TRANSPORTATION PROBLEMS

Transcription

1 Chapter 5 TRANSPORTATION PROBLEMS 5.1. Classic transportation problem The problem The standard form of the transportation problem is as follows: inf m n c ij x ij, where: i=1 j=1 n x ij = a i, 1 i m, j=1 m x ij = b j, 1 j n, i=1 x ij 0, 1 i m, 1 j n. a i 0, 1 i m, b j 0, 1 j n, m a i = n b j. i=1 j=1 (5.1) (5.2) Theorem The necessary condition for the equation system considered in 5.1 to have a non-negative solution is that the relations 5.2 are fulfilled. 67

2 68 TRANSPORTATION PROBLEMS The transportation algorithm (simplex algorithm adaptation) Theoretical fundamentals The transportation problem can be written in matrix form: inf c t x Ax = b x 0 To a transportation problem we can associate a transportation table which contains the problems numerical data. The transportation table 6.1 has m lines and n columns. To each cell (i, j), 1 i m, 1 j n, we can associate a single cost c ij, with the unknown variable x ij, with the value x ij of the variable x ij from a certain solution (x ij ) of the equation system 6.1. TABLE 6.1 c 11 c c 1n a 1 c 21 c c 2n a c m1 c m2... c mn a m b 1 b 2... b n The transportation table contains also an extra column in which are usually written the available quantities a i, 1 i m, and an extra line in which are written the requests b j, 1 j n. In the cell (i, j) is written the unit cost c ij ; if we want to write also the value x ij of the variable x ij, we will split the cell (i, j) by a diagonally in two parts and we will write c ij in the superior part and x ij in the inferior part. The transportation table T has an undirected graph Γ associated to it, having as verticesş the cell set (i, j), 1 i m, 1 j n, of the transportation table T and as edges the sets composed by the two cells situated in the same T line or column. In the following paragraphs we consider only the partial subgraphs from Γ which are obtained by joining neighbor cell edges, meaning the cell situated on the same T line or column between which it does not exist other cell from the set of subgraph vertices. A partial Γ graph is called a π-graph. A π-graph which has an elementary path having at most an edge in each line or column of the transportation table T is called a µ-path or a µ-loop.

3 CLASSIC TRANSPORTATION PROBLEM 69 The transportation algorithm STEP 0. We determine a solution x ij (we will see next how) corresponding to a matrix B formed by m + n 1 columns, linearly independent of the A matrix and after that the set B of the base cells, after which we advance to the next step. STEP 1. We resolve the following equation system: u i + v j = c ij, (i, j) B, we obtain a particular solution (u i, v j ) of this system and we compute the values z ij c ij = u i + v j c ij for which (i, j) R (the set of the cell that are not in the base). If z ij c ij 0 for all cells (i, j) R, we stop (STOP): the base solution x ij is optimal.. Contrarily we determine (s, k) R using the base entry criteria: max{z ij c ij } = z sk c sk and we advance to the next step. STEP 2. We determine the µ-path formed by (s, k) R using a part of the B cells, we adopt a direction and we number its cells starting with the cell (s, k). We then determine the cell (r, t) B using the base exit criteria: min{x ij } =x rt, where the minimum is taken in rapport with all the cells (i, j) having even rank within the µ-path previously determined. We then advance to the next step. STEP 3. We consider the B matrix obtained from the B matrix by replacing the column a rt with the column a sk. We determine the admissible base solution x ij corresponding to B with the help of the base changing formula: xij = x ij x rt, if (i, j) has even rank in the µ path, x ij + x rt, if (i, j) had odd rank in the µ path, x ij, if (i, j) does not exist within the µ path. We then advance to STEP 1, replacing B with B and the solution x ij with x ij. Remark. If, after the algorithm has ended, we have z ij c ij < 0 and for all cells (i, j) R,, then the transportation problem solution is unique.

4 70 TRANSPORTATION PROBLEMS Determining an initial base program In order to apply the transportation algorithm mentioned before it is necessary that we have an initial base program. The general method for obtaining a base program is as follows. We choose a cell (i, j) and we set to the variable x ij the minimum value that it can have, meaning the value: x ij = min{a i, b j }. If x ij = a i, then we eliminate from the transportation table the i rank line and we replace b j by b j = b j x ij. If x ij = b j, then we eliminate from the transportation table the j rank table and we replace a i with a i = a i x ij. If x ij = a i = b j, then we can apply one of the previously mentioned algorithms. In all the three situations presented we get a reduced transportation table, each having a line or column less that the initial one. Following the described steps on all the reduced tables which can be obtained, we determine a transportation problem program. The general method for obtaining a base program from the transportation problem can be personalized by giving various selection rules of the cell (i, j) and giving its value to the x ij variable. From the particular methods for determining an initial base program we mention the following: i) North-West corner method. This method consists in choosing the cell (i, j) situated in the first line and the first column of the used transportation tables. ii) The minimum cost method. In each step we choose the cell (i, j) corresponding to the minimum cost c ij from the used transportation tables Degeneration and cycling As in the general case of the linear programming problems, the base programs degeneration can make the cycling phenomena appear within the transportation algorithm. Obviously, the base solution for the transportation problem in the standard form is non-degenerate if and only if the number of its not null components are equal to m + n 1 and it is degenerate in the contrary. In order to recognize the degenerate problems, it is useful to use the following criteria. Sentence A transportation problem in standard form is degenerate if and only if there is a set of indexes M {1,..., m} and a set of indexes N {1,..., n} so that: i M a i = j N b j.

5 CLASSIC TRANSPORTATION PROBLEM Variants of the transportation problem Transportation problem with surplus request. The transportation problem form with surplus request is the following: inf m n c ij x ij, where: i=1 j=1 n x ij = a i, 1 i m, j=1 m x ij b j, 1 j n, i=1 x ij 0, 1 i m, 1 j n. a i 0, 1 i m, b j 0, 1 j n, m a i n b j. i=1 j=1 In order to solve this problem we introduce a fictive deposit in which the available resource is: n m a m+1 = b j a i, j=1 meaning the total request surplus. The transportation costs from the deposit m + 1 can be taken null or equal to the unitary penalties determined by contracts with the recipient for not honoring the requests, if such penalties exist. The problem becomes a standard transportation problem, which is resolved using a known algorithm. Transportation problem with surplus offer. The transportation problem form with surplus offer is the following: inf m n c ij x ij, where: i=1 j=1 n x ij a i, 1 i m, j=1 m x ij = b j, 1 j n, i=1 i=1 x ij 0, 1 i m, 1 j n. a i 0, 1 i m, b j 0, 1 j n, m a i n b j. i=1 j=1

6 72 TRANSPORTATION PROBLEMS In order to solve this problem we introduce a fictive consumer in which the request is: m n b n+1 = a i b j, i=1 meaning the total request surplus. The transportation costs to the n + 1 recipient can be taken null or equal to the storage costs to the i deposits, 1 i m. The problem becomes a standard transportation problem, which is resolved using a known algorithm. The transportation problem with limited capacities. The transportation problem with limited capacities is the following: where: inf m i=1 j=1 n c ij x ij, n x ij = a i, 1 i m, j=1 m x ij = b j, 1 j n, i=1 j=1 d ij x ij 0, 1 i m, 1 j n. a i 0, 1 i m, b j 0, 1 j n, d ij 0, 1 i m, 1 j n, m a i = n b j. i=1 j=1 m a i b j, 1 j n, n b j a i, 1 i m. i=1 j=1 Solving this problem can be made by using the simplex algorithm that is presented in the following subparagraph. We mention that the problem can be solved as a minimum flux problem Modified simplex algorithm In order to determine an initial program we choose randomly a cell (i, j) and we set to the x ij a value: xij = min{a i, b j, d ij }.

7 CLASSIC TRANSPORTATION PROBLEM 73 If x ij = a i or x ij = b j, then we proceed the normal way. If x ij = d ij < min{a i, b j }, the variable x ij is not considered a base variable. If x ij = d ij = min{a i, b j },, the variable x ij is not considered a base variable. The values ( x ij ) obtained in this way does not form a program for this problem: it can happen that some availabilities are not completely exhausted, and some requests are not entirely satisfied. Leaving from the values ( x ij ) we can obtain the base program for the transportation problem with limited capacities on the following algorithm. Let us assume that a deposit r exists, 1 r m, in which we have the a r resource quantity unsent and the recipients s and k, in which the requests are unsatisfied with the quantities b s and b k ; obviously, we have the following: a r = b s+b k. The algorithm that transforms ( x ij ) in the base program (x ij ) is the following: STEP 1. We add to the r line, an extra cell in which the corresponding value of the x r0 variable is a r. In the same way, we add to the s and k columns an extra cell in which the corresponding values x 0s and x 0k are b s and b k. We replace the initial unitary costs c ij, 1 i m, 1 j n, with c ij = 0 and we set to the newly introduced cells some costs equal to the unit, more exactly c r0 = c 0s = c 0k = 1. STEP 2. We solve the following equations: u r = 1, v s = 1, v k = 1, u i + v j = c ij, (i, j) B, i r, j s, k. STEP 3. We search using the regular transportation algorithm the optimal solution for the problem from STEP 1; obviously in the optimum solution we will have: x r0 = x 0s = x 0k = 0. We omit the extra cells and we come back to the initial c ij costs. The obtained solution is a base program. According to the general result obtained from the simplex algorithm, the optimal test is the following: the program (x ij ) is optimal if z ij c ij 0 for all the secondary (i, j) cells for which x ij = 0 and z ij c ij 0 for all the secondary (i, j) cells for which x ij = d ij.

8 74 TRANSPORTATION PROBLEMS Applications Exercise Using the N-W corner method and the minimum cost method find a base solution for the transportation problem having the data from table 6.2. For these solutions find the corresponding values of the object function. What solution is better and why? TABLE Solution. We first compute an initial base solution using the N-W method: x 11 = min{11, 5} = 5, we eliminate column 1, x 12 = min{6, 9} = 6, we eliminate row 1, x 22 = min{11, 3} = 3, we eliminate column 2, x 23 = min{8, 9} = 8, we eliminate row 2, x 33 = min{1, 8} = 1, we eliminate column 3, x 34 = min{7, 7} = 7. The value of the object funtion is: = 150. We determine a base solution by using the minimum cost method: x 33 = min{8, 9} = 8, we eliminate row 3, x 21 = min{11, 5} = 5, we eliminate column 1, x 14 = min{11, 7} = 7, we eliminate column 4, x 22 = min{6, 9} = 6, we eliminate row 2, x 13 = min{4, 1} = 1, we eliminate column 3, x 12 = min{3, 3} = 3. The value of the object function is: = 92. Obviously the base solution obtained by using the minimum cost method is better because the corresponding cost is smaller.

9 CLASSIC TRANSPORTATION PROBLEM 75 Exercise Find a base solution for the balanced transportation problem having the data from table 6.3. Find the corresponding value of the object function. Find the optimal problem solution. TABLE Solution. STEP 0. The initial base solution determined by the N-W corner method is: x 11 = min{5, 10} = 5, we eliminate column 1, x 12 = min{3, 5} = 3, we eliminate column 2, x 13 = min{7, 2} = 2, we eliminate row 1, x 23 = min{5, 5} = 5. The cost of this solution is C = 28. Fig. 5.1: The initial solution and the µ-loop cell marking. Iteration 1, STEP 1. obtain by setting u 1 = 0): i) We solve the system (a particular solution that is u i + v j = c ij (i, j) B, The solutions of this system are going to be written on the edge of the transportation table (on the vertically we write the u i values, and on the horizontally we write the v j values). ii) We compute the elements: z ij c ij = u i + v j c ij (i, j) R, which are written in the upper right corner of each secondary cell.

10 76 TRANSPORTATION PROBLEMS iii) We compute (the optimum criteria): max (z ij c ij ) = z 22 c 22 = 2 > 0 (i,j) R The optimum criteria is not fulfilled, meaning that the cell (2, 2) goes into the base. Iteration 1, STEP 2. In order to determine the cell that is outside the base, we determine the µ-loop formed by the cell (2, 2) using a part of the base cells: (2, 2), (2, 3), (1, 3), (1, 2), (2, 2). We adopt a way to go of this µ-loop marking with the even ranked cells. The exit condition from the base is: min x ij =x 12, which means that the cell (1, 2) is outside the base. Iteration 1, STEP 3. By using the base changing formulas we get the new solution. The associated cost is: C 1 = 22, which is lower than the previous cost C 0. We repeat the previous steps: Iteration 2, STEP 1. i) We solve the system (a particular solution that is obtain by setting u 1 = 0): u i + v j = c ij (i, j) B, The solutions of this system are going to be written on the edge of the transportation table (on the vertically we write the u i values, and on the horizontally we write the v j values). ii) We compute the elements: z ij c ij = u i + v j c ij (i, j) R, which are written in the upper right corner of each secondary cell. iii) We compute (the optimum criteria): max (z ij c ij ) = z 12 c 12 = 2 < 0, (i,j) R the solution is optimum and it is equal to: x 11 = 5, x 13 = 5, x 22 = 3, x 23 = 2.

11 CLASSIC TRANSPORTATION PROBLEM 77 Fig. 5.2: The final solution. Exercise Find a base solution for the balanced transportation problem having the data from table 6.4. Find the corresponding value of the object function. Find the optimal problem solution. TABLE 6.4 2α 3α 4α α 2β 3α 6α 2α 4α 3β α 4α 5α 3α λ β 7β β 6β Solution. The available quantity must be equal to the request, meaning that 15β = 5β + λ from where we get λ = 10β. STEP 0. We determine a base solution by using the minimum cost method. We have: x 31 = min{β; 10β} = β, we eliminate column 1 from the transportation table, x 14 = min{7β; 2β} = 2β, we eliminate row 1 from the transportation table, x 23 = min{β; 3β} = β, we eliminate column 3 from the transportation table, x 34 = min{4β; 9β} = 4β, we eliminate column 4 from the transportation table, x 32 = min{7β; 5β} = 5β, we eliminate row 3 from the transportation table, x 22 = min{2β; 2β} = 2β, we eliminate column 1 from the transportation table. The transportation cost of this solution is: C 0 = αβ + 2αβ + 2αβ + 12αβ + 20αβ + 12αβ = 49αβ. Iteration 1, STEP 1. obtain by setting u 1 = 0): i) We solve the system (a particular solution that is u i + v j = c ij (i, j) B,

12 78 TRANSPORTATION PROBLEMS The solutions of this system are going to be written on the edge of the transportation table (on the vertically we write the u i values, and on the horizontally we write the v j values). ii) We compute the elements: z ij c ij = u i + v j c ij (i, j) R, which are written in the upper right corner of each secondary cell. iii) We compute (the optimum criteria): max (z ij c ij ) = z 24 c 24 = α > 0. (i,j) R The optimum criteria is not fulfilled, which means that the cell (2, 4) is in the base. Fig. 5.3: µ-loop formed by the cell (2,4) with a part of the base cells, the even ranked cells are marked with. Iteration 1, STEP 2. In order to determine the cell that is outside the base, we determine the µ-loop formed by the cell (2, 4) using a part of the base cells: (2, 4), (3, 4), (3, 2), (2, 2), (2, 4). We adopt a way to go of this µ-loop marking with the even ranked cells. The exit condition from the base is: min x ij =x 24, which means that the cell (2, 2) is outside the base. Iteration 1, STEP 3. By using the base changing formulas we get the new solution. The associated cost is: C 1 = 47αβ,

13 CLASSIC TRANSPORTATION PROBLEM 79 which is lower than the previous cost C 0. We repeat the previous steps: Iteration 2, STEP 1. i) We solve the system (a particular solution that is obtain by setting u 1 = 0): u i + v j = c ij (i, j) B, The solutions of this system are going to be written on the edge of the transportation table (on the vertically we write the u i values, and on the horizontally we write the v j values). Fig. 5.4: The new base solution. This is optimal. ii) We compute the elements: z ij c ij = u i + v j c ij (i, j) R, which are written in the upper right corner of each secondary cell. iii) We compute (the optimum criteria): max (z ij c ij ) = z 21 c 21 = α < 0, (i,j) R the solution is optimum and it is equal to: The minimum cost is 47αβ. x 14 = 2β, x 23 = β, x 14 = 2β, x 31 = β, x 32 = 7β, x 34 = 2β.

14 80 TRANSPORTATION PROBLEMS Exercise Find using the N-W corner method and by the minimum cost method, a base solution for the transportation problem that has the data in table 6.5. For these solutions find the corresponding values of the object function. Which solution is better and why? Find the optimal solution for the problem. TABLE Exercise Find a base solution for the balanced transportation problem having the data from table 6.6. Find the corresponding value of the object function. Find the optimal problem solution. TABLE 6.6 8α 3α 5α 2α 10β 4α α 6α 7α 15β α 9α 4α 3α λ 5β 10β 20β 15β Solution. We compute λ = 25β. The minimum cost value is 140αβ, and the optimal solution is: x 14 = 10β, x 22 = 10β, x 23 = 5β, x 31 = 5β, x 33 = = 15β, x 34 = 5β. Exercise Find a base solution for the transportation problem having the values from table 6.7. The penalty unit costs for not fulfilling the requests are equal to 2, and the storage costs are null. Find the corresponding value of the object function. Find the problem optimal solution. TABLE λ Solution. For λ < 10, the problem is a problem of surplus request (the request is bigger than the offer). We will use a fictive deposit of 10 λ. The transportation table is table 6.8:

15 CLASSIC TRANSPORTATION PROBLEM 81 TABLE λ λ The new transportation problem is balanced and it can be solved using the known algorithm. For λ = 10 the problem is a balanced transportation problem and it can be solved using the known algorithm. For λ > 10 the problem is a transportation problem with surplus offer (the offer is bigger than the request). We will introduce an fictive consumer with a request equal to λ 10. The new transportation table is table 6.9: TABLE λ λ 10 The new transportation problem is balanced and it can be solved using the known algorithm. Exercise For what value of the unitary cost c, the problem having the data from the table 6.10 has the optimal value equal to 140? TABLE c

16 82 TRANSPORTATION PROBLEMS 5.2. Maximum flow in a network Networks In this section we shall see how the transportation problem can be solved using graph theory. Definition A network it is a directed graph G = (X, Γ) in which each edge (u, v) Γ has a capacity c(u, v) 0. If (u, v) / Γ, then c(u, v) = 0. Thus, the capacity represents the maximum amount of flow that can pass through an edge. Let be s the source vertex (node) and t the final destination vertex. Definition The flow ϕ : X X R is a function that satisfy one of the following properties: i) capacity constraints: for every u, v X, ϕ(u, v) c(u, v) (an edge for which we have equality is called saturated edge); ii) antisymmetry: for each u, v X, ϕ(u, v) = ϕ(v, u); iii) flow conservation: for each u X {s, t} we have: ϕ(u, v) = 0. v X The quantity ϕ(u, v), can be positive or negative and is called network flow from u to v. Definition The value of the flow Φ is: Φ = v X ϕ(s, v) or, in other words, the value of the flow is the total flow from the source. The problem of the maximum flow in a directed graph G, from a source vertex s to a destination vertex t, in a transportation network, is to determine the maximum flow from s to t.

17 MAXIMUM FLOW IN A NETWORK Ford-Fulkerson algorithm In this chapter we shall present the Ford-Fulkerson technique for solving the problem of maximum flow, respectively the minimum flow in a transportation network. It is rather technique then an algorithm because it supports several implementation algorithms with different running time. The Ford-Fulkerson uses three important notions from the graph theory: compatible flow, path and cut of minimum capacity. Solving a maximum flow problem STEP 1. In this step we construct a compatible flow. For this we shall consider all unsaturated path from the s to the destination t. Let µ i be a unsaturated path. The flow propagated thus this path is given by: ϕ i = min (u,v) µ i [c(u, v) ϕ(u, v)]. We obtain a new flow which saturates at least one edge with the relation: { ϕ(u, v) + ϕi când (u, v) µ ϕ(u, v) = i ϕ(u, v) când (u, v) / µ i. If the transportation network contains undirected edges in the identification of this path we shall take care to use the undirected edges only in one way. Thus, when we identify such a path we also direct the routes. For each edge we compute the sum of the propagated flow. The flow which is propagated thus the network it is equal with the sum of all the flows from the identified path: Φ = ϕ i. i STEP 2. In this step we determine the maximum value flow using the following assignation procedure: i) mark the source s with [+]; ii) if the vertex i is marked and (i, j) Γ cu ϕ(i, j) < c(i, j), the the vertex j will be marked with [+i]; iii) if the vertex j is marked and (i, j) Γ cu ϕ(i, j) > 0, then the vertex i will be marked with [ j]; iv) if i is a marked vertex and (i, j) is a saturated edge, then the vertex j shall be not marked. If the output t will be marked, then we have a path from s to t. Through this path or chain denoted with µ k will be propagated a flow given by the relation: ϕ k = min( min (c(u, v) ϕ(u, v)), min ϕ(u, v)), ϕ k > 0, (u,v) B (u,v) C

18 84 TRANSPORTATION PROBLEMS where B is the set of edges which are marked by type ii) process, and C is the set of edges which are marked by type iii) process. We obtain a better flow, with using the relation: ϕ(u, v) + ϕ k when (u, v) B ϕ(u, v) = ϕ(u, v) ϕ k when (u, v) C ϕ(u, v) otherwise. If we can not mark the output vertex t of the network, then the algorithm will stop. The sum of all propagated flows through the paths and chains that connects the input vertex s with the output vertex t, represents the maximum flow in the transportation network. STEP 3. Determination the minimum capacity cut. Let A be the set of unmarked vertices of the network, after the process applied in step 2. The set of edges (i, j) with i / A and j A denoted with A is called the minimum capacity cut of graph. This cut represents the border between marked and unmarked vertices of the graph. The capacity of the cut is denoted by c( A) and is defined the sum of the capacity of the edges: c( A) = c(u, v). (u,v) A The Ford-Fulkerson theorem, in a transportation network states that the, maximum flow is equal with the value of cut of minimum value, thus: max Φ(ϕ) = min c( A). ϕ s/ A,t A The cut of minimum values represents the congestion in the network. Because its edges are saturated, for increasing the flow we must increase the capacity of one or more edges of the minim cut. Remarks. i) The marking process is a decision on optimality. ii) Minimum capacity cut has a dual role: -checking the result thus Ford-Fulkerson theorem; -indicates what edges capacity must be increased to increase the capacity of the flow. Solving a minimum flow problem For obtaining a minimum flow we induce in the network an arbitrary flow with the constraints ϕ(u, v) c(u, v), for each edge (u, v).

19 MAXIMUM FLOW IN A NETWORK 85 For the beginning we shall consider the path from the final vertex t to the starting vertexs using the following marking procedure: i) the final vertex t is marked with [+t]; ii) if j is a marked vertex, (i, j) edge and ϕ(i, j) > c(i, j), then the vertex i shall be marked with [ j]; iii) if i is a marked vertex and (i, j) edge, then the vertex j shall be marked with [+i]. If we can mark the starting vertex s then we have a chain µ of over the flow can by reduced. If B is the set of edges from the path µ which belongs to category ii) and θ = min [ϕ(u, v) c(u, v)], (u,v) B then the reduced flow is: ϕ(u, v) + θ ϕ(u, v) = ϕ(u, v) θ ϕ(u, v) when (u, v) B when (u, v) µ, (u, v) / B otherwise If we are unable to mark the starting vertex s, this means that we donţ have any chain on which the flow can be reduced thus we have the minimum flow Transportation problem and maximum flow problem The classical transportation problem can be modeled through the maximum flow problem. Suppose that we have m warehouses and n shops, the transportation from the warehouses to the shops is done using some directed constraints d ij. We construct a directed from the m ware houses to the n shops, the value of each edge is setup with the value d ij, a input vertex X input and a output vertex X output with connections to every warehouses respectively shops, the value of each edge being equal with a i respectively b j. The problem which must be solved it is maximum flow problem.

20 86 TRANSPORTATION PROBLEMS Applications Exercise An economic process has the flow indicated in figure 5.5. The transportation capacities are indicated on the figure. Which is the maximum flow in the system between vertex 1 and 9? Which is the flow between vertex 3 and vertex 5? Fig. 5.5: Flow between two nodes. Solution. We induce a maximum flow between 1 and 9. For this we generates paths from 1 to 9 (oriented routes in the same direction) which are saturated. In identification of this paths we take care to use a unoriented route only in one way. Thus, we indicate this paths and the flow propagated. In the same time we direct the edges. For each edge we compute the propagated flow. Thus, we obtain the paths: µ 1 = ( ), ϕ 1 = 5. µ 2 = ( ), ϕ 2 = 4. µ 3 = ( ), ϕ 3 = 1. µ 4 = ( ), ϕ 4 = 2. µ 5 = ( ), ϕ 5 = 2. µ 6 = ( ), ϕ 6 = 3. µ 7 = ( ), ϕ 7 = 1. µ 8 = ( ), ϕ 8 = 5.

21 MAXIMUM FLOW IN A NETWORK 87 At this time all the paths from vertex 1 to vertex 9 contains only oriented routes (fig. 5.6) and each path has at least one saturated edge. The propagated flow through these paths is equal with: Φ = 8 ϕ i = 23. i=1 Fig. 5.6: Oriented routes. In the next step we apply the marking procedure. Thus, the following vertices are marked like in figure 5.7. Fig. 5.7: Vertex marking.

22 88 TRANSPORTATION PROBLEMS The final vertex 9 can not be marked, thus the propagated flow is maximum. The set of unmarked vertices is: A = {5, 6, 7, 8, 9}, the minimum capacity cut is: A= {(3, 5), (4, 5), (2, 6), (2, 7), (3, 7)}, with the value of the capacity: c( A) = = 23. We remark that the minimum capacity cut is equal with the maximum value of the flow that confirms, through Ford-Fulkerson theorem, that the flow is maximum. Through the route [3, 5] it is propagated a flow of maximum capacity 5. MAPLE code for solving this problem is: > with(networks) : > new(g) : > addvertex(1, 2, 3, 4, 5, 6, 7, 8, 9, G) : > addedge([{1, 2}, {1, 3}, {1, 4}, {2, 6}, {2, 7}, {2, 3}, {3, 4}, {3, 7}, {3, 5}, {6, 7}, {6, 8}, {4, 5}, {5, 7}, {5, 9}, {7, 8}, {7, 9}, {8, 9}]); > weights = [10, 8, 10, 5, 4, 2, 5, 3, 5, 6, 10, 6, 2, 10, 3, 8, 10], G); > flow(g, 1, 9); Every additional quantity of flow must pass through one or more of the edges of the cut. Let us assume that we replace the edge between 2 şi 7 of capacity 4 with capacity 7. Using the marking procedure we obtain the chain from the figure 5.8. Fig. 5.8: Chain obtain after the marking process. The propagated flow is presented in figure 5.9. Edges [2, 3] şi [3, 4] are not used.

23 MAXIMUM FLOW IN A NETWORK 89 Fig. 5.9: Propagated flow. In the figure we observe the path: µ = ( ), ϕ = 2. Situation is presented in figure 5.10, and the propagated flow is equal with 26. Fig. 5.10: Propagated flow and the minimal capacity cut. The new cut has the capacity 26, has the minimum value and is composed from

24 90 TRANSPORTATION PROBLEMS the edges: {(1, 2), (3, 2), (3, 7), (3, 5), (4, 5)} Exercise Let us consider the transportation network from the figure Which is the value of the maximum flow propagated between the vertex 1 and vertex 5? (α > 0 real parameter). Fig. 5.11: The propagated flow. It is shown the minimum capacity cut. Solution. At the first step we generate orientated routes, which saturates, from vertex 1 to vertex 5. µ 1 : 1 40α 2 15α 3 22α 5, ϕ 1 = min{40α, 15α, 22α} = 15α; µ 2 : 1 25α 2 8α 5, ϕ 2 = min{25α, 8α} = 8α; µ 3 : 1 17α 2 12α 4 12α 3 7α 5, ϕ 3 = min{17α, 12α, 12α, 7α} = 7α; µ 4 : 1 18α 4 18α 5, ϕ 4 = min{18α, 18α} = 18α; The flow is Φ = 4 ϕ i = 48α. i=1 In step 2 we mark the vertices. The final vertex can be not marked thus, the flow is maximum (fig.5.12). The set of unmarked nodes is: A = {5}.

25 MAXIMUM FLOW IN A NETWORK 91 Fig. 5.12: Final vertex 5 can not be marked thus, thus the flow is maximum. The minimum capacity cut is: A= {(4, 5), (2, 5), (3, 5)}, and its value is c( A ) = 48α and Ford-Fulkerson theorem confirms that the flow is maximum. To increase the flow we must increase the capacity of one or more of the edges from the minimum capacity cut. Exercise A product is available in three warehouses in the quantities of 30, 40 respectively 50 tones. This product is requested by two consumers in the quantities of 40 respectively 80 t. Between the warehouses and the consumers exist a direct way the transportation is made by trucks with limited capacities (tones) in accordance with table TABLE 6.11 c 1 c 2 d d 2 20 d Determine the optimum transportation plan. Solution. We codificate the problem in accordance with the figure For this we have introduced the node d which is the input in the transportation network and node c which is the output from the transportation network. The problem will be solved with Ford-Fulkerson method. We propagate a maximum flow in the network from the input node d and output node c. For this we generates paths from the node d to the node c (oriented routes in the same way) which are saturated. In the identification of these paths we take care to use the

26 92 TRANSPORTATION PROBLEMS Fig. 5.13: Codification of the maximum flow problem. unoriented route only in one way. Thus, we indicate the direction of the routes and the propagated flow. When we identify a path we also direct the routes. On each edge we compute the total propagated flows. Thus, we obtain have the paths: µ 1 : d 30 d 1 10 c 1 40 c, ϕ 1 = min{30, 10, 40} = 10, µ 2 : d 20 d 1 2 c 2 80 c, ϕ 2 = min{20, 2, 80} = 2, µ 3 : d 40 d 2 20 c 1 30 c, ϕ 3 = min{40, 20, 30} = 20, µ 4 : d 50 d 3 10 c 1 10 c, ϕ 4 = min{50, 10, 10} = 10, µ 5 : d 40 d 3 20 c 2 78 c, ϕ 5 = min{40, 20, 78} = 20. At this time, all the paths from the node c to node d contains only oriented routes and each path has at least one saturated route. The flow propagated through these chains is equal with: Φ = 5 ϕ i = 62. i=1 In the next step we apply the marking procedure. Thus, the following nodes are marked like in figure 5.14.

27 MAXIMUM FLOW IN A NETWORK 93 Fig. 5.14: Marking the nodes. The final node can not be marked thus the flow is maximum. The final node d can not be marked, thus, the flow is maximum. unmarked nodes is: The set of A = {d 1, d 2, d}, and the minimum capacity cut is composed from the routes: A= {(c 1, d 1 ), (c 1, d 2 ), (c 2, d 1 ), (c 3, d 1 ), (c 3, d 2 )}, with the value of the flow: c( A) = = 62. We see that the value of the minimum capacity cut it is equal with the maximum value of the flow, which is a Ford-Fulkerson confirmation that the flow is optimum (maximum) (fig. 5.15).

28 94 TRANSPORTATION PROBLEMS Fig. 5.15: Minimum capacity cut and the propagated flow. Observe that the request can not be satisfied. For increasing the capacity we must increase the capacity of one of the edges which belong to the minimum capacity cut. Exercise Write a MAPLE code for solving the maximum flow problem in a transportation network. Check the result using Ford-Fulkerson theorem. Solution. For a better understanding we shall write the code for problem Initialization of the variables: list of vertex (nodes), list of edges (routes), list of capacities, starting node s destination node t. > nodes := {1, 2, 3, 4, 5, 6, 7, 8, 9}; > routes := [{1, 2}, {1, 3}, {1, 4}, {2, 6}, {2, 7}, {2, 3}, {3, 4}, {3, 7}, {3, 5}, {6, 7}, {6, 8}, {4, 5}, {5, 7}, {5, 9}, {7, 8}, {7, 9}, {8, 9}]; > capacities := [10, 8, 10, 5, 4, 2, 5, 3, 5, 6, 10, 6, 2, 10, 3, 8, 10]; > s := 1; > t := 9; Creation of the network. > with(networks) : > new(g) : > addvertex(nodes, G) : > addedge(routes, weights = capacities, G); Maximum flow. > flow(g, s, t); Minimum capacity cut. > mincut(g, s, t, vf); Minimum capacity cut must be equal with the maximum value of the flow (Ford- Fulkerson theorem).

29 MAXIMUM FLOW IN A NETWORK 95 > vf; Graph G. > draw(g); Exercise For which value of the parameter a the graph from the figure 5.16 has a maximum flow, between nodes 1 and 3, is 40. Fig. 5.16: Maximum flow between nodes 1 and 3. Solution. We apply the Ford-Fulkerson algorithm, and find the value of a from the equation Φ max = 40. Thus, we perform the following steps: a) generate oriented routes from node 1 to node 3: µ 1 : 1 2a 2 8a 5 2a 3, ϕ 1 = 2a, µ 2 : 1 3a 5 3a 6 a 3, ϕ 2 = a, µ 3 : 1 8a 6 3a 4 4a 3, ϕ 3 = 3a. Flow in the network is Φ = 6a; b) mark the nodes: {1, 2, 5, 6}. c) minimum capacity cut is: {(5, 3), (6, 3), (6, 4)}, the value of the flow through it is 6a, thus, the flow is maxim. Equation Φ max = 40 gives a = 20/3. Exercise Solve the above problem using simplex algorithm. Exercise Consider the transportation network from the figure 5.17 with the input 0 and the output 7. Determine the following: i) capacity of the cut of the sets A 1 = {1, 4, 6, 7} and A 2 = {5, 6, 7};

30 96 TRANSPORTATION PROBLEMS Fig. 5.17: Transportation network. ii) maximum flow through the network; iii) what transport capacity must be increased and what it is its value to achieve a maximum flow in the network of 30? Solution is unique? iv) what is the value of the maximum flow in the transportation network if we remove routes 1 4 respectively 5 1? Solution. The capacities of the cuts of the sets A 1 and A 2 are 50 respectively 35. We apply Ford-Fulkerson algorithm. Maximum flow in the network will be equal with 23. To increase the flow we need to increase the minimum capacity cut.