PowerPoint

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50% between 24 and 48 hours;
20% between 48 and 72 hours.
The chances of survival over 72 hours are quite rare.
: http://baike.baidu.com
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18118
P2Pnet

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Satellite Communications Module (optional)


Core Network
EC Module
QoS
Intranet

CCN
:QOS
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admission control BW:BW: , BW
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Bandwidth Allocation
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CCN forwarding tree
Given CCN forwarding tree T(V,E), where
, v1 is the root of T
C={ci | i=1,2,3,…,n } is the set of uplink capacity of vi
ci : the uplink capacity of vi
ci ≤ cj , if vj is the parent node of vi
c2
c4
c3
c5
c6
v1
BSC/RNC
c7
v2
v3
v4
v5
v6
v7
c1
If tree1,, sort , top-down or button-up
,,,
subtree
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CCN channel class
B={bj | j=1,2,3,…,m } is the set of bandwidth required per channel class
bj : the bandwidth required of channel class j
Channel class
Bandwidth/Channel (bj)
is the set of the profit of channel assigned
fij : the profit of the first channel of class j assigned to vi
g(k) : the profit attenuation function of the k-th same channel class assigned to the same node, for example, g(k)= ,
Model of Bandwidth Allocation for CCN–
CCN profit function for bandwidth allocation
fij & g(k) are user designed
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is the set of the amount of basic channel needs
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CCN channel assigned
aij : the amount of channel class j assigned to vi
Amax objective function
Objective function
such that
is maximized
subject to
node +tree forwaringupstream capacity , aijdij
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W (w1, w2, …, wn) P (p1, p2, …, pn)
C
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0-1 Knapsack Problem is NP-Hard
Richard M. Karp (1972). "Reducibility Among Combinatorial Problems“. In R. E. Miller and J. W. Thatcher (editors). Complexity of Computer Computations. New York: Plenum. pp. 85–103.
Garey, M.; D. Johnson. Computers and Intractability; A Guide to the Theory of NP-Completeness. 1979. ISBN 0-7167-1045-5.
Problem Complexity—Knapsack Problem
A similar dynamic programming solution for the 0-1 knapsack problem also runs in  pseudo-polynomial time .()
BWAnm, tree depth 3full tree15 node, 3channel class, 45??? tree,(nm)
,,model,KP, (, NPC)
Channel class (b1, b2, …, bm) (fi1, fi2, …, fim)
uplink
(ci)
channel?
(vi)

CCN-BA
Problem Complexity—NP-Hard
Given an instance X:[C,W,P] in 0-1 knapsack problem, we can find an instance Y:[V,T,C’,B,F,D,G] in CCN-BA such that an optimal solution ay for Y is also an optimal solution for X. Let instance Y be a one level tree, where V=T={v1}, C’={C}, B=W, F=P, D={0}, and G={1}. Denote the total profit of a solution a for X and Y to be px(a) and py(a), respectively. Because F=P, we can easily prove px(a) and py(a) are equal. For simplicity, both px(a) and py(a) are denoted as p (a).
First, we prove ay is a valid allocation for X. Since B=W, F=P, and C’={C}, any solution of Y whose total bandwidth must less than or equal to the given limit C so that ay must be a valid allocation for X.
Since instance Y is a one level tree, for a similar argument, we can prove that any valid allocation ax for X is also a valid allocation for Y.
Ref. scheduling v1.2 p.36 T=T’(vroot ,Ø), C’=C, B=W, F’=P, D’=0.instance
1. [ay] BWroot,rootBW[ay],(subtree), [ay] is a valid allocation for X (valid X,less than or equal to a given limit C)
2. CCNrootKP
Problem Complexity—NP-Hard
Next, we prove that an optimal solution ay for Y is also an optimal solution for X by contradiction. As we have proved, ay is also a valid allocation for X, whose total profit is p(ay). Assume ay is not an optimal allocation for X, there must be another allocation ax, whose total profit p(ax) is greater than p(ay). And any valid ax is also a valid allocation for Y, whose total profit is p(ax), which is greater than p(ay). This contradicts to the assumption that ay is an optimal solution for Y. As a result, ay must be an optimal solution for X. The reduction of CCN-BA to 0-1 Knapsack Problem is done. The proof of NP-hardness of CCN-BA is straightforward. Q.E.D.
Bandwidth Allocation for CCN is a nested 0-1 knapsack problem and NP-Hard
Px, Py function , Px([a]) = Py([a])
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profit densityF/bj
CCN-BA(T,C,B,F,D)
Set A={} /* the amount of channel class j assigned to vi*/
Set V={} /*channelcandidate node*/
end if
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sort F of V by profit density
let fij ·g(k) be a highest profit density assign of F subject to min_ci(from
vi to v1) ≥ bj
plus 1 to aij
update fij ·g(k+1) to F
if any ci < minimum bj
delete vi & descent(vi) from V
end if
end while
CodingCFstate, debug
let fij(x) be a highest profit density assign of F : def: profit densityBWleveli
plus fij(x) to p
Basic needs (dij)
CCN forwarding tree
Attenuation Function g(k)=
channel class vi
c1=58
c3=20
c2=30
c4=20
c5=20
*
c1=23
c3=23
c2=9
c4=13
c5=13
*
c1=18
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Profit of channel assigned
v1
16
1.33
9.19
1.31
5
1.00
v2
15
1.25
7.07
1.01
7
1.40
v3
25
2.08
14.14
2.02
10
2.00
v4
18
1.50
8.49
1.21
8
1.60
v5
29
2.42
16.26
2.32
10.61
2.12
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c1=6
*
c1=1
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Profit of channel assigned
v3
*
Allocation (aij)
Total profit = basic needs profit + channel assigned by profit density =(13+10+20+12+23 )+15+25+10 =128
channel class vi
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the remaining uplink capacity (ci) per node (vi)
minimum capacity from per node to root
(minimum ci from vi to v1)
candidate channel class can be allocated per node
candidate allocation sorted by profit density
ProfitAttenuationUpdate( )
BAG Solution
6~10

2.00 GB

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BAG
Amount of base station
6
6
3
9
7
7
4
11
8
6
6
12
9
5
3
8
10
4
3
7
28
19
47
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CCN
CCN()CCN