April 2012 Q8 Part iv

[StartedThinking]

I am unable to understand the contribution to losses in the ASET (second table) Actual loss to layer 1 etc.

Especially I did not understand,
For year 468, unlimited loss is 15 Mn and actual loss is 10 Mn because already we have seen that the loss to layer 1 is 15 Mn after considering the layer limits.
Why again did we reduce the 15 Mn to 10 Mn. Can someone help me understanding this part iv of the question?

Thanks in advance.


PS: I have seen earlier doubt about this question, but it is about the typo error in revision notes.

 

[Darren Michaels]

I am unable to understand the contribution to losses in the ASET (second table) Actual loss to layer 1 etc.

Especially I did not understand,
For year 468, unlimited loss is 15 Mn and actual loss is 10 Mn because already we have seen that the loss to layer 1 is 15 Mn after considering the layer limits.
Why again did we reduce the 15 Mn to 10 Mn. Can someone help me understanding this part iv of the question?

Thanks in advance.


PS: I have seen earlier doubt about this question, but it is about the typo error in revision notes.

Hi there
The first table shows what the recoveries from the layers would have been assuming unlimited free reinstatements. The second table shows what the recoveries actually are allowing for the fact that there is only one free reinstatement on each layer.

Also bear in mind that the 2 layers are alternatives to each other (ie you buy one or the other) and are not part of a program.

So for year 468, the first three losses would all recover 5m from layer 1 (hence the unlimited loss is 15m) but as there is only one reinstatement the maximum recoverable is 2x5m=10m which is why the actual loss is only 10m.

For layer 2 the recoveries ignoring reinstatements are 5m, 20m, 20m and 0m, giving an unlimited loss of 45m. But with one free reinstatement the maximum recoverable is 2x20m=40m which is why the actual loss is only 40m.

 

[StartedThinking]

Hi there
The first table shows what the recoveries from the layers would have been assuming unlimited free reinstatements. The second table shows what the recoveries actually are allowing for the fact that there is only one free reinstatement on each layer.

Also bear in mind that the 2 layers are alternatives to each other (ie you buy one or the other) and are not part of a program.

So for year 468, the first three losses would all recover 5m from layer 1 (hence the unlimited loss is 15m) but as there is only one reinstatement the maximum recoverable is 2x5m=10m which is why the actual loss is only 10m.

For layer 2 the recoveries ignoring reinstatements are 5m, 20m, 20m and 0m, giving an unlimited loss of 45m. But with one free reinstatement the maximum recoverable is 2x20m=40m which is why the actual loss is only 40m.

Thank you and it is clear

 

[Jack Payne]

For this question, is it acceptable to assume that reinstatements are made when part of the cover is burned through? So for Layer 2 and year 468 losses will be 5m for first event, cover reinstated, then a further 20m for the next event but no other losses as cover has already been reinstated. Giving a total loss cost of 25m for year 468 for layer 2.

The answer goes on to mention that this is what happens in practice in part vi (from my interpretation).

 

[Ian Senator]

See Darren's post above for the calculations. Reinstatements are made when part of the cover is burned through, yes.