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Readings Readings Chapter 6 Distribution and Network Models BA 452 Lesson B.2 Transshipment and Shortest Route 1 Overview Overview BA 452 Lesson B.2 Transshipment and Shortest Route 2 Overview Transshipment Problems are Transportation Problems extended so that a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node. Transshipment Problems with Transshipment Origins are Transshipment Problems where goods from one origin may move through other origins before reaching a destination. Shortest Route Problems are Transshipment Problems where there is one origin, one destination, one unit supply, and one unit demand, and where that unit is indivisible, as in driving through cities to work. BA 452 Lesson B.2 Transshipment and Shortest Route 3 Overview Tool Summary Define decision variable xij = units moving from origin i to destination j. Write origin constraints (with < or =): n x ij si i 1, 2, ,m Supply j 1 Write destination constraints (with < or =): m x ij dj j 1, 2, ,n Demand i 1 Write transshipment constraints (with < or =): arcs out xij xij 0 Transhipment nodes arcs in BA 452 Lesson B.2 Transshipment and Shortest Route 4 Overview Tool Summary Identify implicit assumptions needed to complete a formulation, such as all agents having an equal value of time. BA 452 Lesson B.2 Transshipment and Shortest Route 5 Transshipment Transshipment BA 452 Lesson B.2 Transshipment and Shortest Route 6 Transshipment Overview Transshipment Problems are Transportation Problems extended so that a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node. BA 452 Lesson B.2 Transshipment and Shortest Route 7 Transshipment This is the network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations: 3 c13 s1 1 c37 c14 s2 d1 6 c46 c47 4 c15 Supply c36 Demand c23 2 Sources c56 c24 c25 5 c57 7 d2 Destinations Intermediate Nodes BA 452 Lesson B.2 Transshipment and Shortest Route 8 Transshipment Notation: xij = number of units shipped from node i to node j cij = cost per unit of shipping from node i to node j si = supply at origin node i dj = demand at destination node j Min c ij xij all arcs s.t. xij arcs out Origin nodes i xij 0 Transhipment nodes xij d j Destination nodes j arcs in xij arcs out arcs in arcs in xij si xij arcs out xij > 0 for all i and j BA 452 Lesson B.2 Transshipment and Shortest Route 9 Transshipment Problem Variations • Minimum shipping guarantee from i to j: xij > Lij • Maximum route capacity from i to j: xij < Lij • Unacceptable route: Remove the corresponding decision variable. BA 452 Lesson B.2 Transshipment and Shortest Route 10 Transshipment Question: The Northside and Southside facilities of Zeron Industries supply three firms (Zrox, Hewes, Rockrite) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently, weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockrite. Both Arnold and Supershelf can supply up to 75 units to its customers. Because of long-standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Zeron N Zeron S Arnold 5 8 Supershelf 7 4 The costs to install the shelving at the various locations are: Zrox Hewes Rockrite Zeron N 1 5 8 Zeron S 3 4 4 BA 452 Lesson B.2 Transshipment and Shortest Route 11 Transshipment Formulate and solve a transshipment linear programming problem for Zeron Industries. BA 452 Lesson B.2 Transshipment and Shortest Route 12 Transshipment Answer: Since demands by the three customer firms (Zrox, Hewes, Rockrite) are fixed, revenue for Zeron Industries is fixed, and so profit maximization is the same as cost minimization. There is data on transportation costs, but there is no data on the cost of ordering from suppliers. Nevertheless, if the unit cost from each supplier is the same, say P per unit, then the cost of ordering supply equal to the fixed demand of 50, 60, and 40 is fixed at 150P. Hence, to minimize cost, we just have to minimize transportation cost. To that end, it may help to draw a network model: BA 452 Lesson B.2 Transshipment and Shortest Route 13 Transshipment 75 ARNOLD Arnold 5 Zeron N 8 75 4 50 Hewes HEWES 60 RockRite 40 1 5 8 3 7 Super Shelf Zrox Zeron WASH BURN S 4 4 BA 452 Lesson B.2 Transshipment and Shortest Route 14 Transshipment Next, Define decision variables: xij = amount shipped from manufacturer i to supplier j xjk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) Problem Features: There will be 1 variable for each manufacturer-supplier pair and each suppliercustomer pair, so 10 variables all together. There will be 1 constraint for each manufacturer, 1 for each supplier, and 1 for each customer, so 7 constraints all together. BA 452 Lesson B.2 Transshipment and Shortest Route 15 Transshipment Define objective function: Minimize total shipping costs. Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 + 3x45 + 4x46 + 4x47 Constrain amount out of Arnold: x13 + x14 < 75 Constrain amount out of Supershelf: x23 + x24 < 75 Constrain amount through Zeron N: x13 + x23 - x35 - x36 - x37 = 0 Constrain amount through Zeron S: x14 + x24 - x45 - x46 - x47 = 0 Constrain amount into Zrox: x35 + x45 = 50 Constrain amount into Hewes: x36 + x46 = 60 Constrain amount into Rockrite: x37 + x47 = 40 BA 452 Lesson B.2 Transshipment and Shortest Route 16 Transshipment Indicies: i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) Minimized shipping costs Out of Arnold through Zeron N Out of Supershelf through Zeron S Through Zeron N into Zrox Through Zeron S into Hewes BA 452 Lesson B.2 Transshipment and Shortest Route 17 Transshipment Indicies: i = 1 (Arnold), 2 (Supershelf) j = 3 (Zeron N), 4 (Zeron S) k = 5 (Zrox), 6 (Hewes), 7 (Rockrite) ZROX 75 ARNOLD Arnold 5 75 Zeron N 8 75 4 50 Hewes HEWES 60 RockRite 40 1 5 8 3 4 7 Super Shelf Zrox Zeron WASH BURN S 4 BA 452 Lesson B.2 Transshipment and Shortest Route 18 Transshipment with Transshipment Origins Transshipment with Transshipment Origins BA 452 Lesson B.2 Transshipment and Shortest Route 19 Transshipment with Transshipment Origins Overview Transshipment Problems with Transshipment Origins seek to minimize the total shipping costs of transporting goods from m origins (each with a supply si) to n destinations (each with a demand dj), where goods from one origin may move through other origins (transshipment nodes) before reaching a particular destination node. BA 452 Lesson B.2 Transshipment and Shortest Route 20 Transshipment with Transshipment Origins Question: Index cities i = 1 (Newbury Park), i = 2 (Thousand Oaks), i = 3 (Westlake Hills), i = 4 (Agoura Hills), i = 5 (Calabasas). Suppose you run rental car lots in each city. Newbury Park has a surplus of 3 cars (it has 3 more cars than it needs), Westlake Hills has a surplus of 2 cars (it has 2 more cars than it needs), and Calabasas has a deficit of 5 cars (it needs 5 more cars than it has). BA 452 Lesson B.2 Transshipment and Shortest Route 21 Transshipment with Transshipment Origins Suppose you calculate the following costs per car of transporting cars between the cities: • transporting between 1 and 2 (that is, either 1 to 2, or 2 to 1) costs $2 • transporting between 1 and 3 costs $3 • transporting between 1 and 4 costs $4 • transporting between 1 and 5 costs $5 • transporting between 2 and 3 costs $2 • transporting between 2 and 4 costs $3 • transporting between 2 and 5 costs $4 • transporting between 3 and 4 costs $2 • transporting between 3 and 5 costs $3 • transporting between 4 and 5 costs $2 How should you move cars between cities? Formulate your rental-car problem as a linear program, but you need not solve for the optimum. Tip: Your written answer should define the decision variables, and formulate the objective and constraints. BA 452 Lesson B.2 Transshipment and Shortest Route 22 Transshipment with Transshipment Origins Answer: Define decision variables: xij = amount of cars moved from City i to City j Define objective function: Minimize total costs. Min 2(x12+x21) + 3(x13+x31) + 4(x14+x41) + 5(x15+x51) + 2(x23+x32) + 3(x24+x42) + 4(x25+x52) + 2(x34+x43) + 3(x35+x53) + 2(x45+x54) Constrain cars from City 1: x12 + x13 + x14 + x15 = 3 + x21 + x31 + x41 + x51 Constrain cars from City 2: x21 + x23 + x24 + x25 = x12 + x32 + x42 + x52 Constrain cars from City 3: x31 + x32 + x34 + x35 = 2 + x13 + x23 + x43 + x53 Constrain cars from City 4: x41 + x42 + x43 + x45 = x14 + x24 + x34 + x54 Constrain cars from City 5: x51 + x52 + x53 + x54 = -5 + x15 + x25 + x35 + x45 (Or constraints can be written with < rather than =. It does not matter since excess supply exactly cancels excess demand.) BA 452 Lesson B.2 Transshipment and Shortest Route 23 Shortest Route Shortest Route BA 452 Lesson B.2 Transshipment and Shortest Route 24 Shortest Route Overview Shortest Route Problems are Transshipment Problems where there is one origin, one destination, one unit supplied, and one unit demanded, and where that unit is indivisible. Shortest Route Problems find the shortest path in a network from one node (or set of nodes) to another node (or set of nodes). The criterion to be minimized in the shortest-route problem is not limited to distance, but can be minimum time or cost. BA 452 Lesson B.2 Transshipment and Shortest Route 25 Shortest Route Notation: xij = 1 if the arc from node i to node j is on the shortest route 0 otherwise cij = distance, time, or cost associated with the arc from node i to node j Min cij xij all arcs s.t. 1 xij Origin node i arcs out arcs out xij xij 0 Transhipment nodes arcs in xij 1 Destination node j arcs in xij > 1 BA 452 Lesson B.2 Transshipment and Shortest Route 26 Shortest Route Question: Susan Winslow has an important business meeting in Paducah this evening. She has a number of alternate routes by which she can travel from the company headquarters in Lewisburg to Paducah. The network of alternate routes and their respective travel time, ticket cost, and transport mode appear on the next two slides. If Susan earns a wage of $15 per hour, what route should she take to minimize the total travel cost? BA 452 Lesson B.2 Transshipment and Shortest Route 27 Shortest Route F 2 5 K A L B C 1 Lewisburg D E G 3 J 6 I H 4 Paducah M BA 452 Lesson B.2 Transshipment and Shortest Route 28 Shortest Route Route (Arc) A B C D E F G H I J K L M Transport Mode Train Plane Bus Taxi Train Bus Bus Taxi Train Bus Taxi Train Bus Time (hours) 4 1 2 6 3 1/3 3 4 2/3 1 2 1/3 6 1/3 3 1/3 1 1/3 4 2/3 Ticket Cost $ 20 $115 $ 10 $ 90 $ 30 $ 15 $ 20 $ 15 $ 15 $ 25 $ 50 $ 10 $ 20 BA 452 Lesson B.2 Transshipment and Shortest Route 29 Shortest Route Using the wage of $15 per hour, compute total cost. Route A B C D E F G H I J K L M Transport Mode Train Plane Bus Taxi Train Bus Bus Taxi Train Bus Taxi Train Bus Time (hours) 4 1 2 6 3 1/3 3 4 2/3 1 2 1/3 6 1/3 3 1/3 1 1/3 4 2/3 Time Cost $60 $15 $30 $90 $50 $45 $70 $15 $35 $95 $50 $20 $70 Ticket Cost $ 20 $115 $ 10 $ 90 $ 30 $ 15 $ 20 $ 15 $ 15 $ 25 $ 50 $ 10 $ 20 BA 452 Lesson B.2 Transshipment and Shortest Route Total Cost $ 80 $130 $ 40 $180 $ 80 $ 60 $ 90 $ 30 $ 50 $120 $100 $ 30 $ 90 30 Shortest Route Define indices: Nodes 1 (origin), 2, …, 6 (destination) Define decision variables: xij = 1 if the route from node i to node j is on the shortest route Problem Features: There is 1 decision variable for each possible route. There is 1 constraint for each node. Cost of route from 1 to 4 Define objective function: Minimize total transportation costs. Min 80x12 + 40x13 + 80x14 + 130x15 + 180x16 + 60x25 + 100x26 + 30x34 + 90x35 + 120x36 + 30x43 + 50x45 + 90x46 + 60x52 + 90x53 + 50x54 + 30x56 BA 452 Lesson B.2 Transshipment and Shortest Route 31 Shortest Route Node flow-conservation constraints: x12 + x13 + x14 + x15 + x16 = 1 (origin) – x12 + x25 + x26 – x52 = 0 (node 2) – x13 + x34 + x35 + x36 – x43 – x53 = 0 (node 3) – x14 – x34 + x43 + x45 + x46 – x54 = 0 (node 4) – x15 – x25 – x35 – x45 + x52 + x53 + x54 + x56 = 0 (node 5) x16 + x26 + x36 + x46 + x56 = 1 (destination) BA 452 Lesson B.2 Transshipment and Shortest Route 32 Shortest Route Minimized transportation cost Variables: xij = 1 if the arc from node i to node j is on the shortest route 0 otherwise Possible arcs: x12, x13, x14, x15, x16, x25, x26, x34, x35, x36, x43, x45, x46, x52, x53, x54, x56 From Origin to Node 3 From Node 3 to Node 4 From Node 4 to Node 5 From Node 5 to Destination BA 452 Lesson B.2 Transshipment and Shortest Route 33 Shortest Route BA 452 Lesson B.2 Transshipment and Shortest Route 34 Shortest Route Input window Cost of Arc from Node 1 to Node 3 Output window Minimized transportation costs BA 452 Lesson B.2 Transshipment and Shortest Route 35 BA 452 Quantitative Analysis End of Lesson B.2 BA 452 Lesson B.2 Transshipment and Shortest Route 36