2. Network flows. Balance of flow: At each node, demand plus total shipments out must equal supply plus total shipments in. Reno. Denver.

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1 . Network flows A network consists of a collection of locations along with connections between them. The locations, called the nodes of the network, can correspond to places of various kinds such as factories, warehouses, stores or customers. The connections, known as arcs, tie together certain pairs of nodes; they may designate roads, cables or flight corridors. A network model is concerned with a flow of something packages, for example, or messages or electricity from node to node along the arcs. To be specific, for this case study we consider a network in which nodes are cities, arcs are highways, and flows are truckloads of manufactured goods. It is convenient to represent a network by a diagram in which the nodes are drawn as circles and the arcs as arrows connecting circles. Thus our cities are circles and our highways are arrows, as in this simple example: Reno Denver Los Angeles Dallas Phoenix Houston The dashed arrows pointing into cities represent amounts produced there and available for shipment out: truckloads at Los Angeles and at Reno. The dashed arrows pointing out of cities represent amounts sold there and hence needed as shipments in: truckloads at Houston and at Dallas. Solid arrows represent possible directions of shipments between cities. The network flow problem seeks seeks to ship all of the supply from Los Angeles and Reno so as to meet all of the demand at Houston and Dallas. A solution to the problem tells, for each city pair represented by an arrow, how many truckloads to ship between the cities in the direction of the arrow. Flow balance equations. The shipment amounts in a solution must satisfy certain linear equations to insure that all truckloads are accounted for. In general terms, these equations say that, at each city, the amount coming in must equal the amount going out. In our particular case, goods can come into a city by being produced there or by being shipped in from other cities; goods can go out from a city by being sold there or by being shipped out to other cities. Thus, in terms of the network, the equations can be stated as follows: Balance of flow: At each node, demand plus total shipments out must equal supply plus total shipments in. 8

2 In our example, supply is at Los Angeles and at Reno, while demand is at Houston and at Dallas. All other supply and demand amounts are. To turn this word description into equations that we can solve, we ll write x ij to represent the flow from node i to node j, where the numbering of nodes is as given in the diagram. Then at node, for example, shipments in are given by x and shipments out by x + x + x. Together with the supply of and demand of zero, this gives us the balance equation + x + x + x = + x. At node, the flow in is x + x and the flow out is x, while both supply and demand are zero, leading us to the balance equation + x = + x + x. Proceeding in this way, and dropping the zeroes, we arrive at the following balance equations for the sixnodes: x + x = x + x + x = + x x = x + x x + x = x + x = x + x + x = x + x The flows through the network are described by the values of nine variables (corresponding to the nine arcs) that must satisfy sixlinear equations (corresponding to the sixnodes). In order to characterize the possible solutions, we first put these equations into matrixform. We move variables to the left of the = sign and constants to the right, and give a coefficient of to each variable that does not appear explicitly. Then the balance of flow can be regarded as a 9 equation system: x x x x x x x x x where (x,x,x,x,x,x,x,x,x ). We have placed the name of each variable under its column of coefficients in the matrix; this list of names has no mathematical significance, but it makes the equations easier to interpret. When Gaussian elimination is applied to these equations, the result is the following equivalent system in echelon form: x x x x x x x x x, 9

3 The pivot (or basic ) variables, corresponding to the columns in which the echelon form steps down, are x, x, x, x and x. The free variables are the other four: x, x, x and x. There are only pivot variables for these equations. One equation at the bottom turns out to be all zeroes. Thus we can conclude that there must be infinitely many solutions. To get a general form for the solutions, we can substitute x = c, x = c, x = c, and x = c for the free variables; then the first five equations become an upper-triangular system in the pivot variables: x + x = x + x + x = c x + x = c x + x = c c x = c These equations are easily solved to give x = c x = c c + c x = + c c x = c + c x = + c c which together with x = c x = c x = c x = c gives a general solution to the network flow equation system. This solution may be rewritten in vector notation as x x x x x = + c + c + c + c. x x x x Several interesting properties of the network can be deduced from this representation. Cycles of flow. Consider first the solution obtained by setting c = c = c = c =. In this case the solution has positive flows only along the five arcs that correspond to the basic (pivot) variables. To depict this solution in a network diagram, we can thicken the basic arcs and show the amount of flow next to each one:

4 The basic arcs form a structure in the network known as a spanning tree: they reach every node (hence are spanning ) yet form no loops (hence are a tree ). It turns out that, for any network flow equations, every subset of basic variables corresponds to some spanning tree. Conversely, every spanning tree gives rise to a triangular system of equations, which has a unique solution when the flows on the other arcs are fixed. Now consider the term that contains c in the general-form solution above. It adds +c to x, c to x, and +c to x. Looking at the network diagram, you can see that x, x and x form a loop or cycle. In fact, exactly this one cycle is formed when nonbasic arc x is added to the spanning tree. If you go around this cycle forward through arc, you will then go backward and forward ; this observation corresponds to the fact that you add c to x, subtract it from x, and add it to x. In summary, you can think of the c term in the general solution as representing an adjustment of flow around a cycle. This adjustment leaves the equations satisfied at all nodes, but changes the solution on the cycle s arcs. The other three terms in the general solution correspond, in an analogous way, to cycles created when the other nonbasic arcs are added to the spanning tree. The addition of arc, for example, gives a cycle with, and. Suppose you start with the solution given by c = c = c = c =, but increase c to a positive value t. Then according to the general-form solution, the flow increases on x and x, while it decreases on x and x : t + t t t

5 Clearly all the flows remain nonnegative for t. At t =, the flow on falls to zero, and the remaining nonzero flows are x = 7, x =, x =, x =, and x =. These five variables make up another basic solution (which we might have arrived at had we ordered the variables differently when performing Gaussian elimination). You can also check that the arcs corresponding to these variables make up another, different spanning tree in the network. Thus, given a nonnegative basic solution to the network flow equations, we have a simple way of moving to another such solution. There are also an infinity of nonnegative solutions between these two, given by setting t to values between and. Minimum-cost solutions. Among all the solutions to the flow balance equations, which should be preferred? There s no good way to tell, from only the information given. Suppose however that we also know a cost per truckload for each arc, expressed as follows in hundreds of dollars: $ $ $ $9 $ $ $ $7 $ Then the cost of all shipments on a particular arc is given by the cost per truckload times the number of truckloads: x for shipments on, 9x for shipments on, and so forth. Ignoring the variables at zero, the total cost for all shipments in our original spanning-tree solution is $x + $x + $x + $9x + $x = $ + $ + $ + $9 + $ = $, while the total cost for the new spanning-tree solution that we derived by increasing the flow on x tois $x + $x + $x + $7x + $x = $ 7 + $ + $ + $7 + $ = $9. The second solution turns out to be the better one. We could have checked whether the second solution would be preferable, before going to the trouble of computing it. Recall that for each truckload added on that is, for each unit of increase in the our parameter t we must remove a truckload from, remove a truckload from, and add a truckload on. Thus each increase of t by one truckload incurs a cost of

6 $7 more on, $9 less on, $ less on, and $ more on for a total of $ less overall. It follows that, the higher you make t, the more you save. The best you can do is to increase t to, which is the highest it can go without forcing some flow to a negative value. At that point you have reached a different basic solution, whose total cost is $ = $ less. These observations suggest a way of finding the lowest possible shipping cost to meet the demands. You start with any basic solution that has nonnegative flows. Then you pick any one nonbasic variable such that the cost per unit of putting flow on the resulting cycle is negative. Finally you increase the flow around this cycle until some basic variable falls to zero, at which point you have found a lower-cost basic solution. You keep repeating this procedure until, eventually, you arrive at some basic solution where none of the associated cycles afford any reduction in cost. At that point, you can stop and declare that you have found the lowest-cost flow that satisfies the balance equations. This approach, with a few refinements, is known as the network simplex method. It really does find a minimum-cost flow for any network. The details of why it works, and how it can be made to run efficiently on large networks, are however beyond the scope of a linear algebra course. You may have noticed that we have not said anything about how a nonnegative basic solution can be found in the first place. When we applied Gaussian elimination, we happened to arrive at a solution that had all variables, but if the variables had been ordered differently we could well have determined a different solution in which some variables had values <. Gaussian elimination has no easy way of forcing the solution to be nonnegative, which is why nonnegativity is ignored in the development of elimination methods. In fact finding a nonnegative solution to an equation system is as hard as finding a minimum-cost solution, and requires the same advanced methods. Existence of solutions. To conclude, we consider the following question: Which other combinations of supplies and demands would allow the network equations to have a solution? Suppose that we represent our original equations,, x x x x x x x x x as A b, so that A stands for the matrixon the left, and b for the vector on the right. You can see that any supply at network node i appears as a positive value of b i, while any demand appears as a negative value. At a node i where there is no supply or demand, b i =. Our question thus amounts to determining the conditions on the vector b

7 such that a solution exists. One way to do this is to apply elimination to b b b b, b b with the result being b b + b b + b + b b + b + b + b. b + b + b + b + b b + b + b + b + b + b Clearly a solution will exist if and only if the constant term for the last equation becomes zero; that is, if and only if b i =. i= Since the supplies are positive b i values, and the demands are negative b i values, this just says that the total supply minus total demand must equal zero. In other words, when total supply equals total demand then there is always a solution to the network equations, while when supply and demand are unequal there is never a solution. This confirms exactly what you would expect. Another way to approach this same result is to observe that every column of A has two nonzeroes, one + and one. Thus the sum over each column of A is zero, or equivalently ea = where e = [] is a row vector of all ones. We can use this fact to reason that A b ea eb = b i. i= That is, if a solution exists, then supplies must equal demands in b, by the same reasoning as before. This approach can also be used to establish the converse, that if supplies minus demands in b are zero then a solution must exist, but somewhat more advanced reasoning is required in that direction.