- The previous analysis
provided us with a
unconditional probability τ on
how likely a station
will transmit a packet at
any given time:
- We can use the
general performance analysis
(See: click here)
technique to compute the
average throughput:
- Definitions and Notations
- S =
normalized
throughput
= the fraction of time that the wireless channel is used to successfully transmit packets
- Ptr =
probability that there is
at least one transmission
of the wireless channel
= probability that the channel is busy
Ptr can be computed as follows:
(1) Ҏ[ a terminal transmit ] = τ (2) Ҏ[ a terminal does not transmit ] = 1 - τ => (3) Ҏ[ all (n) terminals do not transmit ] = (1 - τ)n (independent events) => (4) Ҏ[ at least 1 terminal transmit ] = 1 - (1 - τ)n
Ptr = Ҏ[ at least 1 terminal transmit ] = 1 - (1 - τ)n ...... (1)
- Ps =
probability that a transmission
is
successful
Note:
- Ps is
not the same as
Ҏ[ there is exactly one transmission ]
(the probability that there is
exactly one transmission)
- The probability
Ҏ[ there is exactly one transmission ] is an
unconditional probability
(i.e., we not not given any information
at all)S
- In order to have a
successful transmission,
we must first have a
transmission !!!
So Ps is a conditional probability !!!
Ps is a conditional probability defined as follows:
Ps = Ҏ[ a transmission is successful ] = Ҏ[ there is 1 transmission | there are at least 1 transmission ]Ps can be computed as follows:
Ps = Ҏ[ a transmission is successful ] = Ҏ [ 1 transmission | ≥ 1 transmission ] (use: defin. of cond. probab.) Ҏ [ 1 transmission AND ≥ 1 transmission ] = ------------------------------------------- (1 AND ≥ 1 <=> 1) Ҏ [ ≥ 1 transmission ] Ҏ [ 1 transmission ] = -------------------------- (Rewrite) Ҏ [ ≥ 1 transmission ] Ҏ [ 1 transmission ] = -------------------------- (Ҏ[transmit] = τ, n stations) 1 - Ҏ [ 0 transmission ] C(n,1) τ1 (1 - τ)n-1 = -------------------------- (Rewrite) 1 - C(n,0) τ0 (1 - τ)n-0 n τ (1 - τ)n-1 = ----------------- 1 - (1 - τ)n
Ps = Ҏ[ a transmission is successful ] n τ (1 - τ)n-1 = ----------------- ........ (2) 1 - (1 - τ)n - Ps is
not the same as
Ҏ[ there is exactly one transmission ]
(the probability that there is
exactly one transmission)
- S =
normalized
throughput
- Note:
n τ (1 - τ)n-1 Ps × Ptr = --------------- × (1 - (1 - τ)n) = n τ (1 - τ)n-1 1 - (1 - τ)nAlso:
Ҏ[ exactly 1 transmission ] = n τ (1 - τ)n-1 = Ptr × Ps ...... (A)This is not a coincidence, because:
Ҏ[ exactly 1 transmission ] = Ҏ[ we have a successful transmission ] = Ҏ[ successful packet transmission | 0 transmission ] × Ҏ[ 0 transmission ] + Ҏ[ successful packet transmission | ≥ 1 transmission ] × Ҏ[ ≥ 1 transmission ] (This is the Law of Total Probability - See: click here)
Since it is impossible to have a successful transmission when there are 0 transmission, we have: Ҏ[ exactly 1 transmission ] = 0 × Ҏ[ 0 transmission ] + Ҏ[ successful packet transmission | ≥ 1 transmission ] × Ҏ[ ≥ 1 transmission ] = Ҏ[ successful packet transmission | ≥ 1 transmission ] × Ҏ[ ≥ 1 transmission ] = Ҏ[ 1 packet transmission | ≥ 1 transmission ] × Ҏ[ ≥ 1 transmission ] = Ps × Ptr
- Notice
that the following 2 events
have different meanings
- A transmission is
successful
- To compute the probability of
this event, we must
excludes the
event
"there are 0 transmissions"
(Because we are given (implicitly) that "there is a transmission")
- To compute the probability of
this event, we must
excludes the
event
"there are 0 transmissions"
- There is one transmission
- To compute the probability of this event, we must includes the event "there are 0 transmissions"
- A transmission is
successful
- In general, a transmission channel
goes through the following cycles:
- Therefore,
the normalized throughput
can always be
expressed as follows:
Total amount of useful time S = -------------------------------- Total amount of timeHowever, the general approach can rarely be applied directly
- The specific approach suitable for
analyzing the 802.11 protocol
consider the time divided
into slots:
Notes:
- Each station makes
one transmission attempt
in one slot
(The probability that a station will transmit in a slot is equal to τ)
- There are slots with
0 (zero) transmission
These slots have length σ (= time needed to detect a transmission)
- A slot containing
transmissions
by multiple stations
is
not useful
- A slot containing the
transmission
by one station
contains a
useful period
Note:
- Only the portion of
a slot that contains
data bits
are consider
useful (for throughput purposes)
- Due to protocol overhead, a portion of the slot with a successful transmission is not counted as throughput
- Only the portion of
a slot that contains
data bits
are consider
useful (for throughput purposes)
- Each station makes
one transmission attempt
in one slot
- Normalized throughput expression using slots:
Avg. length of the payload in a slot × Ҏ[ slot contains exactly one transmission ] S = ---------------------------------------------------------------------------------------- Avg. length of a slot Avg. length of the payload in a slot × C(n,1) τ1 (1 - τ)n-1 = ------------------------------------------------------------- Avg. length of a slot Avg. length of the payload in a slot × n τ (1 - τ)n-1 = ------------------------------------------------------ (See note: click here) Avg. length of a slot Avg. length of the payload in a slot × Ptr × Ps = ------------------------------------------------- Avg. length of a slot E[P] × Ptr × Ps = ------------------------- ........ (3) Avg. length of a slot (E[P] = Avg. length of the payload in a slot)
- First some more notations....
- E[P] =
Avg. length (= expected value) of the
payload in a slot
- Ts =
Avg. length of a slot when
the slot contains a
successful transmission
- Tc =
Avg. length of a slot when
the slot contains a
unsuccessful transmission
(collision)
- σ =
"empty slot time",
time
needed for a station to detect
the transmission
(of a packet) from
any other station
(This value is like the end-to-end propagation delay on an Ethernet network)
- E[P] =
Avg. length (= expected value) of the
payload in a slot
- We can express
the
"avg. length of a slot"
in terms of
Ts,
Tc and
σ.
- Components needed to compute the avg. length of a slot:
- There are 3 kinds of slots
- slot that contains 0 transmissions (empty slot)
- slot that contains 1 transmission (a successful transmission)
- slot that contains > 1 transmissions (collission)
- The probablilty that a
slot contains
0 transmission is:
Ҏ[ 0 transmission ] = 1 - Ҏ[ at least 1 transmission ] = 1 - Ptr
- The probablilty that a
slot contains
1 transmission is:
Ҏ[ 1 transmission ] = C(n,1) τ1 (1 - τ)n-1 = n × τ × (1 - τ)n-1 = Ptr × Ps
- The probablilty that a
slot contains
> 1 transmission is:
Ҏ[ > 1 transmission ] = 1 - Ҏ[ ≤ 1 transmission ] = 1 - (1 - Ptr) - Ptr × Ps = Ptr - Ptr × Ps = Ptr × (1 - Ps)
- There are 3 kinds of slots
- Avg. length of a slot:
Avg. length of a slot = Weighted average of the length of the 3 diff. kinds of slot = Ҏ[ 0 transmission ] × σ + Ҏ[ 1 transmission ] × Ts + Ҏ[ > 1 transmission ] × Tc = (1 - Ptr)×σ + (PtrPs)×Ts + (Ptr(1-Ps))×Tc .......... (4)
- From Equations (3) and (4), we
have:
E[P] × Ptr × Ps S = ----------------------- ........ (3) (click here) Avg. length of a slot Avg. length of a slot = (1 - Ptr)×σ + (PtrPs)×Ts + (Ptr(1-Ps))×Tc ........ (4)
E[P] × Ptr × Ps S = ----------------------------------------- ........ (5) (1 - Ptr)×σ + (PtrPs)×Ts + (Ptr(1-Ps))×Tc
- The
the avg. length of a slot that contains
a successful or
an unsuccessful transmission
depends on the
transmission method used.
- Recal there are 2 transmission
methods in 802.11:
- The basic access method
(transmits the packet)
- The RTS/CTS method (sender/receiver exchange RTS/CTS before transmitting the actual packet)
- The basic access method
(transmits the packet)
- The following figure shows the
timing in a
packet transmission:
Note:
- H = time to transmit the header of the packet
- P = time to transmit the payload of the packet
- δ =
the propagation delay
- The packet is received after H + P + δ sec !!!
- The following figure shows a
successful and
an unsuccessful transmission
using the
basic access method:
- Therefore:
Ts = ( H + E[P] + SIFT + δ ) + ( ACK + DIFS + δ ) ..... (6) Tc = H + E[P] + DIFT + δ ..... (7)
- The following figure shows a
successful and
an unsuccessful transmission
using the
RTS/CTS access method:
- Therefore:
Ts = ( RTS + SIFT + δ ) + ( CTS + SIFT + δ ) + ( H + E[P] + SIFT + δ ) + ( ACK + DIFS + δ ) ..... (8) Tc = RTS + DIFT + δ ..... (9)
- For variable length payload,
you need to used the
distribution of the packet length
to compute the
average packet length
Since I am not trying to teach you probability, I will skip this analysis
- For simplicity, we will use:
- E[P] = P (constant)
- Believe it or not, we have found
all the expressions to
compute the throughput
for 802.11 !!!
- Summary:
E[P] × Ptr × Ps S = ------------------------- ........ (3) Avg. length of a slot
Avg. length of a slot = (1 - Ptr)×σ + (PtrPs)×Ts + (Ptr(1-Ps))×Tc .......... (4) Ptr = Ҏ[ at least 1 terminal transmit ] = 1 - (1 - τ)n ...... (1) Ps = Ҏ[ a transmission is successful ] n τ (1 - τ)n-1 = ----------------- ........ (2) 1 - (1 - τ)n σ is a constant depending on the physical layer:
Basic: Ts = ( H + E[P] + SIFT + δ ) + ( ACK + DIFS + δ ) ..... (6) Tc = H + E[P] + DIFT + δ ..... (7)RTS/CTS: Ts = ( RTS + SIFT + δ ) + ( CTS + SIFT + δ ) + ( H + E[P] + SIFT + δ ) + ( ACK + DIFS + δ ) ..... (8) Tc = RTS + DIFT + δ ..... (9)
- The authors did a simulation study and compared the
throughput with the
analytic results:
Notes:
- The lines represent the
analytical results
- The discrete points represent the throughput obtained through simulation
- The lines represent the
analytical results
- Another set of comparison,
when using very few stations:
