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## Network Optimization

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**Network Optimization**Chapter 2 Transshipment Problem**2.1 The Network Simplex Method**• G = (V, E) is a digraph. |V|=n, |E|=m. • A type of commodity is shipped on the network. • For each arc, is the costof sending one unit of commodity from vertex i to vertex j . • Vector c = has m components, and is called the cost vector. • There are three kinds of vertices:**2.1 The Network Simplex Method**(1) Source If the supply at source i is ( >0 ),then let bi =- . (2) Sink If the demand at sink j is ( >0 ),then let bj = . (3) Intermediate vertex Neither a source, nor a sink. If k is an intermediate vertex, then let bk = 0.**2.1 The Network Simplex Method**S denotes supply and D denotes demand**2.1 The Network Simplex Method**• Balanced transshipment problem • Total supply = total demand: , i.e.**2.1 The Network Simplex Method**• inflow at vertex i~the total commodity sent to i; • outflow at vertexi ~the total commodity sent out from i; • netflow at vertex i~inflow at i - outflow at i; • flow vector x~, where is the amount of commodity from i to j ; • feasible flow x~a flow x that makes the netflow at each vertex i equal bi.**2.1 The Network Simplex Method**A Feasible Flow**2.1 The Network Simplex Method**• Single commodity uncapacitated transshipment problem • Find a feasible flow such that its cost is as low as possible. • Let A be the incidence matrix of the network. • The supply-demand can be expressed by Ax=b**2.1 The Network Simplex Method**• For example, if at vertex i: e1 e2 e3 then A x b netflow at i**2.1 The Network Simplex Method**• The transshipment problem is now formulated as min cx s.t. Ax=b • c ~ row vector, x & b ~ column vector • c, , • uncapacitated ~ all xij have no upper bound**2.1 The Network Simplex Method**Example 2.1 • Consider the network shown below with 7 vertices and 10 arcs in which any flow vector x is of the form xT = [x13 x21 x32 x51 x52x61x64 x67 x74x75]. • The cost vector is given by c = [1 1 1 1 5 11 20 5 5 6] and the supply-demand vector by bT = [0 2 3 5 -2 -3 -5].**2.1 The Network Simplex Method**• Note that the rows of A are linearly dependent because the sum of all rows of A is a zero vector. • So, rank (A) < n. • On the other hand, we will see later that A contains annonsingular submatrix. • Therefore, rank (A) = n-1. • We may delete one equation from Ax = b without changing the solution set.**Do you still remember?**• What is a basic? • What is a basic solution? • What is a basic feasible solution? • What is the main idea of Simplex method? From a vertex(feasible basic sol) to another vertex, obj. value keep decreasing.**2.1 The Network Simplex Method**• Feasible tree solution (FTS) • A feasible flow x is called anFTS if there is a spanning tree T such that if . • As each spanning tree has n-1 edges, an FTS x has at most n-1 nonzero components xij.**2.1 The Network Simplex Method**• If an FTS x has n-1 nonzero components, then the associated spanning tree is unique. It must be the tree consisting of the n-1 edges eij that correspond to the n-1 nonzero xij. • We call such an FTS a nondegenerate FTS. • An FTS x is called a degenerate FTS if the number of nonzero components is less than n-1. • A degenerate FTS may associate with more than one spanning tree.**2.1 The Network Simplex Method**Example b = (-3, 3, 0, 0)T It is easy to see that x = ( 3, 0, 0, 0 )T is a feasible solution of the equation Ax = b.**2.1 The Network Simplex Method**• In fact x is a degenerateFTS, and the associated spanning trees are not unique. They can be one of the following trees: Conversely, every spanning tree T is associated with at most one feasible tree solution.**The above few pages told you**• FS can be classified into FTS/not FTS. • FTS can be further classified into DE/non-DE. • Given non-DE FTS, we can defined a unique ST. • Given DE FTS, there will be many ST. • The following few pages tell: Given a ST, we can solve for a unique solution. The solution may be a FS or not FS.**2.1 The Network Simplex Method**• By relabeling vertices and arcs, we may let the part of A corresponding to the arcs in T contain a nonsingular matrix B. • For example, if we consider the following spanning tree T (formed by solid lines) in Example 2.1:**2.1 The Network Simplex Method**• It is not clear if in A, the columns corresponding to the arcs in T contain annonsingular B.**2.1 The Network Simplex Method**• Relabel vertices and arcs as follows:**2.1 The Network Simplex Method**• How to relabel? First pick up an arbitrary vertex and label it as the first one. In general suppose i vertices and i-1 arcs in the tree are already relabeled ( ). Then from all unlabeled vertices, choose one (say w) which is adjacent to a relabeled vertex (say u), and label w as the i+1th vertex, and also label the arc between w and u in the tree (either (u,w), or (w,u)) as the ith arc. Repeat the procedure until we have labeled all n vertices and n-1 arcs in the tree.**2.1 The Network Simplex Method**• a7, a8, a9, a10 are the four arcs not in T, now the incidence matrix**2.1 The Network Simplex Method**• Let x = (x1, … , x10)T, where xi corresponds to arc ai. Rows 2-7 and columns 1-6 form a matrix B, which is upper triangular, and the diagonal elements are 1 or -1. So, B is a nonsingular matrix.**2.1 The Network Simplex Method**• The feasible tree solution corresponding to T must have x7 = x8 = x9 = x10 = 0. • For xB= ( x1, x2, x3, x4, x5, x6 )T, it can be decided by • bi ~ the supply or demand, or zero corresponding to vi.**2.1 The Network Simplex Method**• Therefore, xB is uniquely determined: (1)If , thenis an FTS, i.e., T is associated with a unique FTS. (2)If some components of xB are negative, then is not a feasible solution, i.e., T is not associated with any FTS.**2.1 The Network Simplex Method**• Summary of the relationship between FTS and ST**2.1 The Network Simplex Method**• From the above analysis, we know that a Tree Solution is actually a Basic Solution of the system Ax = b, and the base B can be determined by an associated spanning tree. An FTS is a feasible basic solution (FBS). • In simplex method we move from one FBS to another FBS. Now in network simplex method, we move from one FTS to another FTS, and geometrically, from one spanning tree to another spanning tree.**2.1 The Network Simplex Method**The dual solution defined by a spanning tree We know that the dual problem of min cx s.t. Ax=b is max yb s.t. where y is a row vector: . Also, the complementarity slackness condition holds.**2.1 The Network Simplex Method**• Now for a nondegenerate FTS x, the associated spanning tree T is unique, and for any arc in T, . Let the column in A which corresponds to arc be, then by the complementarity slackness condition, i.e. (*)**2.1 The Network Simplex Method**• The above set (*) has n-1 equations for nvariables. So, the solutions are not unique. If we fix yn = 0, then y1, … , yn-1 can be uniquely determined. (If we fix yn to another value t, then in the solution, each shall increase its value by t.) • We call the row vector , determined by the equations (*), the dual solution defined by the spanning tree T. (*)**2.1 The Network Simplex Method**• Profitable arcs • Suppose x is an FTS that defines a spanning tree T(x) and a dual solution y. An arc (i, j) which is not in T(x) is said to be profitable with respect to the FTS x if . • Note that if we change the value of yn in deciding y, as all shall change by the same value, the value of every shall remain unchanged. So, the determination of profitable arcs is independent of the value of yn chosen.**2.1 The Network Simplex Method**Theorem 2.2 An FTS u with no profitable arcs is an optimal solution. Proof Suppose u is associated with tree T(u) and dual solution y. Define . We know that . In fact (1)If arc , then by the definition of y, (2)If arc , then . So, for any arc, .**2.1 The Network Simplex Method**• Now for any feasible solution v: Av= b, , . (1) • In particular, when v = u, . (2) • (1) - (2): .**2.1 The Network Simplex Method**• Since there is no profitable arc, for any arc , . • But , and hence u is an optimal solution.**2.1 The network simplex method**• The above theorem provides an optimality criterion. • Now we consider what happens if there is a profitable arc . Of course e is not in T, and . If we can increase the flow on from to ,meanwhile keep for other arcs not in T, then as , we may have (will explain in detail later), i.e. we obtain a better solution v.**2.1 The Network Simplex Method**• When arc e is added to the spanning tree T, a unique cycle C(e) will be obtained.**2.1 The Network Simplex Method**• In the cycle, we call each arc with the same direction as e a forward arc (arcs e1& e2 in the above graph) and each arc with the opposite direction a backward arc (arcs e3 & e4 in the above graph). • If the flow along arc e is increased from 0 to t, then the flow along each forward arc ei is from ui to ui+t, and the flow along each backward arc ej is from uj to uj-t.**2.1 The Network Simplex Method**• There must be some backward arcs in the cycle. Why? Because if not, we may increase tinfinitely (Can this problem be unbounded?). • But when , the new feasible flow v shall let , which is impossible as and .**2.1 The Network Simplex Method**• Now since cycle C(e) has backward arcs, when t is increased to certain amount, the flow along a backward arc becomes zero, and we cannot further increase t (for otherwise some flow shall become negative, violating the feasibility).**2.1 The Network Simplex Method**• In the above example, the largest value of t to maintain a non-negative flow is . • Suppose t = u3. In this case, call e3 the leaving arc. • Generally, if an arc f in C(e) is the leaving arc, we obtain a new tree by deleting and adjoining . • The updated flow vector v is associated with , i.e., v is a new FTS, and the associated spanning tree is .**2.1 The Network Simplex Method**Theorem 2.3 If u is an FTS with a profitable arc e and if v is the updated flow using this arc e, then v is an FTS and . Proof We first show that v is still an FTS, because (1)When we change from flow u to v, the net-flow at each vertex remains unaffected, and hence all supply / demand requirements are still satisfied, i.e. Av = b. (2)Obviously, .**2.1 The Network Simplex Method**• We next show that. • Let , where y is the dual solution of u. • By Theorem 2.2, . • Consider two types of arcs: (a)If arc is in the tree T(u) (b)If arc is not in the T(u) and other than the profitable arc .**2.1 The Network Simplex Method**• Therefore, . • In particular, if u is a nondegenerate FTS , arc t = min { uij | is a backward arc of T(u) }>0 . • If all calculated FTS are non-degenerate, the computation must be terminated after a finite number of iterations (i.e. no cycling).**2.1 The Network Simplex Method**• We now summarize the steps. • Initial Step:find an initial FTS and an associated spanning tree. Let k = 1. • Iterative Steps: for k = 1, 2, … Step 1Calculate dual solution on the tree . Step 2 If , i.e., the network has no profitable arc, terminate, is the optimal FTS.Otherwise obtain aprofitable arc with .**2.1 The Network Simplex Method**Step 3Form the cycle and determine t = min {| is a backward arc in C(e) } • Suppose the minimum is attained at arc f. • Let If (i, j) is a forward arc of C(e) If (i, j) is a backward arc of C(e) If (i, j) = (p, q) For other arcs and . • and return to Step 1.**2.1 The Network Simplex Method**• Let us now use Example 2.1 to explain the algorithm. S = 5 S = 2 D = 2 D = 5 D = 3 S = 3**Tradition about arc labeling**• In the above example, there are totally 10 arcs: (1,3), (2,1), (3,2) (5,1), (5,2), (6,1), (6,4) (6,7) (7,4) (7,5). • We normally, label them according to the first number and then the second. • e1 = (1,3), e2 = (2,1), …