Guys, Oh! Guys !
I'm quite disappointed in you :
One of the uses of the transport factor is really about profitability of tankering.
Let's go back to the initial def : K = ðTOW / ðLW,
which means in layman's terms that for each landed ton of transported fuel, one needs to uplift K tons of extra fuel.
The price of that landed fuel will be Po * ðTOW = Po * ðLW * K...... ("o" for origin ).
Let's call Pd the fuel price at destination : the transported fuel, ðLW, would have been paid Pd * ðLW, had one not transported any fuel.
So, tankering will be profitable if :
Pd * ðLW > Po * ðLW * K, in other words if the cost of the extra fuel at departure is lower than the cost of the transported ( meaning "landed") fuel at destination.
Simplifying the above equation :
Tankering is profitable if : Pd/Po > K
The beauty of it is that it works, whatever the type of aircraft, whatever the distance....be it a 744 or a Cessna 150.
It allows to have a ball park figure while waiting for one's dispatch to finalise the computations.
Beware, though : for greater quantities of extra fuel on long sectors, one could alter significantly the flight profile through delayed steps... and find that one would spend some extra time in the air, adding the effects of DOC to the economics of the flight.
Mutt,
From your CFP, I only have the final burn-off of 77,499 kg.
The final figures of TOW = 286.15 T ; LW = 208.65 T give a K= 1.371
Of the extra fuel = 19.500 T you'd keep at destination 19.500 / 1.371 = 14.223 T, thus using 5.277 kg extra burn-off to land 14.223 t at destination.