Discrete Dynamic Programming Example 91 Littleville Suppose that you are the city traffic engineer for the town of Littleville Figure 91a depicts the arrangement of one and twoway streets in a proposed improvement plan for ID: 760161
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Slide1
Chapter
9
Shortest Paths and
Discrete Dynamic Programming
Slide2Example 9.1 Littleville
Suppose
that you are the city traffic engineer for the town of
Littleville
. Figure
9.1(a
) depicts the arrangement of one- and two-way streets in a proposed improvement plan for
Littleville’s
downtown, including the estimated average time in seconds that a car will require to transit each block.
From
survey and other data we can estimate how many driver trips originate at various origin points in the town, and the destination for which each trip is bound. But such survey data cannot indicate what streets will be selected by motorists to move from origin to destination in a street network that does not yet exist
.
Slide3Example 9.1 Littleville
One
of the tasks of a traffic engineer is to project the route that drivers will elect, so that city leaders can evaluate whether flows will concentrate where they hope. A good starting point is to assume that drivers will do the most rational thing-follow the shortest time path from their origin to their destination. We need to compute all such shortest paths.
Slide4Example 9.1 Littleville
20
18
12
18
13
28
38
32
30
18
28
36
25
49
21
40
Slide5Example 9.1 Littleville
1
2
3
4
10
9
7
6
5
8
20
12
18
13
25
21
28
49
18
28
36
40
18
38
32
30
Slide69.1 Shortest Path Models
Urban traffic, Satellite communications, or surface of a microchip.
Mathematical
graphs
model travel, flow, and adjacency patterns in a
network
. [9.1]
The
nodes
or
vertices
of a graph represent entities, intersections, and transfer points of the network. [9.2]
The
arcs
of
a graph
model available directed (one-way) links between nodes. Edges represent undirected (two-way
)
links.
[
9.3]
Slide79.1 Shortest Path Models
A path is a sequence of arcs or edges connecting two specified nodes in a graph. Each arc or edge must have exactly one node in common with its predecessors in the sequence, any arcs must passed in the forward direction, and no node may be visited more than once. [9.4]
Slide8Shortest Path Problems
Shortest
path problems seek minimum total length paths between specified pairs of nodes in a graph. [9.5]
Slide9Example 9.1 Littleville
1
2
3
4
10
9
7
6
5
8
20
12
18
13
25
21
28
49
18
28
36
40
18
38
Name:
Littleville
Graph
: arcs and edges
Costs
: nonnegative
Output
: shortest paths
Pairs
: all nodes to all others
Slide10Example 9.2 Texas Transfer
Texas Transfer, a major trucker in the southwest, ships goods from its hub warehouse in Ft. Worth to all the cities shown. Trucks leave the hub and proceed directly to their destination city, with no intermediate
dropoffs
or pickups.
Texas
Transfer drivers are allowed to choose their own route from Ft. Worth to their destination. However, management’s proposal in current labor negotiations calls for drivers to be paid on the basis of shortest standard mileage to the location. That is, they will be paid according to the length of the shortest path from Ft. Worth to their destination city in the network of Figure 9.3. To see the impact of this proposal, we need to compute such shortest path distances for all cities
.
Slide11Example 9.2 Texas Transfer
Notice
that the character of this shortest path task differs significantly from the
Littleville
example. Here we need only optimal path lengths, not the paths themselves.
Also
, we need shortest path lengths for only one origin or source—the Ft. Worth hub.
Slide12Example 9.2 Texas Transfer
1
2
3
4
10
9
7
6
5
8
359
122
345
443
92
199
246
213
195
79
215
153
180
167
415
El Paso
Lubbock
Amarillo
Ft. Worth
Abilene
San Angelo
San Antonio
Corpus Christi
Houston
Austin
Name: Texas Transfer
Graph
:
edges only
Costs
: nonnegative
Output
: shortest
path lengths
Pairs
:
one node
to all others
Slide13Example 9.3 Two Ring Circus
The Two Ring Circus is nearing the end of its season and planning a return to winter headquarters near Tallahassee, Florida. Present commitments will end in Lincoln, Nebraska, but there are still some opportunities for bookings in cities along the route home.
Figure
9.4(a) shows the travel routes available and the estimated costs (in $1000) of moving the circus over those routes. It also designates the cities where bookings have been offered and the anticipated net receipts (in thousands of dollars).
We
want to compute the optimal path home for Two Ring. Notice that the cost of a path is the difference of travel costs and net receipts from performances along the way. But receipts occur at nodes of the network, not on edges as shortest path models require.
Slide14Example 9.3 Two Ring Circus
Chattanooga(3.0)
1
3
2
9
8
7
5
4
6
3.6
2.8
3.8
4.5
2.1
2.6
5.5
4.5
4.5
2.4
2.2
5.1
Wichita
(3.2)
Lincoln
(0.0)
Davenport (1.6)
Little Rock(3.0)
Jackson(1.0)
Tallahassee(0.0)
Montgomery (1.5)
Springfield (1.5)
4.6
1.7
4.6
Slide15Undirected and Directed Graph (Digraph)
A shortest path problems including edges (i, j) of cost ci,j can be converted to an equivalent one on a digraph by replacing each edge with a pair of opposed arcs as [9.6]
i
j
c
i,j
i
j
c
i,j
c
j,i
Slide16Example 9.3 Two Ring Circus
Chattanooga (3.0)
1
3
2
9
8
7
5
4
6
3.6
2.8
3.8
4.5
2.1
2.6
5.5
4.5
4.5
2.4
2.2
5.1
Wichita
(3.2)
Lincoln
(0.0)
Davenport (1.6)
Little Rock(3.0)
Jackson(1.0)
Tallahassee(0.0)
Montgomery (1.5)
Springfield (1.5)
4.6
1.7
4.6
3.6
1.7
4.6
2.8
5.1
4.5
4.5
5.5
4.6
2.1
3.8
2.4
2.6
2.2
4.5
Slide17Two Ring Example Model
Chattanooga
1
3
2
9
8
7
5
4
6
3.6
-0.4
0.8
1.5
1.1
1.1
2.5
3.5
3.0
0.9
2.2
3.6
Wichita
Lincoln
Davenport
Little Rock
Jackson
Tallahassee
Montgomery
Springfield
4.6
0.2
1.6
2.0
0.1
3.1
2.8
2.3
1.3
1.5
4.0
1.6
-0.9
2.8
-0.6
1.6
0.7
4.5
Name: Two Ring Circus
Graph
:
directed
Costs
:
arbitrary
Output
: shortest
path Pairs
:
one source to one destination
Slide189.2 Dynamic Programming Approach to Shortest Paths
Dynamic programming methods exploit the fact that it is sometimes easiest to solve one optimization problem by taking on an entire family.
Shortest path algorithms exploit relationships among optimal paths (and path lengths) for different pairs on the nodes. [9.7]
Slide19Functional Notationfor One Source to All Other Nodes
[k] length of a shortest path from the source node to the node k (= + if no path exists)xi,j[k] 1 if arc/edge (i, j) is part of the optimal path from the source node to node k 0 otherwise
Slide20Functional Notationfor All Nodes to All Other Nodes
[k, l] length of a shortest path from node k to the node l (= + if no path exists)xi,j[k, l] 1 if arc/edge (i,j) is part of the optimal path from the node k to node l 0 otherwise
Slide21Optimal Paths and Subpaths
If the highlighted path (through Austin and San Antonio) is the shortest from Ft. Worth to Corpus Christi, the subpath from Ft. Worth to San Antonio must also be the shortest. (No negative dicycle)
1
2
3
4
10
9
7
6
5
8
359
122
345
443
92
199
246
213
195
79
215
153
180
167
415
El Paso
Lubbock
Amarillo
Ft. Worth
Abilene
San Angelo
San Antonio
Corpus Christi
Houston
Austin
Slide22Negative Dicycles Exception
A dicycle is a path that begins and ends at the same node, and a negative dicycle is a dicycle of negative total length. [9.8]Shortest path models with negative dicycles are much less tractable than other cases because dynamic programming methods usually do not apply. [9.9]
1
2
3
4
2
10
-25
3
12
[1]=0
[2]=-3
[3]=5
[4]=17
Slide23Principle of Optimality
In a graph with no negative
dicycles
, optimal paths must have optimal
subpaths
. [9.10]
Functional equations of a dynamic program encode the recursive connections among optimal solution values that are implied by a principle of optimality.
[9.11]
Slide24Functional Equationsfor One Node to All Others
The
functional equations for shortest path problems from a single source
s
in a graph with no negative
dicycles
are [9.12]
[
s
] = 0
[
k
] =
min
{
[
i
] +
c
i,k
:
(
i,
k
) exists
}
all
k≠s
Slide25Sufficiency of Functional Equationsin One to All Case
Node
value [
k
]
in a graph with no negative
dicycles
are lengths of shortest paths from a given source
s
if and only if they satisfy functional equations in [9.12]. [9.13]
Slide26Sufficiency of Functional Equationsin One to All Case
[4]=[5]+ c5,4[5]=[3]+ c3,5[4]+[5]=[5]+ c5,4+ [3]+ c3,5[4]=cs,5+ c5,4[4] cs,1+ c1,2+c2,4
1
2
3
4
10
9
7
6
5
8
359
122
345
443
92
199
246
213
195
79
215
153
180
167
415
El Paso
[
4
]
=623
Lubbock
[
2
]
=347
Amarillo
[
1
]
=359
Ft. Worth
[
3
]
=0
Abilene
[
5
]
=180
San Angelo
[
6
]
=272
San Antonio
[
7
]
=274
Corpus Christi
[
10
]=427
Houston[9]=246
Austin
[
8
]
=195
Slide27Functional Equationsfor All Nodes to All Others
The
functional equations for shortest path problems from a single source
s
in a graph with no negative
dicycles
are [9.14]
[
k,k
] = 0 all
k
[
k,l
] =
min
{
c
k,l
,
{
[
k
,
i
] +
[
i
,
l
] :
i
≠
k, l
}
}
all
k≠l
Node
pair values
[
k,l
]
in a graph with no negative
dicycles
are lengths of shortest paths
from
k
to
l
if
and only if they satisfy functional equations in [
9.14].
[
9.15]
Slide28Solving Shortest Path Problems byLinear Programming
Although shortest path problems over graphs with no negative
dicycles
can be solved by linear programming, dynamic programming methods based on the principle of optimality are far more efficient. [9.16]
Slide299.3 Shortest Paths from One Node to All Others: Bellman-Ford
Solving the functional equations can be carried out by calculating the single value
[
k
] (one-pass evaluation).
Dicycles
introduce circular dependencies in functional equations that preclude their solution by
one-pass
evaluation, even if the length of all
dicycles
is non-negative.
[9.17]
Slide30Repeated Evaluation Algorithm: Bellman-Ford
(t)
[
k
] value of
[
k
] obtained on the
t
th
iteration
d
[
k
]node preceding
k
in the best known path from
s
to
k
.
Slide31Algorithm 9A: One to All (no Negative Dicycles) Bellman-Ford
Step 0: Initiation
. With
s
the source node, initialize optimal path lengths
(0)
[
k
] = 0 if
k
=
s
= + otherwise
and set iteration counter t
1
Step
1: Evaluation.
For each
k
evaluate
(t)
[
k
]
min
{
(
t-1)
[
i
] +
c
i,k
: (
i, k
) exists
}
if
(t)
[
k
]
<
(
t-1)
[
k
], also set
d
[
k
]
the number of a neighboring node
i
achieving the minimum
(t)
[
k
]
.
Slide32Algorithm 9A: One to All (no Negative Dicycles) Bellman-Ford
Step 2: Stopping
.
Terminate if
(t)
[
k
]
<
(t-1)
[
k
] for every
k
, or if t= the number of nodes in the graph. Value
(t)
[
k
]
then equal the required shortest path lengths unless some
(t)
[
k
]
changed at the last t, in which case the graph contains a negative
dicycle
.
Step 3: Advance.
If some [
k
]
changed and t < the number of nodes, increment t
t+1 and return to Step 1.
Slide33Two Ring Example Model
Chattanooga
1
3
2
9
8
7
5
4
6
3.6
-0.4
0.8
1.5
1.1
1.1
2.5
3.5
3.0
0.9
2.2
3.6
Wichita
Lincoln
Davenport
Little Rock
Jackson
Tallahassee
Montgomery
Springfield
4.6
0.2
1.6
2.0
0.1
3.1
2.8
2.3
1.3
1.5
4.0
1.6
-0.9
2.8
-0.6
1.6
0.7
4.5
Slide34Bellman-Ford Solution of the Two Ring Circus Example
t
(t)
[
1
]
(t)
[
2
]
(t)
[
3
]
(t)
[4]
(t)
[
5
]
(t)
[
6
]
(t)
[
7
]
(t)
[
8
]
(t)
[
9
]
0
0
+
+
+
+
+
+
+
+
1
3.6
2.8
4.6
2
3.7
4.1
3
5.7
3.2
4
4.8
6.7
5
5.5
t
d
[
1
]
d[
2
]
d[
3
]
d[4]
d[
5
]
d[
6
]
d[
7
]
d[
8
]
d[
9
]
1
1
1
1
2
2
3
3
5
5
4
7
7
5
8
Slide35Bellman-Ford Algorithm
At the completion of Bellman-Ford Algorithm, a shortest path from source
s
to any other node
k
can be recovered by starting at
k
, backtracking to neighboring node
d
[
k
], and continuing with an optimal path from
s
to the neighbor until source s is encountered. [9.18]
If Bellman-Ford Algorithm encounters negative
dicycles
, it will demonstrate their presence by continuing to change
(t)
[
k
] on iteration t = the number of nodes. Any node
k
with such changing
(t)
[
k
] belongs to a negative
dicycle
. [9.19]
Slide36Two Ring Example Model
Chattanooga
1
3
2
9
8
7
5
4
6
3.6
-0.4
0.8
1.5
1.1
1.1
2.5
3.5
3.0
0.9
2.2
3.6
Wichita
Lincoln
Davenport
Little Rock
Jackson
Tallahassee
Montgomery
Springfield
4.6
0.2
1.6
2.0
0.1
3.1
2.8
2.3
1.3
1.5
4.0
1.6
-0.9
2.8
-0.6
1.6
0.7
4.5
Slide379.4 Shortest Paths from All Nodes to All Others: Floyd-Warshall
(t)
[
k, l
] length of a shortest path for
k
to
l
using only intermediate nodes numbered less than or equal to t.
d
[
k, l
] node just before
l
on the current path from
k
to
l
.
Slide38Algorithm 9B: All to All (no Negative Dicycles) Floyd-Warshall
Step 0: Initiation
. All nodes should have consecutive positive numbers starting with 1. For all arcs and edges (
k, l)
in the graph, initialize
(0)
[
k, l
]
c
k,l
d
[
k, l
]
k
For
k
, l
pairs with no arc/edge (
k,
l
), assign
(0)
[
k, l
]
= 0 if
k
=
l =
+ otherwise
also set iteration counter t
1
Step
1: Evaluation.
For all
k, l
≠ t update
(t)
[
k, l
]
min {
(
t-1)
[
k, l
],
(t-1)
[
k,
t
]
+
(t-1)
[t
, l
]
}
if
(t)
[
k, l
]
<
(
t-1)
[
k, l
], also set
d
[
k, l
]
d
[t
,
l
]
Slide39Algorithm 9B: All to All (no Negative Dicycles) Floyd-Warshall
Step 2: Stopping
.
Terminate
if t = the number of nodes in the graph, or if
(t)
[
k, k
]
<
0
for any node
k
. Value
(t)
[
k, l
]
then equal the required shortest path lengths unless some
(t)
[
k, k
]
is negative, in which case the graph contains a negative
dicycle
through
k
.
Step 3: Advance.
If t < the
number of nodes and
(t)
[
k,k
]0
, increment t
t+1 and return to Step 1.
Slide40Example 9.1 Littleville
1
2
3
4
10
9
7
6
5
8
20
12
18
13
25
21
28
49
18
28
36
40
18
38
32
30
Slide41Floyd-Warshall Solution of the Littleville Example
Nu
l=1
l=2
l=3
l=4
l=5
l=6
l=7
l=8
l=9
l=10
k=1
0
12
¥
¥
¥
¥
¥
¥
¥
¥
k=2
¥
0
18
¥
¥
32
¥
¥
¥
¥
k=3
¥
¥
0
13
¥
¥
30
¥
¥
¥
k=4
¥
¥
¥
0
¥
¥
¥
¥
¥
38
k=5
20
¥
¥
¥
0
18
¥
¥
¥
¥
k=6
¥
32
¥
¥
18
0
¥
¥
25
¥
k=7
¥
¥
30
¥
¥
28
0
¥
21
49
k=8
¥
¥
¥
¥
18
¥
¥
0
36
¥
k=9
¥
¥
¥
¥
¥
25
21
36
0
40
k=10
¥
¥
¥
¥
¥
¥
49
28
40
0
d
k=1
1
k=2
2
2
k=3
3
3
k=4
4
k=5
5
5
k=6
6
6
6
k=7
7
7
7
7
k=8
8
8
k=9
9
9
9
9
k=10
10
10
10
Slide42Floyd-Warshall Solution of the Littleville Example
Nu
l=1
l=2
l=3
l=4
l=5
l=6
l=7
l=8
l=9
l=10
k=1
0
12
¥
¥
¥
¥
¥
¥
¥
¥
k=2
¥
0
18
¥
¥
32
¥
¥
¥
¥
k=3
¥
¥
0
13
¥
¥
30
¥
¥
¥
k=4
¥
¥
¥
0
¥
¥
¥
¥
¥
38
k=5
20
32
¥
¥
0
18
¥
¥
¥
¥
k=6
¥
32
¥
¥
18
0
¥
¥
25
¥
k=7
¥
¥
30
¥
¥
28
0
¥
21
49
k=8
¥
¥
¥
¥
18
¥
¥
0
36
¥
k=9
¥
¥
¥
¥
¥
25
21
36
0
40
k=10
¥
¥
¥
¥
¥
¥
49
28
40
0
d
k=1
1
k=2
2
2
k=3
3
3
k=4
4
k=5
5
1
5
k=6
6
6
6
k=7
7
7
7
7
k=8
8
8
k=9
9
9
9
9
k=10
10
10
10
Slide43Floyd-Warshall Solution of the Littleville Example
Nu
l=1
l=2
l=3
l=4
l=5
l=6
l=7
l=8
l=9
l=10
k=1
0
12
30
¥
¥
44
¥
¥
¥
¥
k=2
¥
0
18
¥
¥
32
¥
¥
¥
¥
k=3
¥
¥
0
13
¥
¥
30
¥
¥
¥
k=4
¥
¥
¥
0
¥
¥
¥
¥
¥
38
k=5
20
32
50
¥
0
18
¥
¥
¥
¥
k=6
¥
32
50
¥
18
0
¥
¥
25
¥
k=7
¥
¥
30
¥
¥
28
0
¥
21
49
k=8
¥
¥
¥
¥
18
¥
¥
0
36
¥
k=9
¥
¥
¥
¥
¥
25
21
36
0
40
k=10
¥
¥
¥
¥
¥
¥
49
28
40
0
d
k=1
1
2
2
k=2
2
2
k=3
3
3
k=4
4
k=5
5
1
2
5
k=6
6
2
6
6
k=7
7
7
7
7
k=8
8
8
k=9
9
9
9
9
k=10
10
10
10
Slide44Floyd-Warshall Solution of the Littleville Example
Nu
l=1
l=2
l=3
l=4
l=5
l=6
l=7
l=8
l=9
l=10
k=1
0
12
30
43
¥
44
60
¥
¥
¥
k=2
¥
0
18
31
¥
32
48
¥
¥
¥
k=3
¥
¥
0
13
¥
¥
30
¥
¥
¥
k=4
¥
¥
¥
0
¥
¥
¥
¥
¥
38
k=5
20
32
50
63
0
18
80
¥
¥
¥
k=6
¥
32
50
63
18
0
80
¥
25
¥
k=7
¥
¥
30
43
¥
28
0
¥
21
49
k=8
¥
¥
¥
¥
18
¥
¥
0
36
¥
k=9
¥
¥
¥
¥
¥
25
21
36
0
40
k=10
¥
¥
¥
¥
¥
¥
49
28
40
0
d
k=1
1
2
3
2
3
k=2
2
3
2
3
k=3
3
3
k=4
4
k=5
5
1
2
3
5
3
k=6
6
2
3
6
3
6
k=7
7
3
7
7
7
k=8
8
8
k=9
9
9
9
9
k=10
10
10
10
Slide45Floyd-Warshall Solution of the Littleville Example
Nu
l=1
l=2
l=3
l=4
l=5
l=6
l=7
l=8
l=9
l=10
k=1
0
12
30
43
62
44
60
105
69
81
k=2
70
0
18
31
50
32
48
93
57
69
k=3
96
90
0
13
76
58
30
79
51
51
k=4
104
116
117
0
84
102
87
66
78
38
k=5
20
32
50
63
0
18
80
79
43
83
k=6
38
32
50
63
18
0
80
61
25
65
k=7
66
60
30
43
46
28
0
57
21
49
k=8
38
50
68
81
18
36
96
0
36
76
k=9
63
57
51
64
43
25
21
36
0
40
k=10
66
78
79
92
46
64
49
28
40
0
d
k=1
1
2
3
6
2
3
9
6
4
k=2
6
2
3
6
2
3
9
6
4
k=3
7
7
3
7
7
3
10
7
4
k=4
10
10
10
10
10
10
10
10
4
k=5
5
1
2
3
5
3
9
6
9
k=6
5
6
2
3
6
3
9
6
9
k=7
6
6
7
3
6
7
9
7
7
k=8
5
5
5
5
8
5
5
8
9
k=9
6
6
7
7
6
9
9
9
9
k=10
8
8
7
7
8
8
10
10
10
Slide46Floyd-Warshall Algorithm
At the completion of
Floyd-
Warshall
Algorithm, a shortest path from any node
k
to any other node
l
can be recovered by starting at
l
, backtracking to neighboring node
d
[
k, l
], and continuing with an optimal path from
k
to the neighbor until source
k
itself is encountered. [9.20]
If
Floyd-
Warshall
Algorithm is applied to a graph with negative
dicycles
, it will demonstrate their presence by making
some
(t)
[
k,k
]
< 0 and terminating. Under those circumstances, the implied negative
dicycle
includes node
k
. [9.21]
Slide479.5 Shortest Path from One Node to All Others with Cost Non-negative: Dijkstra
The bellman-Ford and Floyd-
Warshall
algorithms can solve any of the shortest path models without negative
dicycles
.
With further assumptions, other algorithms may be more efficient.
Slide48Algorithm 9C: One to All (Non-Negative Costs): Dijkstra
Step 0: Initiation
. With
s
the source node, initialize optimal path lengths
[
i
] = 0 if
i
=
s
= + otherwise
Then mark all nodes temporary, and choose
p
s
as the next permanently labeled node.
Step
1: Processing.
Mark node
p
permanent, and for every arc/edge (
p
,
i
) leading from
p
to a temporary
node, update [
i
]
min
{
[
i
],
[
p
]
+
c
p,i
}
if [
i
] changed in value, also set
d
[
k
]
p
.
Slide49Step 2: Stopping. If no temporary nodes remain, stop; values [i] now reflect the required shortest path lengths. Step 3: Next Permanent. Choose as next permanently labeled node p a temporary node with least current value [i], that is, [p] = min {[i] : i temporary } Then return to Step 1.
Algorithm 9C: One to All
(Non-Negative Costs):
Dijkstra
Slide50Once a node is classified permanent by Dijkstra Algorithm, its [p] and d[p] labels never change again. Nodes that are not yet permanently labeled are classified temporary. [9.22]Dijkstra Algorithm is the most efficient method available for computing shortest paths from one node to all others in (general) graphs and digraphs having all are/edge costs non-negative. [9.23]Each iteration of Dijkstra’s Algorithm selects as the new permanently labeled node p a temporary node of minimum [i]. [9.24]
Permanently and Temporary Labeled Nodes
Slide51Dijkstra Algorithm Solution of the Texas Transfer Example
1
2
3
4
10
9
7
6
5
8
359
122
345
443
92
199
246
213
195
79
215
153
180
167
415
El Paso
Lubbock
Amarillo
Ft. Worth
Abilene
San Angelo
San Antonio
Corpus Christi
Houston
Austin
Slide52Dijkstra Algorithm Solution of the Texas Transfer Example
p
n[1]
n[2]
n[3]
n[4]
n[5]
n[6]
n[7]
n[8]
n[9]
n[10]
init.
¥
¥
0
¥
¥
¥
¥
¥
¥
¥
3
359
Perm
180
195
246
5
347
623
Perm
272
8
274
Perm
9
Perm
461
6
Perm
7
Perm
427
2
Perm
1
Perm
10
Perm
4
Perm
p
d[1]
d[2]
d[3]
d[4]
d[5]
d[6]
d[7]
d[8]
d[9]
d[10]
3
3
3
3
3
5
5
5
5
8
8
9
9
6
7
7
2
1
10
4
Slide53After each major iteration, Dijkstra Algorithm values [i] represent lengths of shortest paths from source s to i that use only permanently labeled nodes. [9.25]
Justification of the Dijkstra Algorithm
s
i
p
All permanent path length
[
p
]
All permanent path length
[
i
]
Non-negative cost
[
i
]
[
p
]
Slide549.6 Shortest Path from One Node to All Others in Acyclic Digraph
Acyclic DigraphsA digraph (completely directed graph) is acyclic if and only if its nodes can be numbered so that every arc (I, j) has I < j. [9.26].
1
2
3
5
4
Acyclic
1
2
3
5
4
Has edges
1
2
3
5
4
Has
dicycles
Slide559.6 Shortest Path from One Node to All Others in Acyclic Digraph
Any acyclic digraph can be numbered as required in principle [9.26] by transiting the graph in depth-first fashion, numbering nodes in decreasing sequence as soon as all their outbound arcs lead to nodes already visited. [9.27].
c
a
d
b
e
1st
2nd
3rd
5
4
3
2
1
c
a
d
b
e
1st
2nd
3rd
4th
5
4
3
2
1
Slide56Algorithm 9D: One to All (Acyclic Digraph)
Step 0: Initiation.
Number nodes so that each arc (
i
,
j
) of the digraph has
i
<
j
. Then set the source
s
optimal path length [
s
]
0
Step 1:
Stopping.
Terminate if all
[
k
] have now been fixed. Otherwise, let
p
be the lowest number of an unprocessed node.
Step
2: Processing.
If there are no inbound arcs at node
p,
set
[
p
]
+. Otherwise, compute
[
p
]
min
{
[
i
] +
c
i,p
: (
i, p
) exists
}
And let d[
p
]
the number of node
i
achieving the minimum. Then return to Step 1.
Slide57Algorithm 9D is the most efficient possible for computing shortest paths from one node to all others in acyclic digraphs. [9.28]
Permanently and Temporary Labeled Nodes
Slide58Figure 9.11 Acyclic Shortest path Example
1
3
2
8
7
5
4
6
5
16
4
-10
1
13
7
18
3
8
9
9
2
5
[
1
]=0
[
2
]=5; d[
2
]=1
[
3
]=-5; d[
3
]=2
[
4
]=-3; d[
4
]=3
[
5
]=;
[
6
]=;
[
7
]=9; d[
7
]=2
[
8
]=22; d[8]=7
[
9
]=4; d[
9
]=4
Slide599.7 CPM Project Scheduling and Longest Paths
Project Management
For planning and control purposes, projects are usually subdivided into a collection of work
activities
. Each activity has an estimated
duration
,
a
k
time required to accomplish activity
k
Activity
j
is a
predecessor
of activity
k
if activity
j
must be completed before activity
k
can begin. [9.29].
Slide60CPM Project Networks
CPM project networks
have special
start
and
finish
nodes, plus one node for each activity. Arcs of zero length connect
start
to all activities without predecessors. Other arcs of length
a
k
connect each activity node
k
to all activities of which it is a predecessor, or to
finish
if there are no such nodes. [9.30]
Slide61Example 9.4: We Build Construction
We Build is developing a 1-story medical office building on available land near a hospital.
kActivityDurationPredecessors1Foundation15--2Rough plumbing5--3Concrete slab41, 24Structural members335Roof746Rough electrical1047Heating & AC132, 48Walls184, 6, 79Interior finish205, 8
To plan materials deliveries and arrange subcontractors to do the various activities, We Build needs a schedule. In particular, they want to know the earliest time after project start that each activity can begin.
Slide62Figure 9.12 We Build Construction Project Network
Start
3
1
9
6
2
8
0
0
4
15
7
13
3
20
10
5
4
5
7
Finish
5
3
3
3
18
Slide63CPM Schedules and Longest Paths
The earliest start time of any activity
k
in
a project equals the length of the longest path from
start
to node
k
in the corresponding network. [9.31]
The
minimum time to complete a project equals the
length of the longest
or critical path
from
start
to
finish
in
the corresponding
project network
. [
9.32]
Slide64Critical Paths
kActivityEarly StartCritical Path1Foundation0Start-12Rough plumbing0Start-23Concrete slab15Start-1-34Structural members19Start-1-3-45Roof22Start-1-3-4-56Rough electrical22Start-1-3-4-67Heating & AC22Start-1-3-4-78Walls35Start-1-3-4-7-89Interior finish53Start-1-3-4-7-8-9
Start
3
1
9
6
2
8
0
0
4
15
7
13
3
20
10
5
4
5
7
Finish
5
3
3
3
18
Slide65Computing an Early Start Schedule for the We Build Construction Example
a
k
time required to accomplish activity
k
[
k
] length of the longest path from
start
to node
k
d[
k
]
immediate predecessor of node
k
in a longest path
The minimum time to complete a project equals the
length of the longest
or critical path
from
start
to
finish
in
the corresponding
project network
. [
9.32]
Slide66Algorithm 9E: CPM Early Start Scheduling
Step 0: Initiation.
Number activity nodes so that each precedence arc (
i
,
j
) of the CPM project network has
i
<
j
. Then initialize the schedule time of the project start node as
[
start
]
0
Step 1:
Stopping.
Terminate if the early start of the project
finish
node has been fixed. Otherwise, let
p
be the lowest number of an unprocessed node.
Step
2: Processing.
Compute the activity
p
early start schedule time
[
p
]
max
{
[
i
] +
a
i
:
i
is a predecessor of
p
}
And let d[
p
]
the number of node
i
achieving the maximum. Then return to Step 1.
Slide67Early Start Schedules
[s]=0;
[1]=0; d[1]=s
[2]=0; d[2]=s
[3]=15; d[3]=1
[4]=19; d[4]=3
[5]=22; d[5]=4
[6]=22; d[6]=4
[7]=22; d[7]=4
[8]=35; d[8]=7
[9]=53; d[9]=8
Start
3
1
9
6
2
8
0
0
4
15
7
13
3
20
10
5
4
5
7
Finish
5
3
3
3
18
[
f
]=73; d[
f
]=
9
Slide68Late Start Schedules and Schedule Slack
The latest start time of any activity
k
in
a project equals the due date less the length path from
k
to
finish
node in the corresponding CPM network. [9.33]
Schedule slack is the difference between the earliest and latest times that an activity can begin. [9.34]
Slide69Late Start Schedules
[s]=0; d[s]=1
[1]=0; d[1]=3
[2]=10; d[2]=3
[3]=15; d[3]=4
[4]=19; d[4]=7
[5]=46; d[5]=9
[6]=25; d[6]=8
[7]=22; d[7]=8
[8]=35; d[8]=9
[9]=53; d[9]=f
Start
3
1
9
6
2
8
0
0
4
15
7
13
3
20
10
5
4
5
7
Finish
5
3
3
3
18
[
f
]=73;
Slide70Schedule Slack
k
Activity
Early Start
Late Start
Slack
1
Foundation
0
0
0
2
Rough plumbing
0
10
10
3
Concrete slab
15
15
0
4
Structural members
19
19
0
5
Roof
22
46
24
6
Rough electrical
22
25
3
7
Heating
& AC
22
22
0
8
Walls
35
35
0
9
Interior finish
53
53
0
Slide71More on CPM Network
Difficulty of General Longest Path Problems
Longest path problems are the maximize analogs of the shortest path problem. The difficult case of negative
dicycle
in shortest path problems becomes the case of
positive
dicycles
in longest path models –
dicycles
of positive total length. Positive
dicycles
are far more common than their negative relatives.
Acyclic Character of Project Network
Every well-formed project network is acyclic. [9.35]
Slide729.8 Discrete Dynamic Programming Models
In a
sequential decision problem
, decision can be arranged in a clear sequence. Then the decisions can be confronted one by one in sequence. That same sequence produces an acyclic digraph on which a shortest or longest path problem can be posed.
States
in dynamic programming characterize conditions of incomplete solution at which decisions should be considered. [9.36]
Slide73Example 9.5: Wagner-Whitin Lot Sizing
Wagner-Whitin lot sizing – production planning in an environment where there is a substantial setup cost incurred with each production run.
Period, k123456Requirement, rk1040205515Setup cost, sk505050505050Production cost, pk133111Holding cost, hk222222
rk product required at time ksk setup cost at time kpk variable prod. cost at time khk unit holding cost through k
An optimal solution must carefully balance production and holding costs.
Slide74States in Dynamic Programming
Produce or hold for period
k
, but not both.
Definition of
states
: The process has reached state
k
when requirements for periods
l
<
k
have been fulfilled and there is no on-hand inventory.
Nodes of the digraph associated with any dynamic program corresponding to state of incomplete solution. [9.37]
Arcs
of the digraph associated with any dynamic program corresponding to
decisions. They link state nodes with a subsequent state to which the decision leads.
[
9.38]
Slide75Figure 9.13 Digraph for Wagner-Whitin Example
3
7
6
2
60
170
525
55
4
5
180
110
1
55
65
360
315
280
135
170
305
100
270
70
305
145
350
515
Period,
k
1
2
3
4
5
6
r
k
10
40
20
5
5
15
s
k
50
50
50
50
50
50
p
k
1
3
3
1
1
1
h
k
2
2
2222
C
3,6
= setup + production + holding
=s
3
+ (r
3
+ r
4
+ r
5
)p
3
+ [(r
4
+ r
5
)h
3
+r
5
h
4
]
=50 + (20+5+5)3 + [(5+5)2 + (5)2]
=170
Slide76Dynamic Programming Solutionsas an Optimal Path
Optimal solutions to elementary dynamic programs correspond to shortest or longest paths from a beginning to an ending state in the associated digraph. [9.39]
Dynamic programming functional equation recursions encode connections among optimal values for different states of incomplete solution.
[
9.40]
Slide77Solution for Wagner-Whitin Example
3
7
6
2
60
170
525
55
4
5
180
110
1
55
65
360
315
280
135
170
305
100
270
70
305
145
350
515
[
1
]=0
[
2
]=60; d[
2
]=1
[
3
]=180; d[
3
]=1
[
4
]=280; d[
4
]=1
[
5
]=315; d[
5
]=1, or 3
[
6
]=350; d[
6
]=3, or 4
[
7
]=415; d[
7
]=5, or 6
Slide78Dynamic Programming Models with Both Stages and States
In dynamic programs with both stages and states, stages delineate the sequence of required decisions and states encode the conditions within which decisions can be considered. [9.41]
Slide79Example 9.6: President’s Library
The retiring president is collecting all his papers in a new presidential library. Materials for its archives are stored in covered cardboard boxes. All are 1.25 feet wide, but their height varies. .
1234567Height (ft.), hi0.250.400.801.001.502.003.00# of boxes (1000), bi1021230864
Boxes in the archives will be placed on metal shelves, with no box stacked on top of another. The issue is how much shelving will be required. Figure 9.14 shows that if, for example, two shelf
spacings
are employed. Whether they choose to provide 1, 2 ,..., 7
spacings
, they
want to
use the space as efficiently as possible.
Slide80Figure 9.14 President’s Library Shelving Face Areas (two shelf spacings)
0 10 20 30 40 50 60 70
Feet of Shelves(000’s)
Box Height(feet)
3
2
1
0
Slide81Dynamic Programming Modeling of the President’s Library Example
stage k kth-to-last choice of a spacing state i having provided shelves for box sizes 0,1,.. i ci,j choice to provide a spacing for box sizes through j
Slide82Figure 9.15 Digraph for President’s Library Example
7
0
0
0
0
0
0
0
6
1
1
1
1
1
1
2
2
2
2
2
3
3
3
3
4
4
4
5
5
K=7
K=6
K=5
K=4
K=3
K=2
K=1
K=0
S=1
S=2
S=3
S=4
S=5
S=6
S=7
S=6
15
S=5
15
37.5
S=4
15
35
67.5
S=3
37.5
71.25
110
180
S=2
12
52.5
93.75
140
225
S=1
1
14
55
97.5
145
232.5
S=0
3.125
6
24
67.5
116.25
170
270
Slide83Backward Solution of Dynamic Programs
Elementary dynamic programs are often most easily conceptualized in backward sequence, that is, proceeding from final to initial conditions. [9.42]Functional equation for the President’s library example:k[i] optimal cost to finish from stage k, state i 7[7] = 0 k[i] = min {ci,j + k-1[j] : j > i } k = 1,2,..,6; i (7 - k)
Slide84Backward Solution of Dynamic Programs
Stage kState ik[i] dk[i] 10270.0071232.5072225.0073180.007467.507537.507615.00720135.0041122.5042120.0043105.004450.006530.006
Stage kState ik[i] dk[i] 30117.5041105.0042102.504387.504445.00540108.1311100.004297.504382.50450103.131196.503294.503
Stage
k
State
i
k
[
i
]
d
k
[
i
]
6
0
99.63
1
1
95.50
2
7
0
98.63
1
Slide85Multiple Problem Solutions Obtained Simultaneously
The dynamic programming strategy of exploiting connections among optimal solutions for a number of related problems often implies that optima for multiple problem scenarios are produced by a single shortest or longest path computation. [9.43]