EE5412
LanQA1.doc
Aloha, maximum through put = 18.4%
Maximum throughput of a CSMA/CD LAN is ρ = 1/[ 1 + a(1 +2e)]
where a = message propagation time/message transmission time,
and e = exponent value
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A group of N stations share a 56kbps pure ALOHA channel. Each station output
a 1000 bits frame on the average in every 100 sec. What is max. value of N?
Ans:
input per station = one frame per 100 sec = 0.01 frame per sec
channel capacity = 56k / 1000 frame per sec = 56 frame per sec
for Aloha, maximum through put = 18.4%
ie max throughput = 18.4% x 56 frame per sec = 10.3 frame per sec
hence max N = 10.3 / 0.01 = 1030
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One Hundred stations are distributed over a 4-km bus and the bus propagation delay is assumed to be 5 μ sec/km. A CSMA/CD protocol is used. The transmission rate is 10 Mbps and data frames are 1000 bits long, on the average. Assuming a homogeneous network, calculate the maximum number of frames per second that each station is allowed to generate.
Note: the maximum throughput of a CSMA/CD LAN is
ρ = 1/[ 1 + a(1 +2e)]
where a = message propagation time/message transmission time,
and e = exponent value
Ans: a = 10 x 4 x 5/1000 = 0.2
ρ = 1/(1+6.44a) = 0.4372
max throughput = 0.4372 x [10 M] /1000 frames per sec
= 4372 frames per sec
frame per station =4372/100 = 43.72 frame/station per sec
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Fifty stations are distributed over a 5-Km bus. A CSMA/CD protocol is used. The transmission rate is 5 Mbps and data frames are 500 bits long, on the average. Calculate the maximum number of frames per second that each station can generate. Use 5 μ sec/km as the bus propagation delay.
No of stations, N = 50
transmission rate, C = 5 Mbps
propagation delay = 5 μ sec/km
bus length = 5 km
frame length = 500 bits
channel capacity :
= 5 Mbps / 500 bits = 10000 frame per sec
total propagation delay of the bus, τ = 5 μ sec/km x 5 km
= 25 μ sec
maximum traffic intensity, Smax = N λ / [channel capacity]
= [no of station] x [frame arrival rate per station] / [ channel capacity in frames per sec]
= 1/[ 1 + a(1 +2e)]
= 1/(1+6.44a)
where a = 25/100 = 0.25 [propagation delay time/ frame transmission time]
hence Smax = 1/(1+6.44a) = 1/(1+6.44 x 0.25) = 0.383
max arrival per station = λ max = Smax x channel capacity / N = 0.383 x 10000 / 50
= 76.62 frame/sec
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[Computer Network,Tanenbaum,Ch3 Q5]
Measurement of an infinite user slotted ALOHA channel show that 10% of the slots are idle.
1. What is the channel load, G?
2. What is the throughput?
3. Is the channel under-loaded or overloaded?
Ans:
P0 = 10% = 0.1
a . From Poisson law, P0=e-G,
G = -lnP0 = ln (0.1) = 2.3
b . S = G e-G
= 2.3 e-G = 2.3 x 0.1 = 0.23
c . G > 1 the system is overloaded
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[Tantenbaurn ch3 Q4]
A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slotted in units of 40 msec.
a. What is the chance of success on the first attempt?
b. What is the probability of exactly k collision and then a success?
c. What is the expected number of transmission attempts needed?
Ans:
a. with G==2; poisson statistic gives prob = e-2
b. ( 1 – e-G)k e-G = 0.135 x 0.865k
d. expected number of retransmission , eG = 7.4
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A 1-km long, 10 Mbps CSMA/CD LAN has a propagation speed of 2 x 108m/s. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel to send a 32-bit acknowledgement frame. What is the effective data rate, excluding overhead, assuming that thee are no collisions.
Ans:
Round trip propagation time of the cable = 10 usec
A complete transmission has four phases:
1. transmitter seizes cable : 10 usec
2. transmit data 25.6 usec
3. receiver sizes cable 10usec
4. acknowledgment sent 3.2 usec
In the period of 48.8 usec, 224 bits are sent, hence the effective data rate is 244/48.8 usec = 4.6 Mbps
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