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Programme in Science
An Adjusting Capacity Analysis on Gupta and Kumar’s
Paper with considering hidden terminals problem
Supervisor: Professor Tay Yong Chiang
Zeng Zhan
U052105J
Department of Mathematics
National University of Singapore
2006/2007
Acknowledgement
I want to thank Professor Tay for his kind supervise and guide not only to the problem
itself, but also what the real and professional mathematics is.
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Catalog
1. Introduction
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2. Definition of hidden terminals problem
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3. Analysis on Gupta and Kumar’s Paper
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4. Collision Probability Approach
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5. Disjointed factor Approach
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6. Discussion
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7. References
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1. Introduction
While the term wireless network may technically be used to refer to any type of
network that is wireless, the term is most commonly use to refer to a
telecommunications network whose interconnections between nodes is implemented
without the use of wires, such as a computer network. In a seminal paper, Gupta and
Kumar analyzed the throughput capacity of a wireless network (IEEE Trans.
Information Theory, Mar. 2000). In their paper, they analyzed the networks which
consist of a group of nodes that communicate with each other over a wireless channel
without any centralized control. One of their main results is that they deduce the
upper bound of transport capacity for Arbitrary Network under the Protocol Model is
of order
bit-meters per second, where n is the number of nodes used in the
network. See Gupta and Kumar [1]. This result is worth considering for designers. It
implies that the throughout furnished to each user diminishes to zero as the number of
users increases since the upper bound of the throughput is of order 1/
bits-meters
per second for each node. However, in their analysis of upper bound of transport
capacity of Arbitrary Network under the Protocol Model, they do not take hidden
terminals into account, which are a long-standing problem in wireless networks.
In this report, firstly, we will briefly describe what hidden terminals problem is and
why Gupta and Kumar do not take hidden terminals into account. Then, we will
address this issue by using two approaches to incorporate the hidden terminals
problem under Gupta and Kumar’s model. Finally, we will discuss the significance of
solving this problem.
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2. Hidden Terminal Problem
The hidden terminal problem occurs because the sender cannot hear as far as the
receiver. Let us say that if the send “thinks” that the channel is idle, it starts to
transmit to the receiver. At the same time, the receiver can hear the transmission of
some other transmitters. Therefore, a collision will occur at the receiver which causes
that the transmission to the receiver is unsuccessful. To understand the hidden
terminals problem, let us consider an example depicted in Figure 1.
Figure 1. Hidden terminals problem
Suppose that node Xr is receiving a transmission from node Xs over the some
subchannel at the same time that node 3 is transmitting a packet to node 4 over the
same subchannel. From Figure 1, we can see that Xs and Xs’ can not hear each other
since they are outside the hearing area of each other, where the areas of two circles
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represent the hearing area. So both of them will think the channel is idle and start to
transmit. However, node Xr can hear the transmission of node Xs’ to node Xr’.
Therefore, a collision occurs on node Xr and node Xr cannot decode the information
sent by node Xs.
3. Why do we say that they do not take hidden terminals into
account?
In their paper, they consider an Arbitrary Network, in which there is no restriction
on the location of the nodes. Further, under their Protocol Model, they say that if node
Xs transmits over the mth subchannel to node Xr, then this transmission is
successfully received by node Xr if
| Xs’ – Xr | ≥ (1 + Δ) | Xs – Xr |
for every other node Xs’ that is simultaneously transmitting over the same subchannel,
where the quantity Δ > 0 models situations where a guard zone is specified by the
protocol to prevent a neighboring node from transmitting on the same subchannel at
the same time. However, look at the case depicted in the Figure 2 below, it satisfies
the inequality above, but the transmission is unsuccessful due to the hidden terminals
collision.
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Figure 2. A counter example for Gupta and Kumar’s analysis
Therefore, their analysis is for the no-hidden-terminal environment.
In the following, I have used two approaches to address this issue.
4. Approach 1: collision probability approach
I have mentioned above that their analysis is based on the no-hidden-terminal
wireless network. The existence of hidden terminals can reduce the probability of
successful transmission between two nodes. So that reduces the transport capacity of
arbitrary network.
Let us denote the collision probability and the transport capacity of arbitrary
network by p and T, respectively. The expected new transport capacity with
considering the hidden terminals
Tnew = (1-p)⋅Told .
This is easy to understand. (
old
) represents the transport capacity due to the time
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spent in resending packets because of hidden terminals collision. Therefore, with
considering the hidden terminals, Told should exclude that fraction. So the expected
new transport capacity Tnew = (1-p)⋅Told.
Now I come to describe how to compute the collision probability p.
Figure 3. Links
From a sending node’s perspective, its sending link, say link 1 in Figure 3, can be
in one of three potential states: transmission state, channel busy state, and channel
idle state.
Let Xi denote the normalized “self” airtime, which includes the successful and
collided transmission time, where the word “normalized” means the total time for the
whole network’s working is 1 unit of time.
Let q be the fraction of time used for transmitting a data packet. For a specific
topology of the wireless network, q is a fixed number. For more details on q, see Yan
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Gao and Dah-Ming Chiu [2].
Therefore, (q⋅Xi) represents the normalized times spent in transmitting data packet
for link i.
Since Xr is in the hearing area of Xs’ from Figure 3, hidden terminals problem
occurs if the time interval of the transmissions of two pairs of nodes overlaps each
other. Let A be the event that link1 overlaps link2, then the overlap probability P (A)
= q⋅X1 + q⋅X2.
Proof:
Referring to the figure 4 below, we can see that once the two time interval
overlaps, the collision happens.
Figure 4. Overlap probability
Let’s fix the time interval (q⋅X1) first. The collision occurs when q⋅X2 located
from position A to position B. Since q⋅X1 and q⋅X2 are normalized time, P(A) = q⋅
X1 + q⋅X2
From above, I have got the collision probability for one link. This can determine
the transport capacity for this link. The throughput capacity is defined as the
minimum link throughput capacity of a path. Therefore, the collision probability for
the whole wireless network equals the maximum of the collision probability of every
link.
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Therefore, the collision probability p = P(A)max
Finally, the expected new transport capacity is bounded above by (1 - P(A)max)⋅
(old upper bound), i.e.
(1 - P(A)max)⋅C⋅
where C is a constant for a specific topology of the wireless network. See Gupta and
Kumar [1].
Furthermore, P(A)max depends on the size of n, the number of nodes in the wireless
network. When n increases, P(A) will follow to increase since nodes are more related
to each other in the sense of distance. So the expected new transport capacity will be
smaller.
5. Approach 2: Disjointed factor approach
Generally speaking, this approach is going to add inequalities to the Gupta and
Kumar’s paper to make transmissions successful with considering the hidden
terminals problem, then adjust their upper bound based on all the inequalities and
restrictions.
In Section 2, I have explained the hidden terminals problem. Let us look back to
the Gupta and Kumar’s paper on deriving the upper bound of transport capacity of
Arbitrary Network under Protocol Model, where they consider node Xs transmitting
to a node Xr. Then this transmission is successfully received by node Xr if
| Xs’ – Xr | ≥ (1 + Δ) | Xs – Xr |
for every other node Xs’ simultaneously transmitting to some other node Xr’. From
Section 3, we have explained this inequality is not enough to make transmission
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successful. We have to adjust it by adding some more inequalities.
Hidden terminals problem occurs when Xr can hear Xs’ but Xs cannot. Therefore,
in order to avoid this problem, Xr must not hear Xs’. In the Protocol Model, this
means
| Xr – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
where Xr’ is receiving the transmission from Xs’.
Other situation which we also need to consider is the case that Xs can hear Xs’. In
this case, Xs thinks the channel is busy and will wait for Xs’s finishing transmitting.
Therefore, in order to let Xs transmit properly, Xs must not hear Xs’. In the Protocol
Model, this means
| Xs – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
Above all, in order to let Xs transmit to Xr properly where Xs’ is transmitting to
Xr’ at the same time and over the same subchannel, we must have three inequalities
to follow, which are stated below.
| Xs’ – Xr | ≥ (1 + Δ) | Xs – Xr |
| Xr – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
| Xs – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
After we have the basic conditions, we can adjust their proof and final results.
From the proof of Gupta and Kumar’s paper, what we need to adjust the disjointed
factor delta/2.
In their paper, they deduce that | Xr – Xr’ | ≥ (Δ /2)⋅( | Xs – Xr | + | Xs’ – Xr’ | ).
Therefore, they say that disks of radius Δ /2, which we call the disjointed factor, times
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the lengths of hops centered at the receivers over the same subchannel in the same
slots are essentially disjoint.
After we apply the new inequalities to the situation that Xr is receiving a
transmission from Xs over the mth subchannel at the same time that Xr’ is receiving a
transmission from Xs’ over the same subchannel, we have six inequalities since both
two pairs must transmit properly.
| Xs’ – Xr | ≥ (1 + Δ) | Xs – Xr |
| Xr – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
| Xs – Xs’ | ≥ (1 + Δ) | Xs’ – Xr’ |
The three inequalities above are to make sure Xs is transmitting to Xr properly.
| Xs’ – Xr | ≥ (1 + Δ) | Xs’ – Xr’ |
| Xr – Xs’ | ≥ (1 + Δ) | Xs – Xr |
| Xs – Xs’ | ≥ (1 + Δ) | Xs – Xr |
The three inequalities above are to make sure Xs’ is transmitting to Xr’ properly.
With more restrictions, the disjointed factor must be bigger.
However, with these equations, we will only can get | Xr – Xr’ | ≥ (Δ /2)⋅( | Xs –
Xr | + | Xs’ – Xr’ | ) in the worst case. See figure 5.
Figure 5.
The worst case
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This does not affect the disjointed factor. However, this kind of case happens rarely
under new conditions Therefore, I will use probability to calculate the expected
disjointed factor since in the most cases, the disjointed factor is larger than Δ /2.
I want to compute the probability that | Xr – Xr’ | ≥ min{ | Xs’ - Xr |, | Xs – Xs’ |,
| Xr – Xs’ |}denoted by P(B), where B is the event that | Xr – Xr’ | ≥ min{ | Xs’ - Xr
|, | Xs – Xs’ |, | Xr – Xs’ |}.
For any two pair, I can fix three of them and move the other one. Let’s say Xs, Xr,
Xs’ are fixed. See figure 6. In order to make event B happen, Xr’ should be out of the
circle with centre Xr, radius equaling min{ | Xs’ - Xr |, | Xs – Xs’ |, | Xr – Xs’ |}. In
figure 6, min{ | Xs’ - Xr |, | Xs – Xs’ |, | Xr – Xs’ |} =
| Xs – Xs’ |
Figure 6. Topology of the two pairs
P(B) = (1 – area of the circle) / 1(the whole area ), where Gupta and Kumar set the
whole area 1 and nodes are uniformly distributed in the area.
When event B happens,
| Xr – Xr’ | ≥ min{ | Xs’ - Xr |, | Xs – Xs’ |, | Xr – Xs’ |}≥ (1 + Δ) | Xs – Xr |
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| Xr – Xr’ | ≥ min{ | Xs’ - Xr |, | Xs – Xs’ |, | Xr – Xs’ |}≥ (1 + Δ) | Xs’ – Xr’ |
Therefore, adding two equations together, we can get | Xr – Xr’ | ≥ ((Δ +1)/2)⋅( |
Xs – Xr | + | Xs’ – Xr’ | ).
If the event B does not happen, the disjointed factor is still Δ /2.
Therefore, the expectation of the disjointed factor
Q=
For the whole wireless network, we just take the minimum of P(B). Once we know
the topology of the wireless network, we can compute P(B)min.
We can deduce that the expected upper bound of the transport capacity is Δ /(Δ +
P(B)min ) times the old one since we use (Δ + P(B)min)/2 instead of Δ/2.
Furthermore, P(B)min actually depends on n. As n increases, r will be larger since
the average distance between two nodes decreases. When n is sufficiently large, r = 1.
6. Discussion
(1) From the two approaches above, the final results tell us that once we take hidden
terminal into account, the transport capacity will be lower. This can be predicted by
our intuition since with more restrictions, the wireless network will be less activated.
(2) The results deduced from two approaches are consistent with Gupta and Kumar’s
notion. That is when the number of users is increased, the throughout furnished to
each user diminishes to zero. This implies that wireless networks connecting smaller
numbers of users, or featuring connections mostly with nearby neighbors are the
really right choice.
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7. References
[1] Piyush Gupta and P. R. Kumar, “The Capacity of Wireless Networks”, IEEE
Transactions on Information Theory, VOL.No. 2, March 2000
[2] Yan Gao, Dah-Ming Chiu and John C. S. Lui “Determing the End-to-end
Throughput Capacity in Multi-Hop Networks: Methodology and Applications”
[3] Jinyang Li, Charles Blake, Douglas S.J.De Couto, Hu Imm Lee and Robert
Morris“Capacity of Ad Hoc Wireless Networks” The Eighth ACM International
Conference on Mobile omputting and Networking (MobiCom’ 02)
[4] Ping Chung Ng and Soung Chang Liew “Offered Load Control in IEEE802.11
Multi-hop Ad-hoc Networks” 0-7803-8815-1-04 2004 IEEE
[5] Y.C. Tay and K.C. CHUA, “The Capacity analysis for the IEEE 802.11 MAC
Protocol”
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