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Uncertainty in the Demand for Service - Faculty of Industrial ...

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Note that (3) applies if and only if √ N(1 − ρN) converges to β (β>0). Indeed, <strong>for</strong>mally <strong>the</strong><br />

Theorem <strong>of</strong> Half<strong>in</strong>-Whitt reads:<br />

As N ↑ ∞, P (W ait > 0) = E2,N → α (0 < α < 1)<br />

iff √ N(1 − ρN) → β (0 < β < ∞)<br />

(equivalently N ≈ RN + β √ RN) .<br />

In practice, this rule makes <strong>the</strong> life <strong>of</strong> a call-center manager easier: he or she can actually<br />

specify <strong>the</strong> desired delay probability and achieve it by follow<strong>in</strong>g <strong>the</strong> square-root safety staff<strong>in</strong>g<br />

rule (3), simply choos<strong>in</strong>g <strong>the</strong> right β.<br />

2.2 Erlang-A: Summary <strong>of</strong> Some Results<br />

Try<strong>in</strong>g to make <strong>the</strong> M/M/N model more realistic and useful, <strong>the</strong> follow<strong>in</strong>g assumption is added:<br />

each customer has limited patience, that is, as <strong>the</strong> wait<strong>in</strong>g time <strong>in</strong> <strong>the</strong> queue grows <strong>the</strong> customer<br />

may abandon. We assume that patience is distributed exponentially with mean 1/θ. This<br />

model is referred to as Erlang-A (A <strong>for</strong> Abandonment). It is also a Birth-Death process, and its<br />

transition rate diagram is depicted below.<br />

<br />

<br />

<br />

. . .<br />

0 1 2 N-1 N N+1<br />

2 <br />

<br />

N<br />

N <br />

Figure 2: Erlang- A as a Birth-Death Process<br />

The steady-state distribution is found by solv<strong>in</strong>g <strong>of</strong> <strong>the</strong> follow<strong>in</strong>g equations:<br />

<br />

λπj = (j + 1)µπj+1<br />

0 ≤ j ≤ N − 1<br />

λπj = (Nµ + (j + 1 − N)θ)πj+1 j ≥ N<br />

The solution <strong>of</strong> <strong>the</strong> steady-state equations is given by:<br />

⎧<br />

⎪⎨<br />

πj =<br />

⎪⎩<br />

(λ/µ) j<br />

π0,<br />

j!<br />

j<br />

<br />

<br />

λ (λ/µ) N<br />

Nµ + (k − N)θ N!<br />

0 ≤ j ≤ N<br />

π0, j ≥ N + 1<br />

where<br />

π0 =<br />

n<br />

j=0<br />

k=n+1<br />

(λ/µ) j<br />

j! +<br />

∞<br />

j<br />

j=N+1 k=N+1<br />

<br />

λ<br />

<br />

(λ/µ) N<br />

Nµ + (k − N)θ N!<br />

N<br />

. . .<br />

2<br />

−1<br />

6<br />

(5)

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