm50)/2$, so the premium-feasible interval is $[50,100]$. Intersecting the two intervals gives the combined continuous feasible range $[55.9488,100]$, approximately $\$55.95$ to $\$100.00$.

Check the binding endpoints. At the exact lower profit root, expected profit is $\$2{,}000$, while premium volume exceeds its minimum because the lower root is above $\$50$ and below $\$100$. At $p=100$, demand is 100, so premium volume is $\$10{,}000$, and profit is $(100-40)(100)-1{,}000=\$5{,}000$. Thus the upper endpoint satisfies both conditions. If prices are restricted to cents, $p=\$55.95$ gives demand $188.1$ and profit $15.95(188.1)-1{,}000=\$2{,}000.195$, so it meets the profit target; $\$100.00$ meets the premium target exactly.

Common mistake avoided. Do not report the union of the two intervals. A price must meet both portfolio requirements, so the answer is the overlap, not every price that satisfies at least one of them.

Practical Interpretation and Exam Checks

These constraints make operational requirements explicit without changing the meaning of the objective being optimized. For example, a chosen objective may favor a price with strong expected profit, while a minimum premium-volume constraint rules out prices that sell too few policies or produce too little total premium. The constraint does not itself say which feasible price is best; it removes choices that fail the requirement.

A threshold is meaningful only in relation to the model’s attainable outcomes. For linear demand, maximum premium volume occurs at $a/(2b)$. For profit $(p-c)(a-bp)-F$, the quadratic’s vertex gives the unconstrained price of maximum modeled profit, provided that price lies in the valid domain. If the required target is above that maximum, there is no feasible price under the assumed model. Changing the demand assumption, costs, or target may change the answer, but the inequality must always reflect the assumptions actually supplied.

State the units. Distinguish premium per policy from total portfolio premium, and state whether fixed expenses are included in profit.
Respect the demand domain. A mathematically calculated root outside the stated range is not automatically a valid price choice.
Check maxima. A target above the maximum modeled outcome has no solution, even if a quadratic formula can be written down.
Verify boundaries. Substitute candidate endpoints into the original revenue or profit expression, not only the rearranged polynomial.
Handle rounding deliberately. A rounded boundary can fail a threshold by a small amount. If prices have fixed increments, test the nearest permitted prices directly.

Key takeaway: A minimum premium-volume or profit requirement is a constraint on the portfolio outcome at a price, not a simple restriction on the premium charged. Derive the outcome using expected demand and costs, solve its inequality, check attainability and domain, and intersect the result with every other applicable requirement. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each hypothetical portfolio, formulate the applicable inequality and determine the requested feasible price set. Treat prices as continuous unless a question says otherwise.

Demand is $Q(p)=180-p$ for $0\le p\le180$. What prices produce at least $\$8{,}000$ in expected premium volume? (Answer: $80\le p\le100$.)
Demand is $Q(p)=160-p$, expected variable cost is $\$40$ per policy, and fixed expense is zero. What prices produce at least $\$2{,}400$ in expected profit? (Answer: approximately $\$65.36\le p\le\$134.64$.)
Demand is $Q(p)=250-2p$ for $0\le p\le125$. What prices produce at least $\$7{,}200$ in premium volume? (Answer: $45\le p\le80$.)
Demand is $Q(p)=120-p$, variable cost is $\$30$ per policy, and fixed expense is zero. Is a minimum expected profit of $\$2{,}100$ attainable? (Answer: No; maximum profit is $(75-30)(120-75)=\$2{,}025$, below the target.)
Demand is $Q(p)=220-p$, variable cost is $\$40$ per policy, and fixed expense is $\$500$. What prices satisfy both a minimum expected profit of $\$4{,}000$ and a minimum premium volume of $\$10{,}000$? (Answer: approximately $\$70.00\le p\le\$155.83$.)

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