Network Architecture Network Design and Analysis

Network Design and Analysis-----Wang Wenjie Notes on Routing: 1

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Network Architecture

Network Design and Analysis

Wang Wenjie

[email protected]


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Notes on Routing


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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


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Formulating a Communication Network

Flow Problem

• A Four-Node Network Example

– Node-Arc Formulation

– Arc-Path Formulation


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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


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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


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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


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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


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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


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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


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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 ）


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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


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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


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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？


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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


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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


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• 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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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• If there is noly gj(x) 0, NO hi(x)=0:

There exists such that:u,x


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• For following problem:

s/t Ax=b

The lagrangean is:

The optimality condition is:

)x(fminx


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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


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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


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• 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


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• 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


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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


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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


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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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