Bob Owens said:
[The cost of Air Travel is not the inhibiting factor it once was...]In that air travel is far cheaper today than it was in the past.
Not true. The cost is
always an inhibiting factor. We've just moved up the demand curve because the leisure fares are cheaper. It's still pretty much the same inhibiting factor.
Sure, if you compare today with 1975, you'll find that there are far more leisure passengers.
Then or now?
Now, of course.
🙄
The elasticity, however, remains relatively unchanged at best.
Which elasticity?
Well, both. But in particular I was referring to leisure elasticity.
Not every observation has to be backed up by a formula. Its the other way around formulas are created to explain observations.
Fine. But you're not even explaining observations. You're speculating on what would happen if airfares went up, with not even empirical evidence to back it up.
Where do I say anything to support your claim that I'm treating demand as completly inelastic??
JS answered that one:
JS said:
You are confusing demand with quantity demanded.
Problems already. This is a complex situation and we are already allowing you to make assumptions, assumptions that are not factual.
Bear with me. I guarantee you that every assumption I make early on will be proven as irrelevant later. Feel free to hang onto all of them and throw 'em at me when I'm done with my explanation of pricing.
Actually I aced Calculus...
Good. So this should be cake for you.
Begin Part 2
R = P(k+aP) [thanks again JS]
which means that R = kP + aP²
If you wish to find the maximum or minimum of a formula, you take the first derivative (in this case with respect to price) and set it equal to zero.
So, maximum revenue would be:
k + 2aP = 0
2aP = -k
P = -(k/2a)
Since you aced calculus, I'm sure you'll recall that this would work for any function Q, linear or not. In the case of airline demand, which is bimodal, you'd end up with two maxima and one minimum. By taking the second derivative, and testing the signs at the three points, you can determine which of the three are maxima and which is the minimum.
Provided that you can accurately discriminate between the two different passenger groups, you can set prices accordingly. If you cannot separate the two groups, then you must test the two price points to determine which produces higher revenue.
Well, that'd all be fine if maximizing revenues were the goal. It's not. Maximizing
profit is the goal.
Again, I'll simplify things a bit here, but I'll explain up front why we can do so. The big assumption I will make is that all costs not directly associated with adding an additional person to a flight are "fixed costs." This is true in the sense that once a schedule is set, all of the variable costs associated with schedule changes are no longer variable. And, since we're talking about setting ticket prices for a particular seat on a particular flight, schedule-related costs
are now fixed.
So, with an established schedule, the marginal profit associated with a passenger can be expressed as
marginal profit = P-c,
where P is the price paid, and c is the marginal cost of carrying the additional passenger. That marginal cost includes several factors, such as the additional fuel burn caused by the additional weight, wear and tear on the aircraft interior, the marginal costs of selling the ticket and addressing any customer service issues (e.g., calls to res to change the seat), etc., etc....the list is long.
The total profit for a particular flight, assuming a single price for every ticket, can be expressed as:
profit = Q(P-c)
Now, if you had different prices, you'd have to express it more like this:
profit = Q1(P1-c) + Q2(P2-c) + Q3(P3-c)......
It makes the math more complicated, so this exercise is left up to the reader if you wish to test it out for yourself. And, yes, you could even break down different marginal costs for different passengers if you have more time on your hands.
😉
Anyway, we'll go back to the simpler single-price model. Recall that Q=k+aP. Substitute it into the single-price profit model and you get
profit = (k+aP)(P-c) = kP-kc+aP²-acP
Take the first derivative to find the maximum, and you'll get
k-ac+2aP=0
2aP=ac-k
P=(ac-k)/2a
Again, if the demand function isn't linear, you still can do exactly the same thing.
Incidentally, a, the slope of the demand curve, represents the degree of elasticity. If a=0, demand is perfectly inelastic (i.e., quantity demanded would remain unchanged regardless of price). If a=∞ (that's infinity, in case your computer doesn't display it properly), demand is perfectly elastic (i.e., a tiny change in price would have an infinite impact in quantity demanded).
Anyway, as before, if you have a more complex polynomial expression for the demand curve, you can find the local maxima and minima by taking the first derivative and setting it equal to zero, and you can determine if they're minima or maxima by taking the second derivative at those points and checking their signs.
Now for the next level of complexity. This is more advanced pricing.
If you want to truly maximize profits, you'd charge exactly the maximum that each person is willing to pay. This would allow you to get much more profit than with a single-price model. Your total profit function would be
profit = §(P-c)Q
where § represents the integral (it's the closest symbol I could find). Since we've already discussed Q as a function of P, I'm not going to go back into the substitution...just understand that we're integrating with respect to P, and Q is a function of P.
When setting your lower bound for the integration, you must set it to the higher of these two values:
1) Available capacity
2) The marginal cost c
in order to maximize profit. In other words, you want to either sell out a flight or be unable to sell any tickets for more than it costs you to serve that one additional passenger. The upper bound could be set to infinity, but realistically it makes no sense to set it any higher than the highest price anyone is willing to pay for the ticket.
To the degree that you approach the integration model (i.e., adding additional fares), your profits should rise.
Of course, with all that I've discussed here, the biggest assumption is that we know what the demand function looks like. At best, we have a visceral sense of what it looks like. One of the jobs of Yield Management is to ascertain as clearly as possible
exactly what the demand function's shape is. If they know the shape, they can determine exactly what pricing to set.
YM has one other big job. As I said above, if you want to truly maximize profits, you'd charge exactly the maximum that each person is willing to pay. In order to do that, you have to figure out what differentiates one customer from another, so as to be able to price accurately. YM gets to figure that stuff out, too. That's how the rules for each fare code are determined.
One final note. The profit maximizing functions mentioned above are simplified to the case where the airline has a monopoly on the route. Naturally, for most routes, that is not the case. However, you can adjust your demand function accordingly by applying the reasonable assumption that if your competitors charge a lower price for the same fare rules, they will get those customers before you do. If you all charge the same price for the same rules, you can apply varying levels of sophistication to determine the market share breakdowns.
OK. If you managed to follow all of that, you should be able to understand why increasing prices will not necessarily translate into increasing profits, or even increasing revenues.
