Split cab fare proportional to saving made across the board Interesting dilemma.
I've figured out how to split a cab fare 3 ways (proportional to time, mileage and equally) but I've found that when the cab fare is split proportional to time and mileage, the savings are not spread equally across board (poor rider B). 
e.g. in this case the total fare is $11, and the total mileage is 13.8 mi.
Riders                      A                 B                 C
=====================================================================
Time on journey             10 min            19 min            7 min 
Mileage on journey          3.8 mi            7.2 mi            2.8 mi

Results below:
Method                        Rider A           Rider B            Rider C
==========================================================================
Original fare (no sharing)    $4.00             $5.00              $4.50
Split % to time               $3.10             $5.80              $2.10      
Split % to mileage            $3.05             $5.75              $2.20
Split equally                 $3.70             $3.70              $3.70

As splitting % to time/mileage gives a similar result, I was just wondering if it was possible to introduce a 4th method that takes into account the savings made by each passenger somehow and then divides the fare, so that savings are made across the board. How would I express this mathematically?
 A: I'm not sure there's any one right answer, but taking the original individual fares as the basis, and subtracting the savings proportionally, may work.  The individual fares, one would hope, would encompass all factors for the cost of the trip for all concerned, and hence would make any joint savings achieved as fair as they could be.
Sum of individual fares:  $\$13.50$
Combined fare:  $\$11.00$
Total savings:  $\$2.50$
Rider A's fraction of the savings:  $4/13.50 \times \$2.50 = \$0.74$
Rider B's fraction of the savings:  $5/13.50 \times \$2.50 = \$0.93$
Rider C's fraction of the savings:  $4.5/13.50 \times \$2.50 = \$0.83$
Rider A's discounted fare:  $\$3.26$
Rider B's discounted fare:  $\$4.07$
Rider C's discounted fare:  $\$3.67$
They all save money, and in proportion to what their trip would have cost individually.  The ordering of the list doesn't change, either; B still pays the most, and A pays the least.
A: Another interpretation of the specification would be simply to split the savings evenly.
Single rides would have cost $4+5+4.50=13.50$, so since the actual price is only $11$ there's a total saving of $2.50$, which is 83 cents per passenger.
So they should pay \$3.17, \$4.17, and \$3.67, respectively.
