1. The number of analog duplex channels are Nch = 250. The ratio between the reuse distance D and the cell radius R must be at least D/R = 7. The cell radius is R = 2 km. During a busy hour each subscriber generates one call of two minutes on average. The system can be modeled as Erlang B. The required maximum blocking probability is 3%.
(a) i. To find the maximum number of subscribers per cell we first need the cluster size, which can be found using Tab. B-17.2. According to the table the smallest valid cluster size of Ncluster= 19 is obtained for a D/R = 7.5. The number of channels per cell is then
Ncell= Nch
Ncluster = 250
19 = 13.1. (17.1)
Figure B-17.2 shows the trunking gain. Since a curve for 13 channels per cell is absent, we have to approximate a value between the curves for Ncell = 10 and Ncell = 15. A blocking probability of 0.03 would then result in a cell capacity Ccell of about 9 Erlang per cell. During a busy hour each subscriber produces an average traffic of Tsub = 2/60 = 0.033 Erlang (recall that one Erlang corresponds to one fully used channel). Hence the maximum number of subscribers per cell is
Nm ax,cell =Ccell Tcell
= 9
0.033 = 270. (17.2)
ii. From the above, the cell capacity is Ccell= 9 Erlang. The area of a cell is
Acell= πR2= π22= 12.6 km2. (17.3) Thus, the network capacity is
Cnet = Ccell
Acell = 9
12.6 = 0.72 Erlang/km2. (17.4)
(b) With digital transmission the number of duplex channels is reduced to Nch,dig = 125. But at the same time, the ratio D/R = 7 can be lowered to Ddig/Rdig = 4. From Tab. B-17.2 we obtain a valid cluster size of Ncluster,dig = 7 for a Ddig/Rdig = 4.6. This results in
Ncell,dig = Nch,dig Ncluster,dig
=125
7 = 17.9 (17.5)
channels per cell. By using the figure on slide 18, a maximum blocking probability of 0.03 and Ncell,dig = 17 channels per cell, a cell capacity of approximately Ccell,dig = 13 Erlang is obtained.
Then, since the cell size is the same, the network capacity is Cnet,dig = Ccell,dig
Acell = 13
12.6 = 1.03 Erlang/km2. (17.6)
(c) By reducing the transmit power, the cell radius can be reduced to Rnew = 1 km. Since the ratio Ddig/Rdig = 4 must be preserved, the cluster size will be the same. The number of channels is also the same, which means that the cell capacity is still Ccell,dig = 13. The cell area, however, will decrease to
Acell,new = πR2new= π km2. (17.7)
Thus, the new network capacity is increased to Cnet = Ccell,dig
Acell,new = 13
π = 4.1 Erlang/km2. (17.8)
Since it takes
Acell
Acell,new = R2
R2new = 4 (17.9)
new smaller cells to cover the same area as one old cell, 4 times as many base stations are required to obtain the same coverage.
2. Erlang B − the most commonly used traffic model. Erlang B is used to work out how many lines are required if the traffic figure during the busiest hour is known. This model assumes that all blocked calls are cleared immediately, and the probability of a blocked call is
Pblo ck= TtrNC NC!PNC
k=0 Tt rk
k!
(17.10)
With one operator the offered traffic is Ttr= 12 hence 0.05 ≤ 12NC
NC!PNC
k=0 12k
k!
(17.11)
Hence, for NC = 17 the blocking probability is 1.3%. With three operators, each operator needs 8 channels for Ttr= 4 to get a blocking probability of 3 %. A total of 24 channels is needed.
3. Erlang C — this model assumes that all blocked calls are queued in the system until they can be handled.
Call centers can use this calculation to determine how many call agents to staff, based on the number of calls per hour, the average duration of class and the amount of time calls are left in the queue.
(a) Channels needed
0.05 ≥ TtrNC
TtrNC+ NC!³ 1 −NTt rC
´ PNC−1 k=0
Tt rk k!
(17.12)
Hence, for NC = 19 the blocking probability is 4.4%. With three operators, each operator needs 9 channels for Ttr = 4 to get a blocking probability of 2.4%. A total of 27 channels is needed.
Erlang C systems require more resources than Erlang B.
(b) What is the average waiting time if the average call duration is 5 min?
twait = Pwait
Tcall
NC− Ttr
(17.13) 0.0238 5
9 − 4 (17.14)
0.0435 5
19 − 12 (17.15)
For one operator the waiting time is 0.031 minutes and for three it is 0.024 minutes.
4. TDMA requires a temporal guard interval.
(a) Hence a MS communicating with the BS could be 3000 m away and with the other very close to the BS. The propagation delay is
3000
3 · 108 = 10µs (17.16)
The ”echo” of the channel is 10 µs, hence the total guard time (propagation delay + excess delay) should be more than 20 µs. A TDMA slot in GSM is 577 µs which includes a guard time that is
≈ 30 µs.
(b) With timing advance, the transmitter knows the propagation delay and the guard time could be reduced.
5. For the downlink we have the approximation µC
We make here the approximation that the distance from all BSs to the MS at the cell boundary is equal, namely D − R. This is a pessimistic approximation, as at least some BSs have a larger distance.
The approximation holds the better the larger the reuse distance is.
For the uplink case, since all transmitter powers are assumed to be the same, C ∝ PTXd−η0 and I ∝P6
i=1PTXd−ηi , where d0 is the distance between MS-0 and BS-0 and di is the distance between MS-i and BS-0. The worst-case scenario is when MS-0 is on the boundary of its cell, i.e., d0= R, and the co-channel mobiles MS-i, i = 1, 2, .., 6, are located on the boundary of their cells, respectively, in the direction of BS-0, i.e., di= D − R, i = 1, 2, .., 6. Hence (since η > 1),
Consequently, the upper bound above also holds for the uplink case.
6. The received power of the user of interest, S, and received power from the interfering user, I, is equally strong at the receiver. The (signal-to-interference ratio) SIR is then
γSIR= S
I (17.19)
The signal to noise ration (SNR) is 10 dB (= 10 in linear scale) and defined as γSNR = S
N = 10 (17.20)
The overlap, α, allowed to keep the signal-to-noise and interference ratio (SNIR) at (7 in dB is equal to 5 in linear scale)
γSNIR = S
where α is the maximum overlap (normalized collision time) to fulfill the SNIR requirements. Hence, a full overlap of one packet is admissible.
The throughout without lost packets is λpTp, where λp is the average transmission rate in packets per second. The effective throughput is the percentage of time during which the channel is used in a meaningful way, i.e., that packets are offered, and transmitted successfully. The probability of zeros or one collision is
p (0, Tp) + p (1, Tp) = e−λpTp+ λpTpe−λpTp (17.23) The effective throughput is then
λpTp¡
e−λpTp+ λpTpe−λpTp¢
(17.24)
The maximum of λpTp is found (derivative and set to zero) to be λpTp = 12 +12√
5 = 1.618, thus the maximum effective throughput is 0.81. Remember that in the case when no interference is tolerated, the effective throughput is
1
e = 0.38 (17.25)
which is less and follows intuition.