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Network Design and Analysis-----Wang Wenjie Notes on Routing: 1
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Network Architecture
Network Design and Analysis
Wang Wenjie
Wangwj@gucas.ac.cn
Network Design and Analysis-----Wang Wenjie Notes on Routing: 2
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Notes on Routing
Network Design and Analysis-----Wang Wenjie Notes on Routing: 3
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Topics
• Router 的基本结构• 路由算法概述• 常用路由算法• 最短路由算法• 自适应最短路由的稳定性分析• 最优化路由
– Formulating a Communication Network Flow Problem
– 最优化路由及其特性– 最优化问题求解
Network Design and Analysis-----Wang Wenjie Notes on Routing: 4
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Formulating a Communication Network
Flow Problem
• A Four-Node Network Example
– Node-Arc Formulation
– Arc-Path Formulation
Network Design and Analysis-----Wang Wenjie Notes on Routing: 5
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Node-Arc Formulation(1)
• Assume:
– Traffic only from node 1 to 4 and it is pps (packets per sec). – Packet length to be exponentially distributed
with mean length bits/1
Network Design and Analysis-----Wang Wenjie Notes on Routing: 6
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Node-Arc Formulation(2)
• Question:
What should be the network objective as far as delay is concerned?
• a possible network objective is to minimize maximum delay on a link
Network Design and Analysis-----Wang Wenjie Notes on Routing: 7
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Arc-Path Formulation(1)
• The key assumption :
– Generate a set of possible path between the origin and the destination node a priori
– Enumerate possible (unknown) flows on paths 1-2-4, 1-2-3-4 and 1-3-4 as y1, y2 and y3, respectively
Network Design and Analysis-----Wang Wenjie Notes on Routing: 8
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Arc-Path Formulation(2)
• Other objective function:
minimize average delay per packet in the network
• Note that other objectives are also possible depending on network objective
Network Design and Analysis-----Wang Wenjie Notes on Routing: 9
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General Formulation
There are N nodes and L links in the network
For the arc-path formulation, suppose, on average, there are p paths for each pair
Node-Arc Arc-Path
Single Commodity
# of constraints N 1
# of variables L p
Single Commodity
# of constraints N2/(N-1)/2 N(N-1)
# of variables LN(N-1)/2 pN(N-1)/2
Network Design and Analysis-----Wang Wenjie Notes on Routing: 10
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最优路由 (1)
• 对于一个网络,设在任一节点对 w= ( S , D )之间可以同时通过多条路径将输入到 S 端的业务流 rw 送到目的节点。
• 设任一节点对之间的所有路径用 Pw 表示,各路径上的流量用 xp 表示,这些流量的集合用 Xw 表示。
• 根据定义:各条路径上的流量之和等于输入流量,并且各条链路上的流量一定不小于 0
• 设链路( i,j )上的流量用 Fij 表示:
0 ,w
p wp P
p w
x r w W
x p P w W
对所有的
对所有的
ij pF x 包含(i , j )的所有路径p
Network Design and Analysis-----Wang Wenjie Notes on Routing: 11
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最优路由 (2)
• 寻找最优路由的目的:使网络的成本最低,就是
• Dij 是一个单调函数,它是每条链路的成本。常用的成本函数有:
Cij 是链路( i , j )的容量, dij 是链路的时延(包括传播时延和处理时延)
( ) ijij ij ij ij
ij ij
FD F d F
C F
( )ij ijD F(i , j )
Network Design and Analysis-----Wang Wenjie Notes on Routing: 12
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最优路由 (3)
• 则最优路由的目标是寻找最佳的 Xw ={xp } ,使得成本函数最小:
S/T:
( , )
min ij pi j p
D x
(i , j ) 包括 的所有路径
0 ,w
p wp P
p w
x r w W
x p P w W
对所有的
对所有的
Network Design and Analysis-----Wang Wenjie Notes on Routing: 13
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最优路由特性 (1)
• 主要讨论如何利用成本函数的一阶导数表示最优路由:假设 Dij是 Fij 是的可微函数,定义在 [0 , Cij )上。
• 令 x 为各路径流量组成的一个矢量,则成本函数为:
对 xp 求偏导得:
如果将 D’ij 取值为链路 (i,j) 的长度,则上面的导数是路径 p 上各链路长度之和,它可以看作是路径 p 的长度。可将该导数称为路径 p 的一个微分长度
( , )
( ) ij pi j p
D x D x
(i , j ) 包括 的所有路径
'
( , )
( )ij
i jp
D xD
x
含有 的所有路径p
Network Design and Analysis-----Wang Wenjie Notes on Routing: 14
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最优路由特性 (2)• 令 x*={X*p} 是最佳流矢量,即成本函数最小。• 如果某一路径 p 上的流量 X*p >0 ,则将路径 p 上很少的流量移到相
同 SD 对的另一条路径 p’ 上,必然不会降低成本,即:
因而
该式为 x* 最佳化的必要条件。即最佳路径的流量仅在具有最小一阶微分长度的路径上为正。此外,在最佳的情况下,如果 SD 对的输入流量是分配在几条路径上,则这几条路径必定具有相同长度。
如果 Dij 是一个凸函数,则上式也是 x* 最佳化的充分条件。
'
* *( ) ( )0
pp
D x D x
x x
* ** ( ) ( )
0'pp p
D x D xx
x x
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最优路由的可行方向 (1)
• 从最佳路由特性知道:仅当输入业务流沿着最小一阶微分长度( MFDL )路径流动时才是最佳的。即,如果给定一组流量不是最佳的,则必然有一部分流量是流经非 MFDL 路径的。如果把一部分非 MFDL 上的流量移到 MDFL 路径上,则性能就会改变,成本函数下降。
• 设 x={xp} 为满足约束条件可行解,沿 x={xp} 的方向改变 x : x=x+x ,使得 D(x+x)<D(x) 。两个问题: x 方向应满足什么条件?– 步长应如何选择?
Network Design and Analysis-----Wang Wenjie Notes on Routing: 16
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最优路由的可行方向 (2)
x 可行的方向:可行方向就是 x 在 x 方向做一个微量的变化,得到的新的 x 矢量仍是一个满足约束条件的可行矢量:
微量调整后:
比较后得到:
并且对所有 xp=0 的路径,应当有 xp 0
w
p wp P
x r
( )w
p p wp P
x x r
0w
pp P
x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 17
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最优路由的可行方向 (3)
x 下降的方向: x 沿着 x 方向变为 x+x 时,其成本应当下降
• 下降迭代法:在搜索方向上所得到的最佳点处的梯度与该搜索方向正交:
左边实际上是 G()=D(x+x) 在 x+x=0 处的一阶导数
( )0
w
pw W p P p
D xx
x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 18
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最优路由的可行方向 (4)
• 满足上述条件的常用算法要求 xp 满足的条件:
1
2 对于所有非最短路径 p ,应当有 xp 0
3 、至少有一个 xp <0 ,否则迭代结束。
0w
pp P
x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 19
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最优化问题求解
Optimization Algorithms
• Single Variable Problem
• Multi-variate unconstrained minimization problem
• Multi-variate constrained optimization problem
• Optimality Condition
• Frank-Wolfe(Flow Deviation) Algorithm
Network Design and Analysis-----Wang Wenjie Notes on Routing: 20
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Single Variable Problem
• Problem:
Optimization Problem with single and continuous variable, i.e., x IR.
• Objective
Building a framework from single variable onward to consider multi-variate problem
Network Design and Analysis-----Wang Wenjie Notes on Routing: 21
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Convex Function
• Definition:convex function
A function f(x), x IR is said to be convex, if for any x and y IR, the following condition is satisfied:
f(x+(1- )y)
f(x) +(1- )f(y)
[0,1]
x y
f(x) +(1- )f(y)
x+(1- )y
Network Design and Analysis-----Wang Wenjie Notes on Routing: 22
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Newton Method(1)
• Objective:– Give an algorithmic solution for some instances that f(x) or f'(x)
may not be “easy" to arrive at the solution.
• Ideas: – Assuming : f is twice differentiable. – If x is an optimal satisfies , then : f’(x)=0
– Linearization of left hand side around a point xk, and set to 0:
f’(xk)+f’’(xk)(x- xk)=0
Rearranging: x= xk - f’(xk)/f’’(xk)
let: xk+1 =x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 23
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Newton Method(2)
Step 0: Start with x0, set k=1,
choose a tolerance >0, and maximum iteration count Kmax
Step 1: Compute:
xk+1= xk - f’(xk)/f’’(xk)
Step 2: if | xk+1 - xk| < or k Kmax stop
else k k+1 and go to step 1
Note: f’’(x) 0
function f(.) has to be twice differentiable
Network Design and Analysis-----Wang Wenjie Notes on Routing: 24
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Golden Section method(1)
• Objective:
Minimizing a unimodal function in a given interval
• The problems is:
• T=0.61803399
)x(fmin]b,a[x
f0
f1f2
f3
x0=a x1 x2 x3=b
Network Design and Analysis-----Wang Wenjie Notes on Routing: 25
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Golden Section method(2)
T=0.61803399
choose >0 for tolerance on stopping
x0=a ; f0=f(x0)
x3=b ; f3=f(x3)
x1= x0+(1-T) (x3 – x0) ; f1=f(x1)
x2= x0+T (x3 – x0) ; f2=f(x2)
while(abs(x3 – x0 )> (abs(x1)+abs(x2 ))) do
if f1 > f2 , then /* x3 remains the same */ x0 = x1 ; f0 = f1 x1 = x2 ; f1 = f2 x2 = x0 + T *(x3 – x0 ) f2 = f (x2 )
else /* x0 remains the same */ x3 = x2 ; f3 = f2 x2 = x1 ; f2 = f1 x1 = x0 + (1- T )*(x3 – x0 ) f1 = f (x1 ) endif
if( f1 < f2 ) then minvalue= f1 xmin= x1 else minvalue= f2 xmin= x2 endif
Network Design and Analysis-----Wang Wenjie Notes on Routing: 26
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Multi-variate unconstrained
minimization problem
• Problem:
Optimization Problem with multiple and continuous variable, i.e., x IRn.
• The general representation for n-dimensional unconstrained optimization problem is
• Note that :
)x(fminnIRx
IRIR:f n
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Necessary and sufficient Optimality
Condition
• A necessary and sufficient condition for x* IRn to be an optimal solution is that:
f(x*)=0 and for y IRn , yTf(x*)y 0
• A positive semi-definite matrix,M, satisfies the condition:
yTMy 0
A positive definite matrix, M, satisfies the stronger condition:
yTMy > 0
Network Design and Analysis-----Wang Wenjie Notes on Routing: 28
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Newton method for multi-variate optimization
problem
• Step 0: Start with x1. set k=1,
choose tolerances 1,2>0, and max iteration count Kmax
• Step 1: Compute
xk+1 xk – [2f(xk)]-1f(xk)
• Step 2: if || xk+1 - xk ||< 1
or ||2f(xk+1)|| < 2
or k Kmax stop
else k+1 k and go to step1
Network Design and Analysis-----Wang Wenjie Notes on Routing: 29
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Multi-variate constrained optimization
problem
• Objective function of a constrained minimization problem :
subject to:
inequality constraints: gj(x) 0, j=1,..,m
equality constraints: hi(x)=0, j=1,…, p
)x(fminx
Network Design and Analysis-----Wang Wenjie Notes on Routing: 30
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Multi-variate constrained optimization
problem(Cont’d)
• If the objective function and the constraints are all LINEAR, then we have a linear programming (LP) or linear optimization problem:
subject to: Bx 0 , x 0
xcmin T
x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 31
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Optimality Condition
• Lagragean for the general non-linear programming(NLP) problem:
• In the second line
u denotes the vector:u=(u1,…, um)
g(x) denotes the vector:g(x)=(g1(x),…, gm(x))
Similarly for v and h(x)
p
jjjj
m
ji hv)x(xgu)x(f)v,u,x(L
11
)x(hv)x(gu)x(f TT
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Optimality Condition(Cont’d)
• Optimality condition for NLP is:
There exists such that(note:vj is unrestricted):v,u,x
Network Design and Analysis-----Wang Wenjie Notes on Routing: 33
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Optimality Condition(Cont’d)
• If there is noly gj(x) 0, NO hi(x)=0:
There exists such that:u,x
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Optimality Condition(Cont’d)
• For following problem:
s/t Ax=b
The lagrangean is:
The optimality condition is:
)x(fminx
Network Design and Analysis-----Wang Wenjie Notes on Routing: 35
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Frank-Wolfe(Flow Deviation) Algorithm
• Problem:
s/t:
Ax=b, x 0
the objective function is non-linear and assumed to be convex, and the constraint set is linear
)x(fminx
Network Design and Analysis-----Wang Wenjie Notes on Routing: 36
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Frank-Wolfe Algorithm(Cont’d)
• Finding a direction that SATISFIES the constraints.
Consider a direction dIRn, such that
Ad=0
so if is a feasible point :
then is also feasible :
x 0 x,bxA
dx
bAdxA)dx(A
Network Design and Analysis-----Wang Wenjie Notes on Routing: 37
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Frank-Wolfe Algorithm(Cont’d)
• Suppose the point is: xk, problem is:
S/t:
here f(xk) is the gradient of the function f(x) evaluated at xk.
y)x(fmin Tk
y
0 y,bAy
Network Design and Analysis-----Wang Wenjie Notes on Routing: 38
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Frank-Wolfe Algorithm(Cont’d)
• Suppose yk,is the optimal solution to the LP, then
dk = yk – xk
we observe that:
Adk =A( yk – xk)=b-b=0
Which satisfies the requirement on the direction we imposed above
• Suppose:
)xy()x(f)x(fLB kkTkk
)x(fUB k
Network Design and Analysis-----Wang Wenjie Notes on Routing: 39
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Frank-Wolfe Algorithm(Cont’d)
Step1: Start with a feasible point x1. Set k=1
Choose tolerance , and
maximum iteration counter Kmax
Step2: Solve the linearized sub-problem :
S/t
to obtain the solution yk
y)x(fmin Tk
y
0 y,bAy
Network Design and Analysis-----Wang Wenjie Notes on Routing: 40
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Frank-Wolfe Algorithm(Cont’d)
Step 3: set dk = yk – xk
Step 4: Solve the line search problem :
to find the step size k.
set
Step 5: Check to see if the bound is ‘small’, i.e.
(UK-LB)/(1+|UB|) < or, k Kmax. Then stop.
slse set k=k+1 and go to step 2.
)dx(fmin kk
0
kk
kk dxx 1
Network Design and Analysis-----Wang Wenjie Notes on Routing: 41
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Algorithm Algorithm(Cont’d)
Note:
if the constraints set
Ax=b
are replaced by
Ax b ,x 0
the alporithmic approach remains the same
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