ECE 307 – Techniques for Engineering Decisions Transshipment and Shortest Path Problems George Gross Department of Electrical and Computer Engineering University of Illinois at Urbana-Champaign © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved TRANSSHIPMENT PROBLEMS We consider the shipment of a homogeneous commodity from a specified point or source to a particular destination or sink In general, the source and the sink need not be directly connected; rather, the flow goes through the transshipment points or the intermediate nodes The objective is to determine the maximal flow from the source to the sink © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved FLOW NETWORK EXAMPLE s t © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved TRANSSHIPMENT PROBLEMS nodes 1, 2, 3, 4, and are the transshipment points arcs of the network are ( s, ), ( s, ), ( 1, ), ( 1, ), ( 2, ), ( 3, ), ( 3, ), ( 4, ), ( 5, ), ( 4, t ), ( 5, t ) ; the existence of an arc from to and from to allows bidirectional flows between the two nodes each arc may be constrained in terms of a limit on the flow through the arc © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved MAX FLOW PROBLEM We denote by f ij the flow from i to j and this equals the amount of the commodity shipped from i to j on an arc ( i , j ) that directly connects the nodes i and j The problem is to determine the maximal flow f from s to t taking into account the flow limits k ij of each arc ( i , j ) The mathematical statement of the problem is © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved MAX FLOW PROBLEM max Z = f s.t ≤ f ij ≤ k ij ∀ arc ( i , j ) that connects nodes i and j ∑ f si = f at source s i ∑i f it = ∑i f ij f = at sink t ∑k f jk conservation of flow relations ⎫⎪ ⎬ at each transshipment node j ⎪⎭ © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved MAX FLOW PROBLEM While the simplex approach can solve the max flow problem, it is possible to construct a highly efficient network method to find f directly We develop such a scheme by making use of network or graph theoretic notions We start by introducing some definitions © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved DEFINITIONS OF NETWORK TERMS Each arc is directed and so for an arc ( i , j ), f ij ≥ A forward arc at a node i is one that leaves the node i to some node j and is denoted by ( i , j ) A backward arc at node i is one that enters node i from some node j and is denoted by ( j , i ) © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved DEFINITIONS OF NETWORK TERMS A path connecting node i to node j is a sequence of arcs that starts at node i and terminates at node j we denote a path by P = { ( i, k ), ( k, l ), , ( m, j ) } in the example network •( 1, ), ( 2, ), ( 5, ) is a path from to •( 1, ), ( 3, ) is another path from to © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved DEFINITIONS OF NETWORK TERMS A cycle is a path with i = j , i.e., P = { ( i, k ), ( k, l ), , ( m, i ) } We denote the set of nodes of the network by N the definition is N = { i : i is a node of the network } In the example network N = { s , 1, 2, 3, 4, 5, t } © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 10 determines the shortest distance from to every other node EXAMPLE : FIVE – NODE NETWORK © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 67 EXAMPLE : FIVE – NODE NETWORK 0 3 2 10 4 © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 68 APPPLICATION : EQUIPMENT REPLACEMENT PROBLEM We consider the problem of replacing old equipment or continuing its maintenance As equipment ages, the level of maintenance required increases and typically, this results in increased operating costs O&M costs may be reduced by replacing aging equipment; however, replacement requires additional capital investment and so higher fixed costs © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 69 APPPLICATION : EQUIPMENT REPLACEMENT PROBLEM The problem is how often to replace equipment so as to minimize the total costs given by total costs = capital costs fixed + O&M costs variable © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 70 EXAMPLE: EQUIPMENT REPLACEMENT Equipment replacement is planned during the next years The cost elements are p j = purchase costs in year j s j = salvage value of original equipment after j years of use c j = O&M costs in year j of operation of equipment with the property that … cj < cj + < cj + < … We formulate this problem as a shortest route problem on a directed network © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 71 EQUIPMENT REPLACEMENT PROBLEM end of d13 d12 d15 d14 period d36 d35 d23 d34 d24 start of planning d16 d45 d25 d56 d46 d26 planning period © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 72 APPPLICATION : EQUIPMENT REPLACEMENT PROBLEM where, the “distances” d ij are defined to be finite if i < j , i.e., year i precedes the year j , with j−i d ij = pi − s j−i + ∑ cτ j > i τ =1 purchase salvage value O&M costs price in after j – i for j – i years year i years of use of operation © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 73 APPPLICATION : EQUIPMENT REPLACEMENT PROBLEM For example, if the purchase is made in year d 16 = p − s + c ∑ τ =1 τ The solution is the shortest distance path from year to year ; if for example the path is { ( 1, 2) , (2, 3) , (3, 4) , (4, 5) , (5, 6) } then the solution is interpreted as the replacement of the equipment each year with total costs = ∑ i =1 p i − s + 5c © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 74 COMPACT BOOK STORAGE IN A LIBRARY This problem concerns the storage of books in a limited size library Books are stored according to their size, in terms of height and thickness, with books placed in groups of same or higher height; the set of book heights { Hi } is arranged in ascending order with H1 < H2 < … < Hn © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 75 COMPACT BOOK STORAGE IN A LIBRARY Any book of height Hi may be shelved on a shelf of height at least Hi , i.e., Hi , Hi+1 , Hi+2 , The length Li of shelving required for height Hi is computed given the thickness of each book; the total shelf area required is ∑ Hi Li i if only height class [ corresponding to the tallest book ] exists, total shelf area required is the total length of the thickness of all books times the height of the tallest book © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 76 COMPACT BOOK STORAGE IN A LIBRARY if or more height classes are considered, the total area required is less than the total area required for a single class The costs of construction of shelf areas for each height class Hi have the components si fixed costs [ independent of shelf area ] ci variable costs / unit area © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 77 COMPACT BOOK STORAGE IN A LIBRARY For example, if we consider the problem with height classes Hm and Hn with Hm < Hn all books of height ≤ Hm are shelved in shelf with the height Hm all the other books are shelved on the shelf with height Hn The corresponding total costs are ⎡ ⎢ sm + c m H m ⎢⎣ ⎤ ⎡ L j ⎥ + ⎢ sn + c n H n ⎥⎦ ⎢⎣ j =1 m ∑ ⎤ Lj ⎥ ⎥⎦ j = m +1 n ∑ © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved 78 ... Gross, University of Illinois at Urbana-Champaign, All Rights Reserved FLOW NETWORK EXAMPLE s t © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign, All Rights Reserved TRANSSHIPMENT... s to t is equal to the minimal cut, i.e., the cut S , T with the smallest capacity The max-flow min-cut theorem allows us, in principle, to find the maximal flow in a network by finding the... Urbana-Champaign, All Rights Reserved 19 EXAMPLE Consider the simple network with the flow capacities on each arc indicated s t © 2006 – 2009 George Gross, University of Illinois at Urbana-Champaign,