- Recall: unslotted Aloha:
- When a message msg
arrives:
it will be transmitted at the same time as its arrival moment:
- When a message msg
arrives:
- Throughput S:
S = p × G
Where: S = throughput G = offered load (arrival rate) p = fraction of transmissions that are successful = probability that a transmission is successful
- When is a transmission
successful:
- Consider a transmission x
that arrives
at time
t:
- A tranmission y
that arrive
during the
transmission
of
x will
collide
with the transmission x:
- A tranmission y
that arrive
1 time unit before t
will
also collide with
the transmission x:
- Transmissions that
arrive
outside the
critical interval
(see figure below)
will not
collide with
the transmission x.
Before:
Or after:
- Consider a transmission x
that arrives
at time
t:
- Conclussion:
- A transmission x
is successful
if and only if:
- There are no (zero) message arrivals during 2 time units
- A transmission x
is successful
if and only if:
- Throughput of
the unslotted Aloha:
Throughput S = #successful transmissions per time unit (= slot) = Total # transmissions × fraction successful = G × fraction successful = p × G
p = fraction of successful transmission = probabability that a transmission in slot is successful = probability that no (0) message arrives in previous slot = Prob[ 0 arrivals in 2 time unit (t.u.) ]
Recall Poisson arrival process: e-Gτ (Gτ)k Prob[ k arrivals in τ t.u. ] = ------------- (G = arrival rate) k! (G × 2)0 e(−G × 2) Prob[ 0 arrivals in 2 t.u. ] = ---------------- 0! = e−2G
Hence: Throughput S = p × G = e−2G × G = G e−2G
- Summary: throughput of the
unslotted Aloha protocol
S = G × e−2G (Unslotted Aloha)
- Graph of Slotted Aloha's throughput (S) vs. offered load (G):
- Previously:
Throughput S = G × e−2G (Slotted Aloha)
- The maximum can be found through
differentiation:
S = G × e−2G S' = e−2G - 2G × e−2G
Solve: S' = 0 ==> e−2G - 2G × e−2G = 0 <==> 1 - 2G = 0 <==> 2G = 1 <==> G = 1/2 = 0.5
- The maximum achievable
throughput in the
unslotted Aloha network is:
S(G) = G × e−2G ; max for G = 0.5 S(0.5) = 0.5 × e-1.0 = 0.5 × 0.368 = 0.18Maximum channel utilization with unslotted Alpoha = 18% (not very good)