In Section 5.3, we provided an outline of the derivation of the efficiency of slotted ALOHA. In this problem we’ll complete the derivation.
a. Recall that when there are N active nodes, the efficiency of slotted ALOHA
is Np(1 – p) N–1 . Find the value of p that maximizes this expression.
b. Using the value of p found in (a), find the efficiency of slotted ALOHA by
letting N approach infinity. Hint: (1 – 1/N) N approaches 1/e as N approaches infinity.
Solution:
a. The efficiency of slotted ALOHA is
E(p) = 𝑁𝑝(1 − 𝑝)𝑁−1
By derivation, we have
E ′ (p) = N(1 − p)N−1 − 𝑁𝑝(𝑁 − 1)(1 − 𝑝)𝑁−2 = 𝑁(1 − 𝑝)𝑁−2 ((1 − 𝑝) − 𝑝(𝑁 − 1))
Let
E ′ (p) = 0
Thus, when
p∗ = 1/𝑁
We have the maximum efficiency.
b. When p equals to 1/N , the efficiency of slotted ALOHA is
E(p∗) = 𝑁*1/𝑁(1 − 1/𝑁 )𝑁−1 = (1 − 1/𝑁 )𝑁−1 = (1 − 1 𝑁 )𝑁 / 1 − 1/N
Since
lim N→∞ (1 − 1/𝑁 ) = 1
then lim N→∞ (1 − 1/𝑁 )𝑁 = 1/𝑒
Thus when n approaching infinity, the efficiency is
lim N→∞ 𝐸(𝑝∗) = 1/𝑒
CIte Reference:
http://course.duruofei.com/wp-content/uploads/2015/05/Networking_HW4.pdf