Network Planning VITMM215 Markosz Maliosz 10/14/2015 Shortest Path - PowerPoint PPT Presentation

Network Planning VITMM215 Markosz Maliosz 10/14/2015

Shortest Path – Link Weights  What should be the link weights? – physical distance? – delay on a link? – hop count (w=1)? – some arbitrary number?  Cisco’s default: – inversely proportional to its capacity – higher capacity links are “shorter” – drives traffic to higher capacity links – can lead to strange routing 2

Link weight setting example  Traffic: – q  t : 1 – 3  t : 1 – s  t : 1 – w  t : 1 3

Link Weights  The correct choice depends on objectives  Common goal: minimize delays – if propagation delay is dominant  minimize physical path distance  weight = link distance – if processing delay is dominant  minimize hop count, weight = 1 – if queuing delay is dominant  measure delays along links  calculate shortest path according to measured delay values 4

Tree Networks  Connects all nodes with minimum number of links  There is one and only one path between any node pairs  Example applications: – cable TV network: broadcast from one source to many – Ethernet Spanning Tree Protocol 5

Spanning Tree  Def.: subgraph, which is a tree and contains all nodes 6

Minimum spanning tree  link cost  edge weights  The minimum spanning tree is the minimum cost spanning tree among all spanning trees  If each link has a distinct weight, there will be a unique MST, otherwise the MST is not unique 7

Kruskal’s Algorithm  Greedy algorithm to find minimum spanning tree – create a forest F (a set of trees), where each node in the graph is a separate tree – At each step, take the minimum-weight edge  if it connects two different trees, then add to the forest  combine the two trees into a single tree  otherwise discard that edge – If there is only one tree, the problem is solved Animation: http://students.ceid.upatras.gr/~papagel/project /kruskal.htm 8

Prim’s Algorithm  Greedy algorithm to find minimum spanning tree – select an arbitrary node as initial element of the set – in each iteration, choose a minimum-weight edge ( u , v ), connecting a node v in the set to the node u outside of the set, then node u is brought into the set – this process is repeated until all the set will contain all nodes  When the algorithm terminates, the edges in the set form a minimum spanning tree Animation: http://students.ceid.upatras.gr/~papagel/project/p rim.htm 9

Calculating Spanning Tree 2 4 E C A 2 6 9 5 10 4 6 B D F Find the min-cost spanning tree in the graph! 10

Ethernet STP  An Ethernet network can have multiple possible paths for reliability  STP is intended to create a tree to avoid loops – broadcast storm  802.1D algorithm based on Dijkstra’s alg. – switches are assigned numerical priorities – switch with lowest priority becomes root switch  tie break is lowest MAC address (are unique) – each switch port is given a cost = link weight  based on bandwidth of the link 11

STP Port Costs  Default costs – old version based on 1 Gbps/(link bandwidth) – new version: arbitrary table  port cost: integer, >0 Link speed Old cost New cost 10 Mbps 100 100 100 Mbps 10 19 1 Gbps 1 4 10 Gbps 1 2 12

STP  Bridge Protocol Data Units are sent regularly – priority, cost information and state – flooded through network  Protocol finds min-cost path to root switch with Dijkstra’s algorithm in a distributed way – ports not on shortest path towards root bridge are blocked 13

STP  Optimization in STP – minimizes path length for each leafs to get to the root – link weights can be arbitrary! – with default weights: pushes traffic to higher bandwidth paths 14

Linear Programming 15

Linear Programming  Linear Programming is the optimization of an objective based on some set of constraints using a linear mathematical model  The word “programming” is used in the sense of “planning”  A Linear Program (LP) is a problem that can be expressed as – a linear objective function, – subject to linear equality and linear inequality constraints  Has many applications – transportation, telecommunications, manufacturing, business  e.g. maximizing profit in a factory that manufactures a number of different products from the same raw material using the same resources – engineering: planning, routing, scheduling, assignment, design  minimizing resources in network planning to satisfy user demands  maximizing performance of the network with a limited cost budget 16

Linear Program  Standard form: find the values of x j variables, that minimize (or maximize) the objective function z  subject to (satisfying the following constraints):  Notation: – objective function: z – variables to be determined, decision variables: x j – number of variables: n – coefficients: c j , a ij, b i – number of constraints: m – bounds for variables 17

Linear Program  Standard matrix form:   x  T max z c x 1     x      c: cost coefficients  T  c c c     x 1 n n – subject to:         a a b   11 1 n 1          A x b     A b                 a a   b m 1 mn m 18

LP Example  A company manufacturing two products: P1, P2  3 manufacturing units are required to manufacture both products: U1, U2, U3  During a week the maximum operating hours for the machines: – U1: 50 hours – U2: 35 hours – U3: 80 hours  To manufacture a product it has to be processed on all of the 3 manufacturing units with the following manufacture times in hours (the sequence does not matter): – U1 U2 U3 – P1 10 5 5 – P2 5 5 15  Price of P1: 100$  Price of P2: 80$  How many pieces should from P1 an P2 to be manufactured to achieve maximum income? 19

LP Example  The task is an optimization problem: find the maximum income  Variables to be determined: amount of P1 and P2 to be manufactured, denoted as: x1 and x2  Objective function – maximize income = max (100 x1 + 80 x2)  The limited working hours of the manufacturing units (U1, U2, U3) restrict the amount of products that can be produced during a week  constraints U1 U2 U3 Constraints: 10 x1 + 5 x2 ≤ 50 P1 10 5 5 5 x1 + 5 x2 ≤ 35 P2 5 5 15 5 x1 + 15 x2 ≤ 80 MWH 50 35 80 x1 ≥ 0 x2 ≥ 0 MWH= max. working hours 20

Finding the Solution  Feasible solutions – x vectors, which satisfy the constraints and bounds  Optimal solution – x is feasible and c T x is minimal (maximal)  Problem can be – infeasible – unbounded 21

Geometrical solution x 2 2D representation: Constraints: set of straight lines 10 F(x) = 100 x 1 + 80 x 2 = 800 (const) F(x) = 800 (a) 10x 1 + 5x 2  50 5 (b) 5x 1 + 5x 2  35 4 P 2 (c) 5x 1 + 15x 2  80 Sol: X 1 = 3 X 2 = 4 Extreme x 1 P 1 point P 0 c 3 5 10 15 a b 22

Geometrical Representation  Ax  b  Optimum: 23

Solving LP  Good general-purpose techniques exist for finding optimal solutions – All polynomial-size linear programs can be solved in polynomial time – Algorithms  Simplex  Barrier or Interior Point Method – Software  CPLEX  GNU Linear Programming Kit (GLPK)  lp_solve  etc. 24

Integer Linear Programming  Pure Integer Problem (IP or ILP) – Restriction: elements of vector x must be integers  searching the best integer solution  Mixed Integer LP (MIP or MILP) – Restriction: some elements of vector x must be integers  Binary or zero-one ILP – Restriction: elements of vector x must be 0 or 1 – many kinds of combinatorial and logical restrictions can be modeled through the use of zero-one variable  Much harder problems: NP-hard in general – solution time is much longer 25

Integer Linear Programming  Many real problems contain integer quantities, e.g.: – Set Covering Problem – Travelling Salesman Problem (TSP) – Knapsack problem  Solving ILP – Relaxation: variables are not constrained to be integers  if the continuous problem (LP) always has an integral optimal value (assuming c is integral), then this optimal value is the solution for the ILP too, and it can be found efficiently (as in the previous example for manufacturing)  if the solution is not integral, then finding the optimum is more difficult 26

ILP Relaxation – 2D Illustration 27

ILP Relaxation – 2D Illustration  Optimal value of LP is far from integral feasible solution: 28

Solving ILP  Branch and bound – Branch:  decomposition into subproblems based on possible variable settings recursively: forming nodes of a tree – Bound:  infeasible subproblems pruned  compute upper and lower bounds – by minimizaton: if the lower bound for some subproblem is greater than the upper bound for some other subproblem, then the former may be safely discarded from the search  Branch and cut – Eliminate non-integer solutions by adding cut constraints to LP 29

Minimum cost ring topology planning 30