AMPL Network Optimization

Transcription

1 AMPL Network Optimization ORLAB Operations Research Laboratory Borzou Rostami Politecnico di Milano, Italy December 14, 2011

2 Network Optimization Shortest path models Consider the problem of finding the shortest path from node 1 to node 4 in the following network: Remember that the first step in formulating an optimization problem is to specify the decision variables. we need to find a certain path each path consists of a set of arc

3 We might think to make a decision variable for each arc, indicating whether that arc is in our path or not. Variables: if arc is in the shortest path, and if not. Objective function: In this case, the cost of a path is simply the cost of its component arcs: Min Constraints: The constraints must represent a path from node 1 to node 4. Excluding the origin and destination r and s, at every node i on the path we must have one arc entering this node, and one arc leaving this node. That is, for some arc, and for some other arc.

4 { # Model (shortestpath1.mod) param numnodes; set Nodes := 1..numNodes; param cost{i in Nodes, j in Nodes}; param demand{i in Nodes}; var flow{i in Nodes, j in Nodes} >= 0; minimize tripcost: sum{i in Nodes, j in Nodes} flow[i,j] * cost[i,j]; subject to conservation{i in Nodes}: sum{j in Nodes} flow[i,j] = demand[i] + sum{h in Nodes} flow[h,i];

5 # Data (shortestpath1.dat) param numnodes = 4; param cost: := ; param demand := ;

6 #Model (shortestpath2.mod) set Nodes; set Arcs within {Nodes, Nodes}; # Vertex set # Arc set param demand{i in Nodes}; param cost{(i,j) in Arcs }; var flow{(i,j) in Arcs} >= 0; minimize tripcost: sum{(i,j) in Arcs } flow[i,j] * cost[i,j]; subject to constraint{i in Nodes}: sum{(i,j) in Arcs } flow[i,j] = demand[i] + sum{(h,i) in Arcs } flow[h,i];

7 #Data (shortestpath2.dat) set Nodes := ; set Arcs := (1,2) (1,3) (2,3) (2,4)(3,4); param demand := ; param cost:= ;

8 Network Optimization : Minimum-cost transshipment models: Imagine that the nodes and arcs in the Figure represent cities and intercity transportation links. A manufacturing plant at the city marked PITT will make 450,000 packages of a certain product in the next week, as indicated by the 450 at the left of the diagram.

9 The cities marked NE and SE are the northeast and southeast distribution centers, which receive packages from the plant and transship them to warehouses at the cities coded as BOS, EWR, BWI, ATL and MCO. These warehouses require 90, 120, 120, 70 and 50 thousand packages, respectively, as indicated by the numbers at the right. For each intercity link there is a shipping cost per thousand packages and an upper limit on the packages that can be shipped, indicated by the two numbers next to the corresponding arrow in the diagram. The optimization problem over this network is to find the lowest-cost plan of shipments that uses only the available links, respects the specified capacities, and meets the requirements at the warehouses. # Model (transshipment.mod) set CITIES; set LINKS within {CITIES,CITIES};

10 param supply {CITIES} >= 0; param demand {CITIES} >= 0; param cost {LINKS} >= 0; param capacity {LINKS} >= 0 ; var Ship {(i,j) in LINKS} >= 0, <= capacity[i,j] ; # amounts available at cities # amounts required at cities # shipment costs/l000 packages # max packages that can be shipped # packages to be shipped minimize Total_Cost: sum {(i,j) in LINKS} cost[i,j] * Ship[i,j]; subject to Balance {k in CITIES} : supply[k] + sum {(i,k) in LINKS} Ship[i,k]= demand[k] + sum {(k,j) in LINKS} Ship[k,j];

11 # Data (transshipment.dat) set CITIES := PITT NE SE BOS EWR BWI ATL MCO ; set LINKS := (PITT,NE) (PITT,SE) (NE, BOS) (NE, EWR) (NE,BWI) (SE,EWR) (SE,BWI) (SE,ATL) (SE,MCO); param supply default 0 := PITT 450; param demand default 0 := BOS 90, EWR 120, BWI 120, ATL 70, MCO 50; param : cost capacity := PITT NE PITT SE NE BOS NE EWR NE BWI SE EWR SE BWI SE ATL SE MCO ;

12 Network Optimization :Maximum flow models The Figure presents a diagram of a simple traffic network. The nodes and arcs represent intersections and roads: capacities, shown as numbers next to the roads, are in cars per hour. We want to find the maximum traffic flow that can enter the network at a and leave at g.

13 You may permit a parameter to represent character string values, by including the keyword symbolic in its declaration. A symbolic parameter's values may be strings or numbers, just like a set's members, but the string values may not participate in arithmetic. A model for this situation begins with a set of intersections, and symbolic parameters to indicate the intersections that serve as entrance and exit to the road network: set INTER; param entr symbolic in INTER; param exit symbolic in INTER, <> entr; The set of roads is defined as a subset or the pairs of intersections: set ROADS within (INTER diff {exit}) cross (INTER diff {entr}) ; This definition ensures that no road begins at the exit or ends at the entrance. param cap {ROADS} >= 0 ; var Traff {(i,j) in ROADS}>= 0, <= cap[i,j]; The constraints say that except for the entrance and exit, flow into each intersection equals flow out:

14 subject to Balance {k in INTER diff {entr,exit}} : sum {(i,k) in ROADS}Traff[i,k]=sum {(k,j) in ROADS} Traff[k,j]; Given these constraints, the flow out of the entrance must be the total flow through the network, which is to be maximized: maximize Entering_Traff: sum {(entr,j) in ROADS} Traff[entr,j]; Model (Maximum flow.mod) set INTER; # intersections param entr symbolic in INTER; param exit symbolic in INTER, <> entr; # entrance to road network # exit from road network set ROADS within (INTER diff {exit}) cross (INTER diff {entr}) ; param cap {ROADS} >= 0; var Traff {(i,j) in ROADS} >= 0, <= cap[i,j]; # capacities # traffic loads

15 maximize Entering_Traff : sum {(entr,j) in ROADS} Traff[entr,j]; subject to Balance {k in INTER diff {entr,exit}}: sum {(i,k) in ROADS} Traff[i,k] = sum {(k,j) in ROADS} Traff[k,j]; Data (Maximum flow.dat) set INTER := a b c d e f g; param entr : = a; param exit : = g; param : ROADS : cap:= a b 50 b d 40 c d 60 d e 50 e g 70 a c 100 b e 20 c f 20 d f 60 f g 70;

16 Exercise: For new year s eve, the police department of Milan expects a massive inflow of people to the Duomo Square around midnight.

17 It is expected that from midnight to 1 a.m. the people will leave the Square and reach the closest underground stations: Cordusio, San Babila, Missori, Montenapoleone, Cairoli, Cadorna, Lanza. It is also supposed that the people will take the shortest path to reach the underground. For safety reasons, access to the Duomo underground station will be restricted. Consider the following, simplified, street map of Milan:

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19 In the following table, we report the expected traveling time (in fraction of hours) between each pair of underground stations (including the Duomo one). Since the street map is itself sparse, we also give the table in a sparse form. Give a linear programming formulation for the problem of finding the shortest paths that the people will take, so as to be able to suggest to the police department which routs will be, most likely, taken.

20 Sets V : nodes T: target nodes A V V : arcs Parameters s T: root node : cost for arc (i, j), (i, j) A Variables : quantity of flow on arc (i, j), (i, j) A Model Max s.t.? A

21 Downloading AMPL Student Edition Software My Address Lab session web site