27K (Applied Probability)

A dispatch cycle contains

N

interarrival times and has mean length

N / μ

. The expected total passenger waiting time in one cycle is

\frac{1}{μ} (1 + 2 + \dots + (N - 1)) = \frac{N ( N - 1 )}{2 μ} .

The renewal-reward average cost rate is therefore

C (N) = \frac{K + c N ( N - 1 ) / ( 2 μ )}{N / μ} = \frac{μ K}{N} + \frac{c}{2} (N - 1) .

Thus the continuous optimum is

2 μ K / c

, and the integer optimum is the positive integer minimizing

C (N)

; equivalently it is the smallest

N

with

N (N + 1) \geq \frac{2 μ K}{c},

with both adjacent values optimal in the equality case.

b

The lifetime

L

is uniform on

[0, 3]

with mean

3/2

. The equilibrium age and excess densities are both

\frac{P ( L > x )}{E L} = \frac{2}{3} (1 - \frac{x}{3}) .

Hence, for

0 \leq x \leq 3

P (A_{t} \leq x), P (E_{t} \leq x) ⟶ \frac{2 x}{3} - \frac{x ^{2}}{9} .

The observed total lifetime is length-biased, so

P (A_{t} + E_{t} \leq x) ⟶ \frac{E [ L 1 _{L \leq x} ]}{E L} = \frac{x ^{2}}{9} .

E S = \int_{R^{d}} g (u) λ (u) d u .

Expanding

S^{2}

separates equal and distinct Poisson points. The factorial-moment formula gives

E S^{2} = \int g (u)^{2} λ (u) d u + (\int g (u) λ (u) d u)^{2} .

Subtracting the square of the mean yields

var S = \int_{R^{d}} g (u)^{2} λ (u) d u .

There are no discussions about this article yet.

There are currently no matching articles.