Decision Science Price Differentiation Homework

Khalil

Lara Akkad

DCSN 211: Report on Chapter 3 & 4

Work in groups of two. Prepare a paper report including answers to the following questions. Submit one report per group at the beginning of the class on the due date. Have a second copy of your report with you to correct mistakes during class time.

1. Consider a linear price-response function d (p)=D-m*p. In Chapter 3 (Basic Price Optimization), we established that the total revenue is maximized when p=D/(2*m).

1. At what price is the total contribution maximized, assuming incremental cost c?

Answer: The price that maximizes revenue is always lower than the price that maximizes total contribution (unless incremental costs are 0).

Revenue maximizing price = P’ = D/2*m where m is the slope of the linear price response function.

Contribution maximizing price = P*

Max m(p*) = Max (p*- c) d(p)

• To maximize total contribution, we take the derivative of m(p) and set it as = to 0 to find p*.

m’(p*) = d’(p*)(p*- c) + d(p*)

d’(p*)(p*- c) + d(p*) = 0

p* - c = - d(p*)/d’(p*)

p* = - d(p*)/d’(p*) + c                    Contribution maximizing price (p*)[pic 1]

1. Find a general formula for any values of D, p, and c. Show all steps of your analysis. Test your formula on one example for which you know the correct solution.

Given: linear price response function d (p) = D – m*p & revenue maximizing price = p’ = D/2*m

• m(p) = (p – c) d(p)
• m(p) = (p – c)(D - m*p)
• m(p) = (p – c)(D – m*(D/2*m)
• m (p) = (p – c)(D – D/2)

Testing of general formula:

Using an example from the power point presentation on chp. 3:

• what price maximizes the total revenue: p = 6.25$ for d (p) = 10,000-800p
• given incremental costs = $5.00
• In the example, m(p) = $6,250

Testing our function with the same parameters:

m(p) = (6.25 – 5)(10,000 – 10,000/2) = ($1.25)(5000) = $6,250

1. Consider a logit price-response function d (p) = C*exp(-(a+b*p))/(1+exp(-(a+b*p))) with parameters C=100, a=-10, and b=0.1. Determine the price at which the total contribution is maximized, assuming incremental cost c=50. Report the optimal price and the total contribution obtained at the optimal price. You may want to use an optimization solver to solve this question.

First step: find ^ = - (a/b) = - (-10/0.1) = $100[pic 2]

Second step: d(p) =  =  =  = 50 [pic 3][pic 4][pic 5]

Third step: d’(p) =  =  =  = -2.5 [pic 6][pic 7][pic 8]

Fourth step: m(p) = (p – c) d(p)

To maximize it, use the derivative:

m’(p*) = d’(p*)(p*- c) + d(p*) = (-2.5) (p* - 50) + 50

(-2.5) (p* - 50) + 50 = 0                  contribution maximizing price (p*) = $70[pic 9]

Replacing p* = optimal price in m(p):

m(p) = (p – c) d(p) = (70 – 50) 50 = $1,000 (total contribution at optimal price)

1. Solve Exercise 1 (Arbitrage) from the end of Chapter 4 in the Phillips textbook. Note that some questions may be solved analytically, while others may require an optimization solver.

1. For Pleasantville:

m(p1) = (p1 – c) d(p1) = (p1 – 5) (10,000 – 800p1) = 14,000p1 – 800p12 – 50,000