University of Southern California
Daniel J. Epstein Department of Industrial and Systems Engineering
ISE 330: Introduction to Operations Research
Final Review
Minimum Spanning Tree Problem
Minimum Spanning Tree Problem and Shortest Path Problem
 start with an undirected, connected network, with positive cost for each link
 choose a subset of links that gives shortest total length
Requirement:
 MST: chosen links must provide a path between each pair of nodes
 SP: chosen links must provide a path between origin and destination
Definition: A spanning tree is a connected network of the n nodes that contains no
undirected cycles.
Property:
A network with n nodes requires only ( n – 1 ) links to provide a path
between each pair of nodes.
A greedy algorithm for the MST problem
1. Select any node arbitrarily, then connect it to the nearest distinct node.
2. Identify unconnected node closest to a connected node, add arc to network.
3. Ties are broken arbitrarily. Ties may indicate multiple optimal solutions.
Maximum Flow Problem
The Maximum Flow Problem:
 All flow through the directed, connected network originates from one node,
(source); and, terminates at one other node ( sink). All remaining nodes are
called transshipment nodes.
 Directed arcs indicate flow direction, maximum flow is given by arc capacity.
At source, all arcs point away; at sink, all arcs point into node.
 Objective: maximize total flow from source to sink. Amount leaving source is
equal to amount entering sink.
In some applications, flow originates at more than one node and terminate at more
than one node. Add dummy source/sink in such cases.
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ISE 330: Final Review
http://www-classes.usc.edu/engr/ise/330
Solution Methods:
 Formulate as an LP and use the Simplex Method.
 Augmented Path Algorithm
Definition 1: After some flows have been assigned to the arcs, the residual network
shows the remaining arc capacities (called residual capacities) for
assigning additional flows.
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Definition 2: An augmented path is a directed path from the source to the sink in
the residual network such that ievery arc on this parth has strictly
positive residual capacity. The minimum of these residual capacities is
called the residual capacity of the augmenting path.
The Augmented Path Algorithm for the Maximum Flow Problem:
1. Identify an augmenting path by finding some directed path from source to
sink in the residual network s.t. every arc on this path has residual capacity>0.
2. Identify the residual capacity c* of this augmenting path by finding the
minimum of the residual capacities of the arcs on this path. Increase flow in
this path by c*.
3. 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 step 1.
Finding an augmenting path: Breadth-first search ( fanning-out procedure ).
How to know when we’re done? -- no more augmenting path exist.
Max-flow min-cut theorem: For any network with a single source and sink, the
maximum feasible flow from the source to the sink equals the minimum
cut value for all cuts of the network.
Minimum Cost Flow Problem
The Minimum Cost Flow Problem:
1. Considers a directed and connected network.
2. Nodes: At least one node is a supply node. At least one other node is a
demand node. Remaining nodes are transshipment nodes.
3. Arcs are directed with capacities. Cost of flow through each arc is
proportional to the amount of flow, cost per unit known.
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ISE 330: Final Review
http://www-classes.usc.edu/engr/ise/330
4. A feasible solution exists: the network has sufficient capacity to enable all the
flow generated at the supply nodes to reach all the demand nodes.
5. Objective: minimize total cost of sending available supply through network to
satisfy demand.
Linear Programming Formulation:
xij = flow through arc (i,j)
cij = cost per unit flow through arc (i,j)
uij = arc capacity for arc (i,j)
bi = net flow generated at node i
Feasible Solutions Property: A necessary condition for a minimum cost flow
problem to have any feasible solutions is that the total flow being
generated at the supply nodes equals the total flow being absorbed at
the demand nodes.
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 (including the optimal one) also have integer values.
The Network Simplex Method
We use the upper bound technique to deal efficiently with upper bound constraints,
xij ≤ uij. Whenever a nonbasic variable is at its upper bound, xij = uij, the slack
variable yij = uij – xij = 0 becomes the nonbasic variable. Reverse flow of arc, and
negate the cost.
Network representation of BFS: Every BFS has (n – 1) basic variables, where each
basic variable xij represents the flow through arc i  j. The (n – 1) arcs are referred
to as basic arcs. The arcs corresponding to the nonbasic variables xij = 0 or yij = 0
are called nonbasic arcs.
Note: Any complete set of (n – 1) basic arcs forms a spanning tree.
To obtain as spanning tree solution:
1. For arcs not in the spanning tree, set xij = 0 or yij = 0.
2. For arcs in the spanning tree, solve for xij or yij in the system of linear equations
derived from the node constraints.
A feasible spanning tree is a spanning tree whose solution from the node
constraints also satisfies all the other constraints ( 0  xij  uij or 0  yij  uij).
Fundamental Theorem for the network simplex method: Basic solutions are
spanning tree solutions and that BFS are solutions for feasible spanning trees. The
converse is also true.
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ISE 330: Final Review
http://www-classes.usc.edu/engr/ise/330
Goal: To select the incoming variable that improves Z at the fastest rate…
Nonbasic Arc
Cycle Created
Z when  = 1
To select the leaving basic variable, find the variable that first reaches its lower or
upper bound as  increases.
A Transportation Problem
Requirements Assumption: Each source has a fixed supply of units, si, where the
entire supply must be distributed to he destinations. Each destination has a
fixed demand for units, dj, and the entire demand must be received from the
sources.
Feasible Solutions Property: A transportation problem will have feasible solutions
if and only if sum of supplies equals sum of demands.
Cost Assumption: The cost of distributing units from any particular source to any
particular destination, cij, is directly proportional to the number of units
distributed, xij.
The model: Any transportation model can be described completely by a parameter
table…
Source
1
2
3
Demand
1
c11
c21
:
cm1
d1
Cost per Unit Distributed
Destination
2
…
c12
…
c22
…
:
cm2
…
d2
…
n
c1n
c2n
:
cmn
dn
Supply
s1
s2
:
sm
The LP Formulation of the Problem…
Integer Solutions Property: For transportation problems where every si and dj have
integer values, all the basic variables in every BFS also have integer values.
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ISE 330: Final Review
http://www-classes.usc.edu/engr/ise/330
The Transportation Simplex Method
Format of a transportation simplex tableau:
If x i j is basic…
cij
If x i j is nonbasic…
cij
xij
cij – ui – vj
Constructing an Initial BFS:
[1] From rows and columns still under consideration, select the next basic variable
(allocation) according to some criterion*.
[2] Make allocation large enough to use up remaining supply in row or remaining
demand in column (whichever is smaller).
[3] Eliminate that row or column from further consideration. (Break ties arbitrarily.)
[4] If only one row or only one column remains under consideration, then complete
procedure by selecting every remaining variable associated with that row or
column to be basic with the only feasible allocation. Otherwise, return to [1].
*For step [1]: the methods are…
Northwest corner rule: Begin by selected x11. Thereafter, if xij was the last basic
variable selected, then next select xi,j+1 (one column to the right) if source i
has any supply remaining. Otherwise, next select xi+1, j (one row down).
Vogel’s approximation method: For each row and column remaining under
consideration, calculate its difference, the arithmetic difference between the
smallest and next-to-the-smallest unit cost cij still reamining in that row or
column. In that row or column having the largest difference, select the
variable having the smallest remaining unit cost.
Russel’s approximation method: For each source row i remaining under
consideration, determine its ui, the largest unit cost cij still remaining in that
row. For each destination column j remaining under consideration,
determine its vj, which is the largest unit cost cij still remaining in that column.
For each variable xij not previously selected in these rows and columns,
calculate ij = cij – ui – vj. Select the variable having the largest (in absolute
terms) negative value of ij.
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ISE 330: Final Review
http://www-classes.usc.edu/engr/ise/330
The Transportation Simplex Method
Initialization: Construct an initial BFS by the procedure outlined above.
Optimality Test: Derive ui and vj by selecting the row having the largest number of
allocations, setting its ui = 0, then solving the set of equations cij = ui + vj for
each (i,j) such that xij is basic. If cij – ui – vj  0 for every (i,j) such that xij is
nonbasic, then the current solution is optimal.
Iteration:
1. Determine the entering basic variable: Select the nonbasic variable xij having
the largest (in absolute terms) negative value of cij – ui – vj.
2. Determine the leaving basic variable: Identify the chain reaction required to
retain feasibility when the entering basic variable is increased. From the
donor cells, select the basic variable having the smallest value.
3. Determine the new BFS: Add the value of the leaving basic variable to the
allocation for each recipient cell. Subtract this value from the allocation for
each donor cell.
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