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Suppose a customer denied a class 2 price of $459 at the Killington upgrades to class 1 30% of the time. Assume a Poisson distribution for class 1 demand with L = 20 (as in #1). Use Littlewood’s 2‐class model with upgrades to calculate the optimal protection level for class 1. (Assume as before that demand for class 2 materializes before class 1. Also, you cannot use the inverse cdf formula given in the notes because the Poisson is a discrete distribution. However, the idea behind that formula—comparing the revenue of accepting a class 2 request versus the value of saving the unit—can be used instead). Repeat for upgrade probabilities q=10%, 20% and 40%. How does the protection level change?
**Littlewood’s Two‐Class Model with Upgrades**
One of the interesting wrinkles one can add to Littlewood’s model is the possibility of upgrades. When a class 2 request is denied in period 2 (because we have hit the booking limit for class 2capacity), a customer may then make a class 1 request. If some customers upgrade to class 1,then protecting capacity has some additional benefit to its expected value. Suppose a customer requests class 1 with probability q if class 2 is closed. Then the payoffs andprobabilities associated with denying a class 2 request in period 2 when capacity is x are
Payoff | Probability |
p1 | q |
p1Prob (D1>= x) | 1-q |
Observe that the payoff is p1 if the customer upgrades and p1Prob(D1 ³ x)the traditional expected payoff without upgrades) if they do not. (Note: D1 is considered “pure” demand for class 1 in period 1 and does not include this upgrade demand). Therefore, the expected value of rejecting a class 2 request when capacity is x is
qp1+(1-q)p1Prob(D1>=x)
Thus one only accepts class 2 requests in period 2 if it exceeds this value. Observe that this
expected value is slightly larger than that used in the two‐class model without upgrades. This
makes sense because a class 2 request that is denied will sometimes (with probability q) be
transformed into a higher priced outcome in this model. After a little algebra, the optimal
protection level is seen to be y*1 = F1-1 (p1-p2/(1-q)p1))
10 years ago 5
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