Allocating Losses Through a Reinsurance Tower

How a Reinsurance Tower Allocates a Loss

A large covered loss can pass through several excess-of-loss contracts in sequence. The insurer retains an initial amount; a first reinsurer covers the next band of loss; and higher layers cover additional bands. To allocate the loss correctly, determine where each layer attaches, how much it can pay, and whether the loss reaches that layer.

The focus here is a tower of consecutive per-occurrence layers applied to the same gross covered loss. Each layer is assumed to respond to the same loss basis, with no gaps or overlapping coverage between layers. Assume each reinsurer pays its contractual recovery in full, and that there are no reinstatements, exclusions, or other adjustments. These assumptions isolate the arithmetic of allocation; actual treaty terms govern whether a payment is recoverable.

Definition: Let $X$ be the gross covered occurrence loss before reinsurance, $R$ the insurer’s initial retention, and $M_i$ the limit of layer $i$. A layer with attachment point $A_i$ and limit $M_i$ pays $$ C_i=\min\{\max(X-A_i,0),M_i\}. $$ For a consecutive tower, the first layer attaches at $A_1=R$, and each higher layer attaches immediately above the prior layer’s top. Thus $A_i=R+\sum_{j=1}^{i-1}M_j$. The insurer’s net loss is $N=X-\sum_i C_i$.

The attachment point is the amount of loss below the layer, not the layer’s limit. For example, a layer of 200,000 excess of 100,000 attaches at 100,000 and has a top of 300,000. It can pay at most 200,000. If the loss is 250,000, the layer pays 150,000: the loss has entered the layer by 150,000, but has not reached its top.

Calculating Recoveries and Exhaustion

For each layer, compare the gross loss with its attachment and top. A loss at or below attachment produces no recovery. A loss above attachment but below the top produces a partial recovery. A loss at or above the top exhausts the layer, so its recovery equals its full limit. At exactly the top, the layer is exhausted even though no loss has yet entered the next layer.

\text{Top of layer }i=A_i+M_i, \qquad C_i= \begin{cases} 0, & X\le A_i,\\ X-A_i, & A_i<X<A_i+M_i,\\ M_i, & X\ge A_i+M_i. \end{cases}

The cases express the same capped recovery formula in operational terms. A layer’s recovery cannot be negative and cannot exceed its limit. In a consecutive tower, a loss within a layer has already passed through every lower band, so all lower layers are exhausted. A loss below the tower leaves some lower layers untouched; a loss above the tower top exhausts every layer and leaves an excess above the tower for the insurer.

Worked Example: Allocating a Loss Across Three Layers

A hypothetical insurer has a retention of 100,000. Its tower consists of 200,000 excess of 100,000, 400,000 excess of 300,000, and 500,000 excess of 700,000. A covered occurrence causes a gross loss of 1,150,000. Calculate every recovery, identify the exhaustion status of each layer, and find the insurer’s net loss.

Method. Establish the bands first. The insurer retains the first 100,000. Layer 1 covers 100,000 to 300,000; layer 2 covers 300,000 to 700,000; and layer 3 covers 700,000 to 1,200,000.

Layer 1. Its recovery is $\min(1{,}150{,}000-100{,}000,200{,}000)=200{,}000$. It pays its full limit and is exhausted.

Layer 2. Its recovery is $\min(1{,}150{,}000-300{,}000,400{,}000)=400{,}000$. It also pays its full limit and is exhausted.

Layer 3. Its recovery is $\min(1{,}150{,}000-700{,}000,500{,}000)=450{,}000$. This is below the 500,000 limit, so the layer is partially used and has 50,000 of limit remaining.

Insurer’s net. The insurer retains the initial 100,000, with no loss above the tower top. Therefore $N=100{,}000$. The conservation check is $100{,}000+200{,}000+400{,}000+450{,}000=1{,}150{,}000$, which matches the gross loss. The key trap avoided is treating the full amount above a layer’s attachment as its recovery without capping it at the layer limit.

Why Gross Loss Equals Retained Loss Plus Recoveries

For any loss and any layer, the recovery formula allocates the portion of loss falling within that layer’s band, up to the band’s limit. The insurer’s net loss is whatever remains after those layer payments. This is a conservation identity, not an additional coverage assumption:

X=N+\sum_i C_i.

For a consecutive tower with total limit $T=\sum_i M_i$, all loss at or below $R$ remains with the insurer. Loss between $R$ and $R+T$ is allocated to the covered bands, while loss above $R+T$ is again retained by the insurer. Therefore the net loss can also be written as

N=\min(X,R)+\max(X-(R+T),0).

To see why, consider the three regions. If $X\le R$, no layer pays, and the formula gives $N=X$. If $R<X\le R+T$, the tower absorbs all loss above the retention, giving $N=R$. If $X>R+T$, all layer limits are used and the insurer retains both $R$ and the excess $X-(R+T)$. In each region, adding the recoveries back to net loss returns $X$.

This result is useful as a check, but it does not replace calculating each layer. A net amount alone will not tell you which reinsurer’s limit has been used or whether a particular layer is exhausted.

Worked Example: A Loss Extending Beyond the Tower

A hypothetical insurer retains 75,000. Its first layer is 125,000 excess of 75,000; its second is 300,000 excess of 200,000; and its third is 600,000 excess of 500,000. The gross occurrence loss is 1,060,000. Calculate each recovery and the insurer’s net loss, and state how much capacity remains in the top layer.

Method. The layer bands are 75,000–200,000, 200,000–500,000, and 500,000–1,100,000. The tower has total limit $125{,}000+300{,}000+600{,}000=1{,}025{,}000$, so its top is $75{,}000+1{,}025{,}000=1{,}100{,}000$.

First layer. The loss extends above its 200,000 top, so the recovery is its full 125,000 limit.

Second layer. The loss extends above its 500,000 top, so the recovery is its full 300,000 limit.

Third layer. Its recovery is $\min(1{,}060{,}000-500{,}000,600{,}000)=560{,}000$. The layer is not exhausted; remaining capacity is $600{,}000-560{,}000=40{,}000$.

Insurer’s net and check. The loss is 40,000 below the tower top, so only the initial 75,000 retention remains net. The allocation checks: $75{,}000+125{,}000+300{,}000+560{,}000=1{,}060{,}000$. Equivalently, ceded recovery is $125{,}000+300{,}000+560{,}000=985{,}000$, and $1{,}060{,}000-985{,}000=75{,}000$. Do not call the third layer exhausted merely because the gross loss is large; compare its recovery with its limit.

Reconstructing a Loss From the Allocation

The same conservation identity works in reverse. If the insurer’s retained amount and all layer recoveries are known, add them to recover the gross loss. The result must also be consistent with the stated attachments and limits. A sum that balances by itself is not enough if it places a recovery outside its layer’s possible range.

Worked Example: Inferring Gross Loss From Recoveries

A hypothetical tower has a 100,000 insurer retention, a 140,000 first layer excess of 100,000, a 260,000 second layer excess of 240,000, and a 500,000 third layer excess of 500,000. The reported recoveries are 140,000, 260,000, and 70,000, respectively, and the insurer’s net loss is 100,000. Find the gross loss and check whether the recovery pattern is feasible.

Method. Add the retained loss and all ceded amounts: $100{,}000+140{,}000+260{,}000+70{,}000=570{,}000$. The implied gross loss is 570,000.

Feasibility check. The first layer covers 100,000–240,000, and the gross loss of 570,000 is above its top; its full 140,000 recovery is consistent. The second covers 240,000–500,000, and is also fully exhausted at 260,000. The third begins at 500,000, so a 570,000 loss penetrates it by $570{,}000-500{,}000=70{,}000$, consistent with the reported recovery. Its limit is not exhausted.

Final check. The tower top is $500{,}000+500{,}000=1{,}000{,}000$; the loss is below it, so the insurer’s net is just the 100,000 retention. The full allocation is $100{,}000+140{,}000+260{,}000+70{,}000=570{,}000$. A common mistake is to add only the recoveries and report 470,000 as gross loss. That omits the insurer’s retained 100,000.

Common Allocation Traps

Using the wrong attachment. A layer’s attachment is the amount of loss below it. For a consecutive tower, add the retention and all lower-layer limits to find a higher layer’s attachment.
Forgetting the cap. The amount $X-A_i$ can exceed a layer’s limit. The recovery is capped at $M_i$, and any further loss enters a higher layer or remains net.
Confusing limit with top. A 300,000 limit excess of 200,000 has a top of 500,000, not 300,000.
Omitting the insurer’s share. Gross loss is not the sum of ceded recoveries alone. Add the net loss to the recoveries and check that the total equals $X$.
Calling a layer exhausted too early. A layer is exhausted only when its recovery equals its full limit. A partial recovery, however large in dollars, leaves capacity unused.

Key takeaway: For each layer, establish its attachment and top, calculate the recovery subject to the layer limit, and then check that insurer net plus all recoveries equals the gross loss. Exhaustion is determined by comparing recovery with limit.

Check Your Understanding

For each hypothetical occurrence, allocate the loss or infer the missing amount. Check both the layer terms and the total allocation.

A tower has a 50,000 retention, a 150,000 layer excess of 50,000, and a 300,000 layer excess of 200,000. Gross loss is 430,000. Find both recoveries, net loss, and unused capacity in the second layer. (Answer: Recoveries 150,000 and 230,000; net loss 50,000; second-layer capacity remaining 70,000.)
A tower has a 125,000 retention; layers of 275,000 excess of 125,000, 400,000 excess of 400,000, and 500,000 excess of 800,000. Gross loss is 950,000. State the three recoveries, net loss, and whether the third layer is exhausted. (Answer: Recoveries 275,000, 400,000, and 150,000; net loss 125,000; third layer is not exhausted and has 350,000 remaining.)
A hypothetical occurrence loss of 65,000 is subject to a 80,000 retention followed by layers of 120,000 excess of 80,000 and 200,000 excess of 200,000. Find the recoveries and insurer net. (Answer: Both recoveries are 0; insurer net is 65,000.)
A tower has a 60,000 retention, a 90,000 layer excess of 60,000, and a 180,000 layer excess of 150,000. The first layer pays its full limit and the second pays 45,000. Find gross loss and determine whether the second layer is exhausted. (Answer: Gross loss is 195,000; the second layer is not exhausted and has 135,000 of capacity remaining.)
A tower has a 100,000 retention and layers of 200,000 excess of 100,000, 300,000 excess of 300,000, and 400,000 excess of 600,000. Gross loss is 1,080,000. Calculate each recovery and the insurer’s net, and identify any exhausted layer. (Answer: Recoveries are 200,000, 300,000, and 400,000; insurer net is 180,000; all three layers are exhausted.)

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