By Yao H., Knessl Ch.
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Extra info for On the Shortest Queue Version of the Erlang Loss Model
Example text
As we go from (y, y) to (a, y) in the sum S 2 we have x > 0 until we reach (X ∗ (y), y) where x = 0. Then along the horizontal segment from (X ∗ (y), y) to (Y −1 (y), y) we have x < 0 so that p(m, n) decreases. From (Y −1 (y), y) to On the Shortest Queue Version of the Erlang Loss Model 193 (a, y) we are in the region R and x < 0 so p(m, n) decreases further. We thus have P(n) ∼ S2 (n) and the major contribution to S 2 comes from where m ≈ X ∗ (y)/ε, which lies in D. Expanding (26) near s = 1/4 or x = X ∗ (y) we have √ √ εK (x, y)e (x,y)/ε ∼ εK (X ∗ (y), y)e (X ∗ (y),y)/ε · exp 1 2ε x x (X ∗ (y), y)(x − X ∗ (y))2 .
A(x) 2x 2 1 (x) 2 a + (4a + 1)2 (x − a) (309) Now we can use either t or x as the parameter along the ray s = a, so that ∂t = [tˆ (x)]−1 ∂x . Using (278) (with s = a, t = tˆ(x)), (307) and (309) in (305) we obtain x −a+ = 1 2 1 2(4a + 1) − 1 (x) 2 1 (x) − 1 1 (4a + 1)2 − 2x 2 a + (4a + 1)2 (x − a) 1 1 1 4a + 1 . − + ˆ t ˆ 2 1 + (4a − 1)(2a + 1)e 4a + 1 − (2a + 1)etˆ 1 (x) t (x) (310) 1 (x) However, tˆ (x) = −2(4a + 1) 1 + 2(4a + 1)(x − a) in view of (277), and then using (277) we can easily establish that (310) is an identity.
346) Asymptotically matching (341) as ζ → −∞ to (337) we conclude that √ 2 πa (4a + 1)3/2 (4a + 1)3 R(ζ ) ∼ √ ζ 2 , ζ → −∞. exp − 2 2 2 16a (4a + 3a + 1) 4a + 3a + 1 (347) Similarly, we match (341) as ζ → + ∞ to the layer L3 (cf. (340)) to conclude that √ (4a + 1)3 (16a 2 + 1) 2 2a 5/2 8a 2 + 2a + 1 exp − R(ζ ) ∼ ζ , ζ → +∞. ζ (4a + 1)3/2 16a 2 (8a 2 + 2a + 1) (348) However, (347) and (348) are not sufficient to determine R (ζ ) for all ζ . To this end we must use a third matching, between (341) and the transition layer L1 .