FlyerTalk Forums - View Single Post - Combining multiple small residual funds given 2 Ticketless Travel funds limit
Old Dec 1, 2007 | 10:52 am
  #1  
pshuang
 
Join Date: Nov 1999
Location: SFO
Programs: UA 1.050MM, PersonalCar 0.275MM
Posts: 1,720
Combining multiple small residual funds given 2 Ticketless Travel funds limit

I find myself currently with 8 different record locators with small amounts of Ticketless Travel funds (anywhere from $5 to one with $100+). With the relatively new limit on being able to specify two old tickets with Ticketless Travel funds as payment toward a new reservation, I am wondering if anybody has come up with a good consolidation strategy.

The old strategy as I understood it would be to look for old tickets A, B, C, and D, whose residual values added up to just over the price of a new ticket Z. (Ticket Z would be chosen carefully specifically for its fare amount.) One would buy ticket Z, then cancel it. Without having to put any additional money in, this successfully consolidates A, B, C, and D into either just Z (if Z's fare amount was perfectly chosen) or a little bit left on D and most of the value on Z. Possible drawbacks are that any amounts on A, B, C, or D that were previously refundable become non-refundable if Z is a non-refundable fare; and Z now has the earliest expiration date out of A, B, C, or D. But instead of having 4 Ticketless Travel funds, you now have either 2 or just 1.

This is now harder to pull off. Starting with old tickets A and B, if Z's fare is perfectly chosen, yes, you can still consolidate 2 Ticketless Travel funds into just 1, but otherwise, you end up with a little left on B and most of the value on Z, which isn't a reduction in the number of Ticketless Travel funds at all. Now, if Z isn't perfect, but is close, then what's left on B may be a tiny amount and really should just be ignored, but man, that's annoying! And this requires a lot more cycles. To reduce my 8 Ticketless Travel funds into 2, which can be used in a single future ticketing transaction, would require 6 iterations, which isn't worth it to me. (The old strategy could as required as few as 2 iterations to reduce 8 to 2.)

Thoughts?
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