An application of Linear Programming to a computer network

Transcription

1 An application of Linear Programming to a computer network

2 Problem formulation (I) Problem: We want to maximize on a local network the rate of file transfer from one computer to another. Hypotheses: the maximum transfer rate along each link is assigned; the links are bidirectional, but cannot be used in both directions at the same time (i.e., half-duplex transmission ); the network nodes are not able to store the received data, but they return them immediately to the next link.

3 Problem formulation (II)

4 Problem formulation as LP problem (I) Choice of the decision variables. Transfer rate along each link: x oa, x ob, x oc, x ab, x ad, x be, x cd, x ce, x dn, x en The sign of each decision variable specifies the actual direction of the transfer rate: x ab > 0 the transfer is from a toward b; x ab < 0 the transfer is from b toward a; x ab = 0 there is neither transfer from a to b, nor viceversa.

5 Problem formulation as LP problem (II) Description of the constraints. The transfer rate along each link cannot exceed the maximum transfer rate along the same link: 3 x oa 3 1 x ob 1 1 x oc 1 1 x ab 1 1 x ad 1 3 x be 3 4 x cd 4 4 x ce 4 4 x dn 4 1 x en 1 These are linear constraints.

6 Problem formulation as LP problem (III) Description of the constraints. Each intermediate node does not have the possibility of storing data. Therefore, the input flow to each intermediate node is equal to the output flow from the same node: x oa = x ab + x ad x ob + x ab = x be x oc = x cd + x ce x ad + x cd = x dn x be + x ce = x en They are constraints of flow conservation for the intermediate nodes.

7 Problem formulation as LP problem (IV) Description of the objective function. We want to maximise the rate of data transfer from computer o to computer n. As a consequence of the flow constraints on intermediate nodes, the output flow of computer o is equal to the input flow of computer n. Being such flows equal, one can choose either one or the other of the objective function. Objective function: z = x oa + x ob + x oc (or: z = x dn + x en )

8 Problem formulation as LP problem (V) max z = x oa + x ob + x oc s.t. 3 x oa 3, 1 x ob 1, 1 x oc 1, 1 x ab 1, 1 x ad 1, 3 x be 3, 4 x cd 4, 4 x ce 4, 4 x dn 4, 1 x en 1, x oa = x ab + x ad, x ob + x ab = x be, x oc = x cd + x ce, x ad + x cd = x dn, x be + x ce = x en.

9 Solution via the Excel Solver (I)

10 Solution via the Excel Solver (II)

11 Solution via the Excel Solver (III)

12 Solution via the Excel Solver (IV)

13 Solution via the Excel Solver (V)

14 Solution via the Excel Solver (VI)

15 Solution via the Excel Solver (VII)

16 Problem solution

17 Problem formulation (I) We perform a change of variables x ij y ij and we introduce the slack variables s ij, one for each link: x oa = y oa 3, y oa + s oa = 6 x ob = y ob 1, y ob + s ob = 2 x oc = y oc 1, y oc + s oc = 2 x ab = y ab 1, y ab + s ab = 2 x ad = y ad 1, y ad + s ad = 2 x be = y be 3, y be + s be = 6 x cd = y cd 4, y cd + s cd = 8 x ce = y ce 4, y ce + s ce = 8 x dn = y dn 4, y dn + s dn = 8 x en = y en 1, y en + s en = 2 The new variables y ij and the slack variables s ij are 0.

18 Problem formulation (II) max z = y oa + y ob + y oc 5 s.t. y oa + s oa = 6, y ob + s ob = 2, y oc + s oc = 2, y ab + s ab = 2, y ad + s ad = 2, y be + s be = 6, y cd + s cd = 8, y ce + s ce = 8, y dn + s dn = 8, y en + s en = 2, y oa y ab y ad = 1, y ob y ab + y be = 1, y oc + y cd + y ce = 7, y ad + y cd y dn = 1, y be + y ce y en = 6, y ij, s ij 0 per ogni arco ij.

19 Problem formulation (III) The problem is expressed as max z = c T y (+c 0 ) s.t. Ay = b, y 0, A, b, c assegnati, b 0. We can eliminate the constant term c 0 and add it to the optimal objective function, once we have found it (e.g., with the simplex method) The simplex method can be applied, e.g., by using the MATLAB routine linprog.

20 Solution with the big-m method We need to find a basic feasible solution. We set ( ) A1 A =, A 2 where A 2 corresponds to the rows where we want to add the auxiliary variables. The big-m method requires to introduce of a set of auxiliary variables y aus to form a basic feasible solution. Problem in the form max z = c T ȳ (+c 0 ) s.t. Āȳ = b, ȳ 0, Ā, b, c assigned, b 0. with Ā = ( ) ( ) ( ) A1 0 y, ȳ =, A 2 I y b c = b, c =. aus M1

21 Problem data (I) Vector of the decision variables ( ȳ = y oa, y ob, y oc, y ab, y ad, y be, y cd, y ce, y dn, y en, s oa, s ob, s oc, s ab, s ad, s be, s cd, s ce, s dn, s en, y aus,1, y aus,2, y aus,3, y aus,4, y aus,5 ) T Vector of the constant terms b = (6, 2, 2, 2, 2, 6, 8, 8, 8, 2, 1, 1, 7, 1, 6) T Vector of the coefficients of the objective function c = (1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, M, M, M, M, M) T

22 Problem data (II) Matrix Ā of the constraints The square sub-matrix corresponding to the blue-colored elements is an identity matrix. Therefore, it identifies a basic feasible solution.

23 Solution with the big-m method (I) Initial Tableau, M = 100: yoa yob yoc yab yad ybe ycd yde ydn yen soa sob soc sab sad sbe scd sde sdn sen yaux,1 yaux,2 yaux,3 yaux,4 yaux,5 sol.corrente z soa sob soc sab sad sbe scd sce sdn sen yaus, yaus, yaus, yaus, yaus,

24 Solution with the big M method (II) Initial Tableau in standard form, M = 100: yoa yob yoc yab yad ybe ycd yde ydn yen soa sob soc sab sad sbe scd sde sdn sen yaux,1 yaux,2 yaux,3 yaux,4 yaux,5 sol.corrente z soa sob soc sab sad sbe scd sce sdn sen yaus, yaus, yaus, yaus, yaus,

25 Solution with the big M method (III) Final Tableau, M = 100: yoa yob yoc yab yad ybe ycd yde ydn yen soa sob soc sab sad sbe scd sde sdn sen yaux,1 yaux,2 yaux,3 yaux,4 yaux,5 sol.corrente z soa yob yoc yab yen sbe scd sce yad sen yoa ybe ydn ycd yce

26 Solution with the big M-method (IV) After 10 iterations of the simplex method (with M = 100) you will find with MATLAB the optimal basic solution ( ȳ = 5, 2, 2, 2, 2, 5, 7, 2, 8, 1, 1, 0, 0, 0, 0, 1, 1, 6, 0, 1, 0, 0, 0, 0, 0 The auxiliary variables are outside the basis. Therefore, they do not affect the optimum. After adding the constant c 0 = 5 to the optimal objective function (9), we obtain again the value already obtained with the EXCEL solver (4). ) T

27 References J. Matusek, B. Gärtner: Understanding and Using Linear Programming, Springer, Excel solver for linear programming: MATLAB linprog routine for linear programming: