network optimization models - citadelmacs.citadel.edu/zhangli/courses-taught/f… · ppt file ·...

Frederick S. Hillier Gerald J. Lieberman

Chapter 10Network Optimization Models

Network Optimization Models

10.1 Prototype Example

• The road system for Seervada Park– Location O: park entrance– Location T: a scenic wonder– Trams transport sightseers from park

entrance to location T and back

10.1 Prototype Example

Prototype Example

• Summary: Park management faces three problems– Determine the route with the smallest total distance

• A shortest-path problem– Determine where telephone lines should be laid

• A minimum spanning tree problem– Determine how to route tram to maximize number of

trips during peak season• A maximum flow problem

• Network consists of a set of points and a set of lines connecting points

• Node: point (vertex) in the network• Lines: links, arcs, edges, or branches

– Labeled by naming the node at each end• From node precedes the to node

– Have a flow of some type through them• Directed arcs have unidirectional flow• Undirected arcs (links) allow bidirectional flow

10.2 The Terminology of Networks

The Terminology of Networks

• Directed network– Network has only directed arcs

• Undirected network– Network has only undirected arcs

• Path between two nodes– A sequence of distinct arcs connecting the

• Directed path from node i to node j– Sequence of connecting arcs toward node j

• Undirected path from node i to node j– Sequence of connecting arcs whose direction can be with

toward or away from node j (note a directed path also satisfies the definition of an undirected path, but not vice versa). Example: in Fig. 10.2, A->B->C->E is a directed path and B->C->A->D is an undirected path

• Cycle: A path begins and ends at the same node; DE-ED is a directed cycle; AB-BC-CA is an undirected cycle

• Connected network: Every pair of nodes in the network has at least one undirected path between them

• Tree (spanning tree): Connected network with no undirected cycles; every spanning tree has n – 1 arcs (the minimum number of arcs needed to have a connected network and the maximum number of arcs possible without having undirected cycles).

Thereare severalalternativechoicesof addingan arc ateach stageof the process.

• Arc capacity– Maximum amount of flow that can be carried

on a directed arc

• Supply node (or source node)– Flow out exceeds flow in

• Demand node (sink node)– Flow in exceeds flow out

• Transshipment node (intermediate node)– Flow in equals flow out

10.3 The Shortest-Path Problem

• Consider an undirected, connected network– Contains origin and destination nodes– Each link has a nonnegative distance

• The problem– Find the shortest path from origin to destination

The Shortest-Path Problem

This is a special type of linear programming problem (strictly speaking it is an integer programming problem, but due to the special structure of the model, the integer restriction on the decision variables can be relaxed).

Each node can be thought of as having a unit flow passing through if it is on the selected path, no flow otherwise. Thus, O is the source node, T is the demand node, and other nodes are transshipment nodes.

Minimize Z = 2xOA + 5xOB + 4XOC + 2XAB + 7xAD + xBC + 4xBD + 3xBE + xCB + 4xCE + xDE + 5xDT + xED + 7xET

s.t. (net flow constraints)

xOA + xOB + xOC = 1 (source node O)

-xDT – xET = -1 (sink node T)

-xOA + xAB + xAD = 0 (node A)

xBC + xBE + xBD – xOB – xAB – xCB = 0 (node B)

xCB + xCE – xOC – xBC = 0 (node C)

xDE + xDT – xAD – xBD – xED = 0 (node D)

xED + xET – xDE – xBE – xCE = 0 (node E)

xij ≥ 0

The optimal solution by Excel Solver (see course website) is xOA = xAB = xBE = xED = xDT = 1 (and other xij = 0) with Z = 13. 15

The Excel Solver uses the general simplex method. However, using the shortest path algorithm to solve the problem is much more efficient.

• Algorithm (using a table or directly applying the algorithm on the graph)– Objective of nth iteration: find the nth nearest

node to the origin• Repeat for n = 1, 2… until destination is reached

– Input for nth iteration: n − 1 nearest nodes to the origin (these nodes are called solved nodes and other nodes are called unsolved nodes), including shortest path and distance from the origin

• Algorithm (cont’d.) – Candidates for nth nearest node: for each n – 1 solved

node, find an unsolved node (if any) with shortest connecting link to the solved node and make this unsolved node a candidate node (ties provide additional candidates)

– Calculation of nth nearest node• For each solved node and its candidate, add the

distance between them and the distance of the shortest path from the origin to this solved node

• Candidate with smallest total distance is the nth nearest node

• Shortest path for the Seervada park problem– Looking at last column in Table 10.2, two

potential shortest paths exist from the destination to the origin

• T→ D → E → B → A → O or T → D → B → A → O

• Total of 13 miles on either path

• Network simplex method– An alternate option for solving shortest-path problems

• Three categories of applications– Minimize total distance traveled– Minimize total cost of a sequence of activities– Minimize total time of a sequence of activities

• Other shortest path problems can be solved by slightly modify the algorithm

For example:

- shortest directed path (only directed arcs should be considered)

- shortest path from the origin to all other nodes (stop the algorithm until all nodes are solved nodes)

10.4 The Minimum Spanning Tree Problem

• Given: nodes of a network, potential links, and positive length of each link if it is inserted into the network– Design the network by inserting links– A path must exist between every pair of nodes

• Problem: minimize total length of links inserted into the network

• Network of n nodes requires only n−1 links– Choose the links to form a spanning tree

The Minimum Spanning Tree Problem

• Applications– Design of telecommunications networks– Design of a lightly-used transportation

network to minimize cost of providing links– Design network of power transmission lines– Electrical equipment wiring– Piping systems

• Algorithm– Select any node arbitrarily and then add a link

to connect it to its nearest node– Identify the unconnected node that is closest

to a connected node, and add a link between them

• Repeat until all nodes have been connected

– Ties may be broken arbitrarily• There may be multiple optimal solutions

• Example of graphical approach to implementing the algorithm– Problem: installing telephone lines in Seervada park

The optimal solution does notdepend on the initial node chosen.

Try the algorithmagain using adifferent initialnode.

10.5 The Maximum Flow Problem

• General problem description– All flow through a directed, connected network

originates at a source, and terminates at a sink

• Remaining nodes are transshipment nodes

– Flow through an arc is allowed in only one direction (indicated by the arrowhead)

• Maximum flow is given by arc capacity

– Objective: maximize total flow from source to sink

The Maximum Flow Problem

• Applications– Maximize flow through company’s distribution

network from factories to customers– Maximize flow through company’s supply

network from vendors to factories– Maximize oil flow through a system of

pipelines– Maximize water flow through aqueducts– Maximize flow of vehicles through a

transportation network30

• Algorithms– Simplex method can be used– Augmenting path algorithm is more efficient

• Residual network– Remaining arc capacities after some flows have been assigned

For example, if 5 units flow was sent from O to B with capacity 7, then the residual capacity of arc from O to B is 2. Note that the residual capacity of arc from B to O is 5 (for cancelling some previously assigned flow from O to B).

• Augmenting path– Directed path from source to sink in residual network such that

every arc on path has positive residual capacity. The minimum of these residual capacities is called the residual capacity of the augmenting path.

• Algorithm (each iteration follows these steps)– Identify an augmenting path

• If none exists, net flows already constitute an optimal flow pattern

– Identify the residual capacity, c* of this augmenting path

• It will equal the minimum residual capacity of the arcs on this path

– Increase the flow in this path by c* 34

• Algorithm (cont’d.)– Decrease by c* the residual capacity of each

arc on this augmenting path– Increase by c* the residual capacity of each

arc in the opposite direction on this augmenting path

– Return to the first step

• Example: Seervada park transportation problem

• Use Excel Solver to solve the problem (see the course website)

10.6 The Minimum Cost Flow Problem

• General description of the minimum cost flow problem– The network is directed and connected– At least one of the nodes is a supply node,

and one of the other nodes is a demand node• All remaining nodes are transshipment nodes

– Flow is only allowed in direction of the arrowhead

• Arc capacity gives maximum allowable flow

The Minimum Cost Flow Problem

• General description (cont’d.)– Network has enough arcs with sufficient

capacity to enable all flow generated at supply nodes to reach all demand nodes

– Cost of flow through each arc is proportional to the amount of flow

– Objective: minimize total cost of sending available supply through the network to meet the given demand

• Linear programming problem formulation

• Feasible solutions property

• Integer solutions property– For minimum cost flow problems where every bi and

uij have integer values, all the basic variables in every basic feasible solution also have integer values

• Example in Section 3.4 (Distribution Unlimited Co.)

One of the constraintsis redundant due tosummation of bi’s = 0.

• Special cases that fit the minimum cost flow problem– The transportation problem– The assignment problem– The transshipment problem– The shortest-path problem– The maximum flow problem

• All the above problems can be modeled as the minimum cost flow problem by slightly adjusting the problems.

• The transportation problem: there are no transshipment nodes and no capacity constraints on xij (i.e. uij = ∞)

• The assignment problem: same as the transportation problem and bi = 1 for each supply node and bi = -1 for each demand node.

• The shortest path problem: one supply node with supply 1 unit at origin and one demand node with demand 1 unit at the destination; replace each link by directed arcs except for the links with one end at the origin or destination; the distance between i and j becomes cij or cji and uij = ∞.

• The maximum flow problem: source becomes supply node and sink becomes demand node; let cij = 0 for all existing arcs; select a quantity F which is a safe upper bound on the maximum feasible flow through the network and assign F to the supply node and –F to the demand node; add an arc going directly from the supply node to the demand node with cij = M (big number) and uij = ∞ (since cij = M, the minimum cost flow problem will send F from the supply node to the demand node through other paths as much as possible).

• Network simplex method: An alternative method to solving the special cases when the special-purpose algorithms are not available

10.7 The Network Simplex Method• Streamlined version of the simplex method

– Same basic steps• Finding the entering basic variable• Determining the leaving basic variable• Solving for the new BF solution

• Incorporate the upper bound technique:– To deal with the arc capacity constraints like xij ≤ uij which are

handled like nonnegativity constraints. For example, for arc A->B with uAB = 10 and cAB = 2 and bA = 50 and bB = 40, suppose after some iterations xAB has become the leaving basic variable by reaching it upper bound 10. Then xAB = 10 is replaced by xAB = 10 – yAB, and yAB becomes the new nonbasic variable. Meanwhile, we replace arc A->B by arc B->A (reverse arc) with yAB as its flow quantity and assign this new arc a capacity of 10 and unit cost -2, i.e. uBA = 10, cBA = -2. 58

The Network Simplex Method

Also, to take xAB = 10 into account, decrease bA from 50 to 40 and increase bB from 40 to 50, i.e. bA = 40 and bB = 50.

• Network representation of BF solutions– Basic arcs: arcs corresponding to basic variables

• Key property: they never form undirected cycles– Nonbasic arcs: arcs corresponding to nonbasic

variables

• BF solutions can be obtained by solving spanning trees– For arcs not in the spanning tree, set the

corresponding variables (xij or yij) equal to zero

– For arcs in the spanning tree, solve for the corresponding variables (xij or yij) in the system of linear equations provided by the node constraints

• Feasible spanning tree– Spanning tree whose solution from the node

constraints also satisfies all the other constraints

• Fundamental theorem for the network simplex method– Basic solutions are spanning tree solutions

(and conversely)– BF solutions are solutions for feasible

spanning trees (and conversely)63

Iterations: Iteration I (1) selecting entering basic: candidate arcs are BA, ED and AC.Consider arc AC: add AC to the initial feasible spanning tree with flow Ө. This will create an undirected cycle. Within the cycle, arcs having the same direction as arc AC increase their net flow by Ө, and arcs having opposite direction as arc AC decrease their flow by Ө. Arcs not in the cycle their net flow is unchanged. As a result, we compute the change in Z: ΔZ = cAB*θ + cCE*θ + cDE*(-θ) + cAD*(-θ) = 4θ + θ -3θ - 9θ = -7θ

ΔZ= -2θ+9θ+3θ-θ-3θ = 6θ

ΔZ = 2θ + 3θ = 5θ

Since it’s a minimizing problem, adding arc AC decreases Z value. Thus,xAC is the entering basic variable.

(2) Selecting the leaving basic variable: By adding arc AC, consider the upper bound of the arcs which have the same direction as AC (sincethe net flow increased by θ) and the lower bound of the arcs which havethe opposite direction as arc AC (the net flow decreased by θ).

• Iteration II

• Iteration III:

Iteration IV:

No improvement in Z, thus the current BF solution is optimal.

10.8 A Network Model for Optimizing a Project’s Time-Cost Trade-off

• Network based OR techniques developed in the 1950s– PERT (Program Evaluation Review

Technique)– CPM (Critical Path Method)– Both are used in project management

• Concepts have merged into PERT/CPM

• CPM method for time-cost tradeoff– Addresses a project with a specific deadline

A Network Model for Optimizing a Project’s Time-Cost Trade-off

• CPM method for time-cost trade-off (cont’d.)– Problem: find optimal plan for expediting

activities to minimize the total cost of completing the project within the deadline

• General approach– Use a network to display the various activities

• And the order in which they need to be performed

– Form optimization model• Solve using linear programming

• Prototype example– The Reliable Construction Co. won the contract to

construct a new plant within a time period of 40 weeks– See Table 10.7

• Project network (a network used to represent a project) options– Activity-on-arc (AOA)

• Each activity is represented by an arc• Nodes separate activities from predecessors• Used by original versions of PERT and CPM

• Project network options (cont’d.)– Activity-on-node (AON)

• Each activity is represented by a node• Arcs show precedence relationships between

activities• Has several advantages over AOA• May become the standard format for project

networks

• Path: One of the routes following the arcs from start to finish

• The critical path (the length of the longest path through the project network)– Equals the estimated project duration– Ties provide more critical paths

• Crashing an activity– Taking special costly measures to reduce an activity’s duration

(by special costly measures such as using overtime, hiring additional temporary help and etc.)

– Crashing the project involves crashing a number of activities (normal point: no crashing activity; crash point: activities are fully crashed)

• Example problem: determine least expensive way to crash activities to reduce overall duration to 40 weeks and the company will be paid $5.4 million (normal point: 44 weeks and cost $4.55 million and crash point: 28 weeks and cost $6.15 million)

• Solution methods– Marginal cost analysis

In each iteration, crash an activity (which is allowed to be crashed) on the critical path with the smallest cost to reduce its duration by a week. If there is a tie, crash the activity on all critical paths that costs the least.

Total crashing cost is $140,000. Figure 10.30 shows the resulting project network, where the darker arrows show the criticalpaths.

If we need go further (for example, the deadline is 39 weeks),the next step would require looking at the activities on allthree critical paths to find the least expensive way of shortening all three critical paths by a week, and that’d be B).

• Solution methods cont’d– Linear programming

• For example, consider activity J

• Excel Solver solution (also see on the course website)

10.9 Conclusions

• Problems addressed with network models– Optimizing an existing network– Designing a new network

• Minimum spanning tree problem

• CPM method of time-cost trade-offs– Powerful way of applying network optimization

to project management

network optimization models - citadelmacs.citadel.edu/zhangli/courses-taught/f… · ppt file ·...

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