Hence, using the expression above, we obtain:
00 _ μd μ Λd Λw ∖ (1 - 2P) P (1 - P)
g = -(μw2 '' - μwJ h1 + λwP(1 - P)i3.
The sign of g00 is ambiguous (and note that without congestion g00 =0). We are
now ready to sign the expression (59).
We first compute the expression -2 (P0)2 + PP00 . We have
-2 (P0)2 + PP"´ =--P2(1—P)-----3
( (μw)2 [1 + ΛWwP (1 - P)]
Λw
1 + 2--P (1 - P)2
μw
< 0.
Furthermore, replacing the expression for g0 and after simplifications, we get:
(1-g0)=
1 + £) P (1 - P )
1 + Λw P (1 - P )
> 0.
A combination of the last two expressions leads to:
-2(P0)2 + PP00 (1-g0)
P2(1 - P)
г ι 4
(μw)2 [1 + ΛWWP (1 - P)]
Λw
1 + 2—P (1 - P)2
μw .
(1+£) + (ΛW+Λd) p (1 - P)
V μw∕ μw
We are ready to compute the second term of (59). After substitution, we
obtain:
P0g00P =
μd
(μw )3
μ Λd Λw ∖
Ud- μw)
(1 - 2P) P3 (1 - P)2
h1 + Λw p (1 - p )]4
Note that Ω = Ω1 - 2Λh (P0)3 < Ω1 (since P0 > 0), with:
Ω1 = (-2 (P0)2 + PP00) (1 - g0) + P0g00P.
We wish to show that Ω2 < O with:
Ω1
P2 (1 - P )
(μw)2 [1 + ΛWWP (1 - P)]4
Using the two expressions derived above, we get:
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