Network Architecture Network Design and Analysis

1. Network Architecture Network Design and Analysis Wang Wenjie Wangwj@gucas.ac.cn

2. Notes on Routing

3. 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

4. Formulating a Communication Network Flow Problem • A Four-Node Network Example • Node-Arc Formulation • Arc-Path Formulation

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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：

11. 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）

12. 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:

13. 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

14. 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。

15. 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？

16. 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

17. 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

18. 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。

19. 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

20. 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

21. 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] f(x) +(1- )f(y) x+(1- )y x y

22. 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

23. 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

24. Golden Section method(1) • Objective: Minimizing a unimodal function in a given interval • The problems is: • T=0.61803399 f0 f3 f2 f1 x0=a x1 x2 x3=b

25. 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 = f1x1 = x2 ;f1 = f2x2 = x0 + T *(x3–x0 ) f2 =f (x2 ) else /* x0 remains the same */ x3 = x2 ;f3 = f2x2 = x1 ;f2 = f1x1 = x0 + (1- T )*(x3–x0 ) f1 =f (x1 ) endif if( f1 < f2 ) then minvalue= f1 xmin= x1 else minvalue= f2 xmin= x2endif Golden Section method(2)

26. 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 :

27. Necessary and sufficient Optimality Condition • A necessary and sufficient condition forx*IRn to be an optimal solution is that: f(x*)=0 and for yIRn , yTf(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

28. 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)]-1f(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

29. 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

30. 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

31. 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)

32. Optimality Condition(Cont’d) • Optimality condition for NLP is: There exists such that(note:vj is unrestricted):

33. Optimality Condition(Cont’d) • If there is noly gj(x)  0, NO hi(x)=0: There exists such that:

34. Optimality Condition(Cont’d) • For following problem: s/t Ax=b The lagrangean is: The optimality condition is:

35. 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

36. Frank-Wolfe Algorithm(Cont’d) • Finding a direction that SATISFIES the constraints. Consider a direction dIRn, such that Ad=0 so if is a feasible point : then is also feasible :

37. 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.

38. 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:

39. 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

40. 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.

41. Algorithm Algorithm(Cont’d) Note: if the constraints set Ax=b are replaced by Ax  b ,x  0 the alporithmic approach remains the same