Calculating Surplus Share Ceded Amounts

How a Surplus Share Cession Is Calculated

A surplus share treaty divides the amount insured on an eligible risk between the ceding insurer and the reinsurer. The insurer first keeps an agreed amount, called its retention. The treaty can then accept some or all of the amount above that retention, subject to its capacity. Calculating the cession therefore requires two separate checks: how much of the risk is above the retention, and how much of that surplus the treaty can accept.

This tutorial concerns the allocation of a policy amount, such as a sum insured. It does not calculate a premium or allocate a claim payment. Those amounts may be allocated in proportion to the cession, but their calculation is a separate step. Unless a problem states otherwise, assume the risk is eligible for the treaty, the stated retention applies to that risk, and no other reinsurance or placement changes the allocation.

Definition: Let $B$ be the amount insured on one eligible policy, $R$ the insurer’s retention per risk, and $n$ the treaty’s capacity in lines. One line is equal to the retention. The maximum amount ceded under the treaty is $nR$. The ceded amount is the part of the policy amount above $R$, limited to $nR$. The retained amount is the part of $B$ not ceded.

Calculate in Two Stages

First determine the surplus above the retention. If $B$ is no greater than $R$, there is no surplus to cede. Otherwise, the surplus is $B-R$. Then compare that surplus with the treaty’s maximum ceded amount, $nR$. The smaller of these two amounts is ceded.

C=\min\bigl(\max(B-R,0),\,nR\bigr), \qquad K=B-C

Here, $C$ is the ceded amount and $K$ is the retained amount. The inner maximum prevents a negative surplus when the policy amount is below the retention. The outer minimum prevents the ceded amount from exceeding treaty capacity. Since $K=B-C$, any amount not ceded remains with the insurer, including any amount above the treaty’s capacity.

For a treaty with $n$ lines, the reinsurer’s capacity is $nR$, not $(n-1)R$. The retention is the insurer’s first line; the treaty capacity counts the additional lines available to the reinsurer. Thus, if the treaty has four lines and the retention is $100,000, the treaty can cede up to $400,000. The total amount that can be accommodated by the retention plus the treaty is $500,000.

Why the Formula Works

The ceded amount must satisfy three restrictions. It cannot be negative, it cannot exceed the surplus above the insurer’s retention, and it cannot exceed the treaty capacity. In inequalities, these requirements are

0\le C,\qquad C\le \max(B-R,0),\qquad C\le nR.

The greatest amount satisfying all three restrictions is the smaller of the surplus and the capacity. That gives $C=\min(\max(B-R,0),nR)$. The retained amount is whatever completes the allocation back to the full policy amount, so $K=B-C$. This also establishes the reconciliation $K+C=B$.

The formula can be read in three cases:

Policy amount	Ceded amount	Retained amount
$B\le R$	0	$B$
$R<B\le R+nR$	$B-R$	$R$
$B>R+nR$	$nR$	$B-nR$

The final case is important: the insurer retains both its original retention and the part of the risk that exceeds the treaty’s capacity. Unless the problem specifies an additional placement, do not assume that this excess is ceded elsewhere.

Worked Example: Policies Below and Above the Retention

A hypothetical insurer has a surplus share treaty with a retention of $100,000 per risk and a capacity of four lines. Calculate the retained and ceded amounts for eligible policies with amounts insured of $70,000, $250,000, and $370,000. Reconcile the allocation for each policy.

Method. First calculate the treaty’s maximum cession: $nR=4(\$100{,}000)=\$400{,}000$. Apply the retention to each policy, then cap the cession at $400,000.

Policy 1: $B=\$70{,}000$. The amount does not exceed the retention, so the surplus is $\max(\$70{,}000-\$100{,}000,0)=\$0$. The cession is $C=\min(\$0,\$400{,}000)=\$0$, and the retained amount is $K=\$70{,}000-\$0=\$70{,}000$. Check: $70,000 + $0 = $70,000.

Policy 2: $B=\$250{,}000$. The surplus is $\$250{,}000-\$100{,}000=\$150{,}000$. The cession is $C=\min(\$150{,}000,\$400{,}000)=\$150{,}000$, and the retained amount is $K=\$250{,}000-\$150{,}000=\$100{,}000$. Check: $100,000 + $150,000 = $250,000.

Policy 3: $B=\$370{,}000$. The surplus is $\$370{,}000-\$100{,}000=\$270{,}000$, which is within capacity. Thus $C=\min(\$270{,}000,\$400{,}000)=\$270{,}000$, and $K=\$370{,}000-\$270{,}000=\$100{,}000$. Check: $100,000 + $270,000 = $370,000.

The allocation is $70,000 retained and nothing ceded on the first policy; $100,000 retained and $150,000 ceded on the second; and $100,000 retained and $270,000 ceded on the third. The common mistake avoided here is treating every policy as if it has a positive surplus: a policy below the retention has no cession.

Worked Example: Applying the Capacity Limit

A hypothetical insurer retains $80,000 per risk under a treaty with a capacity of three lines. A covered policy has an amount insured of $420,000. Find the amount ceded and the amount retained, and explain why the retention is not the final retained amount.

Method. Convert the line capacity to dollars before allocating the risk. The maximum cession is $3(\$80{,}000)=\$240{,}000$.

The surplus above the retention is $\$420{,}000-\$80{,}000=\$340{,}000$. The treaty cannot accept all of that surplus because $340,000 exceeds its $240,000 capacity. Therefore,

C=\min(\$340{,}000,\$240{,}000)=\$240{,}000.

The retained amount is $K=\$420{,}000-\$240{,}000=\$180{,}000$. This is $80,000 of original retention plus $100,000 of surplus beyond the treaty capacity. Check: $180,000 + $240,000 = $420,000.

The common mistake avoided is setting the retained amount equal to the retention even when the surplus exceeds treaty capacity. The insurer retains the excess as well, unless some other arrangement is stated.

Portfolio Calculations

For a portfolio, calculate each policy’s cession using the treaty terms for that risk, then add the individual amounts. A policy below the retention can have a different allocation from a policy above the retention or above capacity. Applying the formula to the portfolio total instead can conceal those differences and produce an incorrect result.

Worked Example: Reconcile a Small Portfolio

A hypothetical insurer has a retention of $120,000 per risk and a three-line surplus share capacity. Four eligible policy amounts are $90,000, $300,000, $510,000, and $620,000. Calculate the cession and retention for each policy, then the portfolio totals.

Method. The maximum cession per risk is $3(\$120{,}000)=\$360{,}000$. Apply the surplus and capacity limits separately to each policy.

For the $90,000 policy, the amount is below retention. Its surplus is $0, its cession is $0, and its retained amount is $90,000.

For the $300,000 policy, the surplus is $\$300{,}000-\$120{,}000=\$180{,}000$. This is within capacity, so the cession is $180,000 and the retained amount is $\$300{,}000-\$180{,}000=\$120{,}000$.

For the $510,000 policy, the surplus is $\$510{,}000-\$120{,}000=\$390{,}000$. Capacity limits the cession to $360,000, leaving $\$510{,}000-\$360{,}000=\$150{,}000$ retained.

For the $620,000 policy, the surplus is $\$620{,}000-\$120{,}000=\$500{,}000$. Again, the cession is limited to $360,000, and the retained amount is $\$620{,}000-\$360{,}000=\$260{,}000$.

The total ceded amount is $\$0+\$180{,}000+\$360{,}000+\$360{,}000=\$900{,}000$. The total retained amount is $\$90{,}000+\$120{,}000+\$150{,}000+\$260{,}000=\$620{,}000$. The total policy amount is $\$90{,}000+\$300{,}000+\$510{,}000+\$620{,}000=\$1{,}520{,}000$. Check: $900,000 + $620,000 = $1,520,000.

The method avoids treating the portfolio as one large risk. Capacity applies per risk here, so each policy receives its own capacity limit before the results are totaled.

Common Exam Traps

Counting the retention as a ceded line. If a treaty has $n$ lines of capacity, its maximum cession is $nR$. The insurer’s retention is additional to those ceded lines.
Ignoring capacity. Ceding all of $B-R$ is correct only when that surplus does not exceed $nR$. When it does, the cession stops at $nR$.
Retaining only $R$ in every case. That is correct only if the treaty can accept the entire surplus. Amounts beyond treaty capacity remain retained under the stated arrangement.
Using the portfolio total as one risk. When the retention and capacity apply per risk, calculate each policy separately before summing.
Failing to reconcile. For every policy, confirm that retained plus ceded equals the original policy amount. A failure to reconcile signals an arithmetic or allocation error.

Key takeaway: Calculate the surplus above the retention, cap the cession at the treaty’s line capacity, and retain everything not ceded. For each risk, verify $K+C=B$.

Check Your Understanding

For each problem, assume the stated treaty applies to the eligible risk and that no other placement is specified. Calculate the requested cession and retention.

A treaty has a $80,000 retention and two lines of capacity. A policy amount is $210,000. What are the ceded and retained amounts? (Answer: $130,000 ceded; $80,000 retained.)
A treaty has a $150,000 retention and three lines of capacity. A policy amount is $700,000. What are the ceded and retained amounts? (Answer: $450,000 ceded; $250,000 retained.)
A treaty has a $90,000 retention and five lines of capacity. A policy amount is $420,000. What are the ceded and retained amounts? (Answer: $330,000 ceded; $90,000 retained.)
A treaty has a $65,000 retention and four lines of capacity. A policy amount is $55,000. What are the ceded and retained amounts? (Answer: $0 ceded; $55,000 retained.)
A treaty has a $200,000 retention and two lines of capacity. Three eligible policy amounts are $140,000, $370,000, and $780,000. What are the portfolio totals ceded and retained? (Answer: $570,000 ceded; $720,000 retained.)

These allocation calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Policy amount	Ceded amount	Retained amount
\(B\le R\)	0	\(B\)
\(R<B\le R+nR\)	\(B-R\)	\(R\)
\(B>R+nR\)	\(nR\)	\(B-nR\)

What You'll Learn