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Inside Collection (Textbook):

Textbook by: Paul E Pfeiffer. E-mail the author

# Cost with Price Breaks

Module by: Paul E Pfeiffer. E-mail the author

Summary: 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 Da1Da1
• A cost of c1 per unit for a1<Da2a1<Da2
• A cost of c2 per unit for a2<Da3a2<Da3
• A cost of c3 per unit for a3<Da3<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 , ) 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 , )
(1)

## Note:

Since (t-ai)=0(t-ai)=0 at t=ait=ai, we could as well have Mi=(ai,)Mi=(ai,).

Special Case. If D is Poisson (μ)(μ), we may obtain an exact solution for E[g(D)]E[g(D)] by using the fact that E[I[m,)(D)]=P(Dm)E[I[m,)(D)]=P(Dm) 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 ) 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 )
(2)

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

## Exercise 1: Example

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 D4D4, a minimal flat fee of $450 will be charged. • If 4<D194<D19, the additional charge is$100 for each person more than 4 (but less than 20).
• If 19<D19<D, the charge is \$80 for each person more than 19.

Thus, C=450,a1=4,a2=19,c1=100C=450,a1=4,a2=19,c1=100, and c2=80c2=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 ) Y = g ( D ) = 450 + 100 I [ 4 , ) ( D ) ( D - 4 ) + ( 80 - 100 ) I [ 19 , ) ( D ) ( D - 19 )
(3)
= 450 + 100 I [ 4 , ) ( D ) D - 20 I [ 19 , ) ( D ) D - 400 I [ 4 , ) ( D ) + 380 I [ 19 , ) ( D ) = 450 + 100 I [ 4 , ) ( D ) D - 20 I [ 19 , ) ( D ) D - 400 I [ 4 , ) ( D ) + 380 I [ 19 , ) ( D )
(4)
1. Obtain the “exact” solution to E[g(D)]E[g(D)].
2. Approximate D by terms up to 40 and obtain E[g(D)]E[g(D)] and P(g(D)v)P(g(D)v) for v=v= 1000, 1500, 2000, 2500.

### Solution

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 = 0.9301
P1500 = dot(G >= 1500,PD)
P1500 = 0.5343
P2000 = dot(G >= 2000,PD)
P2000 = 0.1248
P2500 = dot(G >= 2500,PD)
P2500 = 0.0062


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