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Assigment 1
February 5, 2016
2 Mbps / 200 Kbps = 10 users
5
n
n=0 0.^01 n^0.^9950 โn
- R ร ddrop = 110^6 (500010^3 / 2. 7108 ) = 1. 85104 bits
- As 1. 85 ร 104 bits < 50000 , then the maximum number of bits is : 1. 85 ร 104 bits. or dprop/(dtrans(1bit)) :
- 85 โ 10 โ 2 / 10 โ6 = 1. 85 โ 104
- (^) [dproplinkLength/dtrans(1bit)] = 5 โ 106 /(1. 85 โ 104 ) = 270m
- (^) [dproplinkLength/dtrans(1bit)] = (^) [( (m/m/s1)) 1 /R ]^
= s/R
โข P 0 = (1 โ P )N
(N
m
pm(1 โ p)N^ โm
โm n=
(N
n
pn(1 โ p)N^ โn
- Pk = (1 โ Ptr)kโ^1 Ptr
- Let random variable T represent the number of times that A has to send the file to get it accepted at B. We have, E[T ] =
k=1 kPk^ =^
k=1 k(1^ โ^ Ptr)kโ^1 Ptr^ =^
1 Ptr.
Figure 1: part 1
- We are in state S 0 at time k = 1, when there is no arrival in the first transition step, and it follows
that: P 0 (1) = 1 โ Pa
- We are in state S 1 at time k = 1, when there is an arrival in the first transition step, and it follows
that: P 1 (1) = Pa
- We can not reach state S 2 from S 0 in 1 transition step, and it follows that: P 2 (1) = 0
- We have:
P 0 (1) = (1 โ Pa)P 0 + PlP 1 P 1 (1) = PaP 0 + (1 โ Pa โ Pl)P 1 + PlP 2 P 2 (1) = PaP 1 + (1 โ Pl)P 2
- We have to solve the system of equation:
P 0 (1) = (1 โ Pa)P 0 + PlP 1 P 1 (1) = PaP 0 + (1 โ Pa โ Pl)P 1 + PlP 2 P 2 (1) = PaP 1 + (1 โ Pl)P 2 Under the condition that:
P 0 + Pl + P 2 = 1 Using the equation
P 0 = (1 โ Pa)P 0 + PlP 1
We obtain that: P 1 = Pa Pl
P 0 = ฯP 0
For ฯ = P Pal Using this result in the equation
P 1 = PaP 0 + (1 โ Pa โ Pl)P 1 + PlP 2
we obtain that P 2 = ฯ^2 P 0
Finally, using the condition that
P 0 = (1 โ Pa)P 0 + PlP 1 We get that
P 0 =
1 + ฯ + ฯ^2 =^
1 โ ฯ 1 โ ฯ^3
- Using results of (f), the steady state probabilities are:
P 0 =
1 + ฯ + ฯ^2 =^
1 โ ฯ 1 โ ฯ^3 P 1 = ฯ
1 + ฯ + ฯ^2 =^
1 โ ฯ 1 โ ฯ^3 P 1 = ฯ^2
1 + ฯ + ฯ^2 =^
1 โ ฯ 1 โ ฯ^3
