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0.1 Cost with price breaks

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A general pattern for costs with price breaks is set up and an application is made in the case demand D has a Poisson distribution.

Let D be the demand random variable. Suppose there is

  • A flat fee of C for D ≤ a 1
  • A cost of c 1 per unit for a 1 < D ≤ a 2
  • A cost of c 2 per unit for a 2 < D ≤ a 3
  • A cost of c 3 per unit for a 3 < D

Then (see figure)

g ( t ) = C + c 1 I M 1 ( t ) ( t - a 1 ) + ( c 2 - c 1 ) I M 2 ( t ) ( t - a 2 ) + ( c 3 - c 2 ) I M 3 ( t ) ( t - a 3 ) where M i = [ a i , ∞ )
Since ( t - a i ) = 0 at t = a i , we could as well have M i = ( a i , ∞ ) .

Special Case. If D is Poisson ( μ ) , we may obtain an exact solution for E [ g ( D ) ] by using the fact that E [ I [ m , ∞ ) ( D ) ] = P ( D ≥ m ) and

E [ I [ m , ∞ ) ( D ) D ] = e - μ ∑ k = m ∞ k μ k k ! = μ e - μ ∑ k = m - 1 ∞ μ k k ! = μ P ( D ≥ m - 1 )

As an alternate approach, we approximate D by an appropriate number of values.

A residential College is planning a camping trip over Spring Recess. The number D of persons planning to go is assumed to be Poisson (15). Arrangements for transportation and food have been made as follows:

  • If D ≤ 4 , a minimal flat fee of $450 will be charged.
  • If 4 < D ≤ 19 , the additional charge is $100 for each person more than 4 (but less than 20).
  • If 19 < D , the charge is $80 for each person more than 19.

Thus, C = 450 , a 1 = 4 , a 2 = 19 , c 1 = 100 , and c 2 = 80 . The distribution for D is Poisson (15). Then

Y = g ( D ) = 450 + 100 I [ 4 , ∞ ) ( D ) ( D - 4 ) + ( 80 - 100 ) I [ 19 , ∞ ) ( D ) ( D - 19 )
= 450 + 100 I [ 4 , ∞ ) ( D ) D - 20 I [ 19 , ∞ ) ( D ) D - 400 I [ 4 , ∞ ) ( D ) + 380 I [ 19 , ∞ ) ( D )
  1. Obtain the “exact” solution to E [ g ( D ) ] .
  2. Approximate D by terms up to 40 and obtain E [ g ( D ) ] and P ( g ( D ) ≥ v ) for v = 1000, 1500, 2000, 2500.
mu = 15; % Part (a)EG = 450 + 100*mu*cpoisson(mu,3) - 20*mu*cpoisson(mu,18)... - 400*cpoisson(mu,4) + 380*cpoisson(mu,19)EG = 1.5433e+03 % Part (b)t = 0:40; M1 = t >= 4; M2 = t >= 19; G = 450 + 100*M1.*(t - 4) - 20*M2.*(t - 19);PD = ipoisson(mu,t); EGa = dot(G,PD)EGa = 1.5433e+03 P1000 = dot(G >= 1000,PD) P1000 = 0.9301P1500 = dot(G >= 1500,PD) P1500 = 0.5343P2000 = dot(G >= 2000,PD) P2000 = 0.1248P2500 = dot(G >= 2500,PD) P2500 = 0.0062

Questions & Answers

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Source:  OpenStax, Topics in applied probability. OpenStax CNX. Sep 04, 2009 Download for free at http://cnx.org/content/col10964/1.2
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