Why Discrete Rating Actions Need Direct Comparison
A pricing decision is not always a choice among every possible price in an interval. A rating plan may permit only specified tiers, a regulator may limit changes to approved increments, or an implementation system may offer only a small set of actions. In these cases, the decision-maker chooses from a finite set of alternatives. The task is to evaluate each permitted alternative against the supplied objective and constraints, rather than assume that a continuous optimum can simply be rounded.
For a single segment, an action might specify a premium change and the corresponding forecast of retained policies and profit. For a portfolio, a decision consists of one action for each segment. A higher-priced action may increase expected profit per retained policy but reduce the number of policies retained. Constraints determine which trade-offs are acceptable; the objective determines which acceptable alternative is best.
Formulating the Choice
Let \(j\) index a segment and \(a\) index a permitted action for that segment. Suppose action \(a\) in segment \(j\) has expected profit \(\pi_{ja}\) and expected retained-policy count \(r_{ja}\). If exactly one action must be selected for each segment, a convenient decision indicator is \(x_{ja}\), equal to 1 when action \(a\) is selected and 0 otherwise. The “select exactly one” condition is:
Here \(A_j\) is the set of permitted actions for segment \(j\). If the objective is total expected profit, the objective is to maximize \(\sum_j\sum_{a\in A_j}\pi_{ja}x_{ja}\). A portfolio retention minimum \(R_{\min}\), for example, adds the condition \(\sum_j\sum_{a\in A_j}r_{ja}x_{ja}\ge R_{\min}\). A rate-change cap or premium floor can be represented by additional conditions using the relevant action values.
The values in an action table are forecasts, not guarantees about future policy counts or profit. A constraint written in terms of expected retention is checked against expected retention unless the problem explicitly defines another basis. Do not silently interpret an expected minimum as a guarantee that the realized count will meet the minimum.
Worked Examples
Worked Example: A Single Segment With a Retention Guardrail
A hypothetical insurer receives 300 quotes in one segment and can choose one of three premium actions. Its forecast assumes expected variable claims and expenses of $760 per retained policy under each action. The table shows the forecast number retained. The objective is to maximize expected contribution, subject to retaining at least 150 policies and limiting the premium increase to 10% above the current $1,000 premium. Which action should be selected?
Method. For each action, expected contribution equals expected retained policies multiplied by premium less expected variable claims and expenses per retained policy. The permitted actions are a 0%, 5%, or 10% increase.
| Action | Premium | Expected retained | Contribution per retained policy | Expected contribution |
|---|---|---|---|---|
| 0% increase | $1,000 | 180 | $1,000 − $760 = $240 | 180 × $240 = $43,200 |
| 5% increase | $1,050 | 160 | $1,050 − $760 = $290 | 160 × $290 = $46,400 |
| 10% increase | $1,100 | 135 | $1,100 − $760 = $340 | 135 × $340 = $45,900 |
The 10% action satisfies the rate-change cap, but it does not satisfy the retention minimum because \(135<150\). The 0% and 5% actions are feasible. Comparing their contributions, $46,400 is greater than $43,200, so select the 5% increase. It is feasible because its premium is below the $1,100 cap and its expected retained count is at least 150.
Common mistake avoided: Selecting the action with the highest contribution per retained policy would select the 10% increase, since $340 is the largest margin. That ignores both total expected contribution and the retention constraint. The 10% action is infeasible, and its total contribution is not the greatest even before applying that constraint.
Worked Example: Comparing Two Segment Actions
A hypothetical portfolio has two segments, C and D, and management must choose one of two actions in each. The figures below are forecast expected retained policies and expected profit, in thousands of dollars. The portfolio must retain at least 110 policies and no more than 120. Find the profit-maximizing feasible combination.
| Segment | Action | Expected retained | Expected profit ($000) |
|---|---|---|---|
| C | Higher-profit action | 62 | 18 |
| C | Lower-profit action | 70 | 15 |
| D | Higher-profit action | 48 | 13 |
| D | Lower-price action | 55 | 10 |
There are four combinations. Add the segment counts and profits for each, then check the combined retained count against both guardrails.
| Segment C choice | Segment D choice | Expected retained | Expected profit ($000) | Status |
|---|---|---|---|---|
| Higher-profit | Higher-profit | 62 + 48 = 110 | 18 + 13 = 31 | Feasible |
| Higher-profit | Lower-price | 62 + 55 = 117 | 18 + 10 = 28 | Feasible |
| Lower-profit | Higher-profit | 70 + 48 = 118 | 15 + 13 = 28 | Feasible |
| Lower-profit | Lower-price | 70 + 55 = 125 | 15 + 10 = 25 | Infeasible: exceeds 120 |
The first combination meets the minimum exactly: \(110\ge110\), and it does not exceed the maximum: \(110\le120\). Its $31,000 expected profit exceeds the $28,000 profits of the other feasible combinations. Therefore, choose the higher-profit action in both segments. The result is 110 expected retained policies and $31,000 expected profit.
Common mistake avoided: Choosing the higher-profit action for C and the lower-price action for D yields 117 retained policies and $28,000 profit. It is feasible, but it is not optimal: choosing the higher-profit action for D instead still meets the 110-policy minimum and raises profit to $31,000. More retention is not itself the objective when the stated objective is profit and the guardrails are already satisfied.
Worked Example: Exhaustive Comparison With a Tie-Break Rule
A hypothetical insurer can select one of two actions in each of three segments. Profits are in thousands of dollars, and retention is an expected policy count. The objective is to maximize total expected profit subject to expected retention of at least 145. If combinations tie on profit, management has specified that it will choose the one with greater expected retention. Find the selected combination.
| Segment | Action | Profit ($000) | Expected retained |
|---|---|---|---|
| A | Low | 12 | 70 |
| A | High | 16 | 58 |
| B | Low | 10 | 55 |
| B | High | 14 | 45 |
| C | Low | 8 | 40 |
| C | High | 11 | 32 |
There are \(2^3=8\) possible combinations. The table below gives each total. “H” means the high action and “L” means the low action.
| A, B, C actions | Expected retained | Profit ($000) | Status |
|---|---|---|---|
| L, L, L | 70 + 55 + 40 = 165 | 12 + 10 + 8 = 30 | Feasible |
| H, L, L | 58 + 55 + 40 = 153 | 16 + 10 + 8 = 34 | Feasible |
| L, H, L | 70 + 45 + 40 = 155 | 12 + 14 + 8 = 34 | Feasible |
| L, L, H | 70 + 55 + 32 = 157 | 12 + 10 + 11 = 33 | Feasible |
| H, H, L | 58 + 45 + 40 = 143 | 16 + 14 + 8 = 38 | Infeasible |
| H, L, H | 58 + 55 + 32 = 145 | 16 + 10 + 11 = 37 | Feasible |
| L, H, H | 70 + 45 + 32 = 147 | 12 + 14 + 11 = 37 | Feasible |
| H, H, H | 58 + 45 + 32 = 135 | 16 + 14 + 11 = 41 | Infeasible |
The largest feasible profit is $37,000, achieved by both H, L, H and L, H, H. Their expected retention totals are 145 and 147, respectively. The stated tie-break rule selects L, H, H: the low action in A, high in B, and high in C. It earns $37,000 and has expected retention of 147, which satisfies the 145 minimum.
Check. The two higher-profit combinations, H, H, L and H, H, H, are not competitors because their expected retentions, 143 and 135, are below the minimum. The selected combination is feasible and uses the tie-break rule only after the profit comparison.
Practical Checks and Common Traps
For a small number of actions, a clear table is often the safest calculation method. For a larger number, a spreadsheet or optimization routine can organize the comparisons, but the model still needs the correct action data, objective, and constraints. The method does not excuse a mismatch between a stated business rule and the mathematical test used to represent it.
- Check feasibility before ranking. An action with the highest objective value cannot be selected if it violates even one hard constraint.
- Do not substitute an unstated objective. If the objective is profit, a higher retention count is not automatically preferable. Apply retention only as specified in the objective or constraints.
- Combine segment values consistently. For additive expected profit and policy counts, sum the corresponding segment forecasts. Do not compare one segment’s per-policy margin with another combination’s total profit.
- Respect action definitions. A permitted tier is a complete action, including its associated forecasts and rate-change amount. Do not create an unapproved intermediate action by interpolation.
- State and apply tie-break rules. If multiple feasible actions have the same objective value, they are all optimal under that objective unless a secondary criterion is supplied.
A useful justification for summing expected profits is linearity of expectation. If portfolio profit is the sum of segment profits, then its expectation equals the sum of their expectations, whether or not segment outcomes are independent. Independence would matter for some measures of portfolio variability, but it is not required for this expected-profit calculation. This additivity does not mean that constraints can be ignored: each proposed combination must still be checked against the portfolio-level limits.
Check Your Understanding
For each problem, apply the stated objective and constraints to the permitted actions only.
- A hypothetical segment has three actions: A gives $9,600 profit and 88 expected retained policies; B gives $11,200 and 81; C gives $12,000 and 76. Maximize profit subject to retaining at least 80 policies. Which action is selected? (Answer: B, with $11,200 expected profit and 81 expected retained policies.)
- Two segments must each use one action. Segment X offers P (profit $8,000, retention 52) or Q ($10,000, retention 45). Segment Y offers R ($7,000, retention 43) or S ($9,000, retention 36). Maximize profit with total expected retention of at least 85. Identify every optimal combination. (Answer: P with S, or Q with R; each earns $17,000 and retains 88 expected policies.)
- Three segments offer these actions, with profit in thousands: A1 (14, 65 retained) or A2 (18, 53); B1 (10, 60) or B2 (13, 50); C1 (9, 45) or C2 (12, 36). Maximize profit subject to expected retention of at least 150. Which combination is optimal? (Answer: A2, B1, C1, with $37,000 profit and 158 expected retained policies.)
- Two segments have actions described by (profit in thousands, rate-index change, retention). Segment A offers low (12, −2, 70) and high (17, +4, 55); segment B offers low (8, −1, 50) and high (13, +3, 43). Maximize profit subject to a combined rate-index change no greater than +5. If profits tie, choose higher retention. Which combination is selected? (Answer: A low and B high; profit is $25,000, rate-index change is +1, and expected retention is 113.)