Chapter 6. Bandwidth Allocation
6.2. Bandwidth Allocation Mathematical Model
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Chapter 6. Bandwidth Allocation
6.1. Considerations of Bandwidth Allocation
Because the information needs to be forwarded through the neighbor station, which occupies its bandwidth, the number of users of each base station needs to be rationally distributed to meet the disaster response demands regarding the number of communications channels of each base station to avoid allocation disequilibrium. Take SiChuan Earthquake [45] for example, the disaster areas have ten-time phone calls than usual in internal areas;
5-to-6-time phone calls than usual in external areas; and 80-time phone calls than usual from Beijing to the disaster areas. Obviously, such a traffic demand is far beyond the capacity of CCN. If the bandwidth is not allocated appropriate, it may result in the degradation of disaster response inefficiency. The worse-case scenario is that the bandwidth may be occupied by the less disastrous area and the more disastrous areas may not receive any bandwidth at all.
Therefore, it is necessary to allocate bandwidth properly to the base stations in order to maximize disaster response inefficiency.
6.2. Bandwidth Allocation Mathematical Model
Assuming a CCN forwarding tree is given, each node has m channel classes to choose.
When a channel of a particular class is assigned to a node, the disaster response profit corresponding to that class is earned. The channels assigned to the node cannot exceed the capacity of upward links of the wireless links. And, an ancestor node has to forward the traffic
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of its descendant nodes. The channels assigned to it and its descendant cannot exceed the bandwidth capacity of it. This makes CCN bandwidth allocation (CCN-BA) problem become a nested 0-1 Knapsack problem.
The input parameter are G’(V,E), B, P, A, F that are defined as follows. The input graph, G’, is the network topology of CCN.
V = [vi]n× 1, where i = 1,2,…,n-1 and vi is the disaster operation efficiency of node i. v1 is the node that has an external link and v0 is a virtual node that represents the core network.
E =[ei,j]n× n. eij is the edge between vi and vj, where eij represents the bandwidth capacity between vi and vj . e0,1 and e1,0 are the external downlink and uplink bandwidth.
B = [bk]m× 1 is the set of channel class, where bk is the required bandwidth of channel class k.
P=[pi,j]n× n. pi,j is the path from vi to vj.
A=[ai,j,k]n× n× m , where ai,j,k represents the amount of channel class k assigned to path pi,j.
F=[fi,j,k]n× n× m. fi,j,k(ai,j,k) is the profit of channel class k assigned to path pi,j.
Objective of the CCN bandwidth allocation (CCN-BA) problem is to find a matrix A, such that
Maximize
∑ fi,j,k(ai,j,k) , for all 0 ≤ i,j < n , 1 ≤ k ≤ m
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Subject to
ei,j ≥ ∑ aq,r,k× bk , for all ei,j ϵ pq,r
6.3. Complexity Analysis
In this section, we will prove that the time complexity of CCN-BA problem is NP Hard.
Some variables are added in order to prove the time complexity of CCN-BA problem. The definitions of these variables are as follows.
C is a positive number that represents the capacity of the knapsack.
W={wi|i=1,2,…,m} represents the weights of items, where wi is the weight of the item that belongs to type i. There are m types of items.
S={si|i=1,2,…,m} represents the values of items, where si is the value of the item that belongs to type i.
Z={zi|i=1,2,…, m} represents the amounts of items filled the knapsack.
X(C,W,S) is a problem instance of 0-1 Knapsack problem. Given a set of items, with weights, W, and values, S, 0-1 Knapsack problem is to select the valuable items to fill the knapsack while maximizing the total value and being constrained by the size of the knapsack, C. 0-1 Knapsack problem is NP Hard.
Y(G’(V,E),B,P,F) is a problem instance of CCN-BA problem. Definitions of the input parameters, G’(V,E),B,P and F, can be referred to Section 6.2.
(A) Lemma 1: For any 0-1 Knapsack problem instance, X(C,W,S), there exist a
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CCN-BA problem instance, Y(G’(V,E),B,P,F), that can be transformed into the problem X in polynomial time.
Proof: Given any 0-1 Knapsack problem instance, X(C,W,S), we can find a dedicated CCN-BA problem instance, Y(G’(V,E),B,P,F) that can be transformed into the problem X in polynomial time. Given a set of channels, with bandwidth consumptions B, and profits, F, CCN-BA problem instance, Y(G’(V,E),B,P,F), is to assign channels to paths, P, while maximizing the total profit and being constrained by bandwidth capacities of paths, P. Since, the bandwidth capacities of P are decided by the edges, E, CCN-BA problem is similar to a nested 0-1 Knapsack problem. The process of transforming problem Y into problem X is showed as follow.
Let the size of G’(V,E) be equal to 1, |E|=1. V={v0,v1}; E={e0,1}; P={p0,1} and F={f0,1,k}. The path, p0,1, is composed by e0,1. The bandwidth capacity of p0,1 is equal to e0,1. Given a set of channels, with bandwidth consumptions B, and profits, F, CCN-BA problem instance, Y(G’(V,E),B,P,F), is to assign channels to the path, p0,1, while maximizing the total profit and being constrained by bandwidth capacities of paths, p0,1. Therefore, the CCN-BA problem is reduced to a 0-1 Knapsack problem by setting the size of G’(V,E) to 1.
Let the bandwidth consumptions of channels, B, be equivalent to the weights of items, W. Let the profit gains, F, be equivalent to the values of item, S. Let the bandwidth capacity of the path, p0,1, be equivalent to the size of knapsack, C. Then, the CCN-BA problem instance, Y(G’(V,E),B,P,F), is equivalent to the 0-1 Knapsack problem instance, X(C,W,S). Obviously, the time complexity of the transformation is polynomial. Q.E.D.
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(B) Lemma 2: If Sa is a valid solution of X(C,W,S), then Sa is also a valid solution of Y(G’(V,E),B,P,F).
Proof: From Lemma 1, for any 0-1 Knapsack problem instance, X(C,W,S), we can find a CCN-BA problem instance, Y(G’(V,E),B,P,F), where |E|=1,B=W, F=S and E=C.
Y(G’(V,E),B,P,F) is equivalent to X(C,W,S).
Sa is a valid solution of X(C,W,S). Now, we prove that Sa is also a valid solution of Y(G’(V,E),B,P,F). Let Sa={sai|i=1,2,…,m}. There are m types of items and sai denotes the amount of type i items fill the knapsack. Since Sa is a solution of X(C,W,S), the formula
∑𝑚𝑖=1𝑠𝑎𝑖 × 𝑤𝑖 ≤ 𝐶 denotes that the total weight of the items is constrained by the capacity of the knapsack. Let sai represents the amount of channel class i assigned to path p0,1. Because B=W, E=C and ∑𝑚𝑖=1𝑠𝑎𝑖 × 𝑤𝑖 ≤ 𝐶, it can conduce ∑𝑚𝑖=1𝑠𝑎𝑖× 𝑏𝑖 ≤ 𝐸 which represents the total bandwidth consumed by the assigned channel is equal to or less then the bandwidth capacity of E. Therefore, Sa is also a valid solution of Y(G’(V,E),B,P,F). Q.E.D.
(C) Lemma 3: If Sb is a valid solution of Y(G’(V,E),B,P,F), then Sb is also a valid solution of X(C,W,S).
Proof: Since Sb is a valid solution of Y(G’(V,E),B,P,F), ∑𝑚𝑖=1𝑠𝑏𝑖 × 𝑏𝑖 ≤ 𝐸. Let |E|=1,B=W, F=S and E=C, Y(G’(V,E),B,P,F) is equivalent to X(C,W,S). The Lemma3 can be proved similar to Lemma2. Q.E.D.
(D) Lemma 4: If Sa is an optimal solution of X(C,W,S), then Sa is also an optimal solution of Y(G’(V,E),B,P,F).
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Since Sa is a solution of X(C,W,S), Sa is also a solution of Y(G’(V,E),B,P,F) where |E|=1,B=W, F=S and E=C by Lemma 2. Let TX(Sa)=∑𝑚𝑖=1𝑠𝑎𝑖 × 𝑠𝑖 denotes the total value of sa in problem X(C,W,S) and TY(Sa)= ∑𝑚𝑖=1𝑠𝑎𝑖× 𝑓0,1,𝑖 denotes the total profit of Sa in problem Y(G’(V,E),B,P,F). Because F=S, f0,1,i = si. It can conduce that TY(Sa)=TX(Sa).
Now, we prove by contradiction. Assume that Sa is an optimal solution of X, but not an optimal solution in Y. Since Sa is not an optimal solution in Y, then there exist a solution Sb
where TY(Sa)<TY(Sb). Because TY(Sb)=TX(Sb), TY(Sa)=TX(Sa) and TY(Sa)<TY(Sb), it implies that TX(Sa)<TX(Sb). This contradicts to the assumption that Sa is an optimal solution of X.
Therefore, Sa must be an optimal solution of Y. Q.E.D.
Let TCX(Sa) be the time complexity of obtaining optimal solution Sa in X. From Lemma 1, For any 0-1 Knapsack problem instance, X(C,W,S), there exist a CCN-BA problem instance, Y(G’(V,E),B,P,F), that can be transformed into the problem X in polynomial time. Assume that the transformation time is TY,X and TCY(Sa) be the time complexity of obtaining the optimal solution Sa in Y. TCX(Sa) = TCY(Sa) - TY,X. Because TY,X is polynomial and TCX(Sa) is NP Hard, TCY(Sa) is NP Hard.
Since Sa is an optimal solution of Y which is proved in Lemma 4, TCY(Sa) is the time complexity of CCN-BA problem and had proved as NP Hard. Q.E.D.