Network Optimization

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Network Optimization

Chapter 3Shortest Path Problems

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In a weighted digraph, a path of minimum weight from vertex v to vertex w is called a shortest path (SP) from v to w, and its length is called the shortest distance (SD) from v to w.

For undirected graph, we can define SP and SD between two vertices.

The shortest path problem can be treated as a transshipment problem.

3.1 Shortest paths from a single

source

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3.1 Shortest paths from a single source

(a) If we want to find SP and SD from v to w ， then:

let v be the only source with a supply of 1 unit; let w be the only sink with a demand of 1 unit; let other vertices be intermediate vertices; let the cost of sending one unit of the

commodity from i to j be the weight of the arc (i , j);

we now use the network simplex method to solve this transshipment problem. A 0-1 solution x* will be obtained, and the arcs (i , j) with

=1 form a shortest path from v to w. x

*

ij

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(b) if we want to find shortest paths from a given vertex v to each of the other n-1 vertices in the digraph, then:

let v be the only source with a supply of n-1 units;

let every other vertex be a sink with a demand of 1 unit;

let the cost of sending one unit of commodity from i to j be the weight of the arc (i , j);

then the shortest path problem is transformed to a transshipment problem, and hence can be solved by the network simplex method.

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We study other two algorithms: Dijkstra’s algorithm to find a SP and

the SD from a specified vertex to every other vertex;

Floyd and Warshall algorithm for all-pairs shortest path problem.

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Main idea about the Dijkstra’s method Suppose the 5 nearest vertices to

v1 are v1,v3,v5,v7 and v9. Then finding the sixth nearest

vertex is easy. Assume the sixth nearest vertex is

v6 and the shortest path is (v1,…v?, v6).

Then v? must be one the 5 nearest vertices. Can you see why?

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Another important idea!!

Suppose 1356 is the SP from 1 to 6. Then, for sure, 135 is the SP from 1 to 5. Can you see why?

As a result, to save the SP from 1 to 6, I just need to write down 5 and the SP from 1 to 5.

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Dijkstra’s algorithm Let the network G = (V, E), V = {1, 2, …,

n}, and the weight of the arc (i , j) be a(i , j) . If there is no arc from i to j (i j), then a(i , j) is taken as a large positive number

M. We want to find the SD and SP from

vertex 1 to all other vertices.

0

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In the Dijkstra’s algorithm, each vertex i is assigned a label which is either permanent, or tentative.

The permanent label L(i) of i is the SD from 1 to i;

The tentative label L’(i) of i is an upper bound for the SD from 1 to i.

At each stage of the procedure, V is partitioned to two sets: P and T, where P is the set of vertices with permanent labels, and T = V \ P is the set of vertices with tentative labels.

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We also need to use an index V(i) to record the vertex immediately before i . This index may be updated after each iteration, and when we complete computation, it shows the vertex immediately before vertex i in the shortest path from vertex 1 to i.

Dijkstra’s algorithm Step 0 (initial step) Set L(1) = 0. Set L’(j) = a(1, j) and V(j)=1 for j = 2, 3,

…, n. Set P = {1}, T = {2, 3, …, n}.

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Step 1 (Designation of a permanent label)

Find k T such that Declare vertex k to be permanently labeled:

Set T = T - k, P = P + k, L(k) = L’(k). If T = Ø (i.e. P = V), stop; the

computation is completed.

(j)}{L' min (k)L'Tj

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Step 2 (Revision of tentative labels) Set L’(j) = min {L’(j), L(k) + a(k, j)} for

all j T, and if now L’(j) = L(k) + a(k, j),

then update V(j) as V(j)=k. Go to step 1.

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Informal steps

L’(1)=0, V’(1)=1, P = {1}, T = {2,…} For all v in T, L(v)=M, V(v)=1, For any v is still in T and adjacent to v*,

Update L’(v), V’(v) Find v*, s.t.

L’(v*) L(v), v in T. L(v*) = L’(v*) and V(v*) = V’(v*) Move v* from T to P.

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In each step 1 → step 2 → step 1 iteration, a vertex is moved from T to P.

So, we need to have n-1 iterations to complete computation, and for all j=2,3,…n, the indexes V(2), …, V(n) gives us n-1 arcs which together with all n vertices form a subgraph H of G(V, E).

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If there exists a path from vertex 1 to any other vertex, then H must be connected. H is acyclic because if V(k)=i, arc (i, k) is in the shortest path from 1 to k, and i must enter P earlier than k does.

So, H is a spanning tree rooted at vertex 1. And H is a shortest distance tree which includes the shortest paths from vertex 1 to other vertices.

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Example 3.1 Obtain the SD from 1 to the remaining

vertices in the directed network shown below, using Dijkstra’s algorithm.

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Iteration 1Step 1. P = {1} L(1)=0

T = {2, 3, 4, 5, 6, 7}L’(2) = 4, L’(3) = 6, L’(4) = 8,L’(5) = L’(6) = L’(7) = M.V(2)=V(3)=V(4)=V(5)=V(6)=V(7)=1.Vertex 2 is assigned a permanent label.

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Step 2. P = {1,

2} L(2) = 4 Record ar

c (1,2)

T = {3, 4, 5, 6, 7}L’(3) = min {6, L(2) + a(2, 3)}=5L’(4) = min {8, L(2) + a(2, 4)}=8L’(5) = min {M, L(2) + a(2, 5)}=11L’(6) = min {M, L(2) + a(2, 6)}=ML’(7) = min {M, L(2) + a(2, 7)}=MV(3)=2, V(5)=2.

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Iteration 2Step 1. P={1,2} L(1) = 0 L(2) = 4

T = {3, 4, 5, 6, 7}Min{L’(i) | i in T}= L’(3) Vertex 3 is assigned a permanent label.

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Step 2 P = {1, 2,

3} L(3) = 5

T = {4, 5, 6, 7}L’(4) = min {8, L(3) + a(3, 4)}=7L’(5) = min {11, L(3) + a(3, 5)}=10L’(6) = min {M, L(3) + a(3, 6)}=9L’(7) = min {M, L(3) + a(3, 7)}=MV(4)=3, V(5)=3, V(6)=3.

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Iteration 3Step 1. P = {1, 2,

3} L(1) = 0 L(2) = 4 L(3) = 5

T = {4, 5, 6, 7}Min {L’(i) | i in T} = L’(4)Vertex 4 is assigned a permanent label.

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Step 2 P = {1, 2, 3,

4} L(4) = 7 T = {5, 6, 7}

L’(5) = min {10, L(4) + a(4, 5)}=10L’(6) = min {9, L(4) + a(4, 6)}=9L’(7) = min {M, L(4) + a(4, 7)}=M

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Iteration 4Step 1 P = {1, 2, 3,

4} L(1) = 0, L(2) = 4, L(3) = 5, L(4) = 7

T = {5, 6, 7}Min {L’(i) | i in T} =L’(6) Vertex 6 is assigned a permanent label.

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Step 2. P = {1, 2, 3, 4, 6} L(6) = 9

T = {5, 7}L’(5) = min {10, L(6) + a(6, 5)}=10L’(7) = min {M, L(6) + a(6, 7)}=17V(7)=6.

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Iteration 5Step 1 P = {1, 2, 3,

4, 6} L(1) = 0, L(2) = 4, L(3) = 5, L(4) = 7, L(6) = 9

T = {5,7}Min {L’(i) | i in T} = L’(5)Vertex 5 is assigned a permanent label.

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Step 2 P = {1, 2, 3, 4, 6, 5} L(5) = 10

T = {7}L’(7) = min {17, L(5) + a(5, 7)}=16V(7)=5

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Iteration 6Step 1. P = {1, 2, 3, 4, 6, 5} L(1) = 0, L(2) = 4, L(3) = 5, L(4) = 7, L(6) = 9, L(5) = 10.

T = {7}L’(7) = 16Vertex 7 gets a permanent label.

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Step 2. P = {1, 2, 3, 4, 6, 5,

7} L(7) = 16

T is empty

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Thus L(1) = 0, L(2) = 4, L(3) = 5, L(4) = 7, L(6) = 9, L(5) = 10 and L(7) = 16, giving the SD from 1 to each vertex.

The indexes V(2)=1, V(3)=2, V(4)=3, V(5)=3, V(6)=3, V(7)=5 show that arcs (1, 2), (2, 3), (3, 4), (3, 5), (3, 6) and (5, 7) constitute a shortest distance tree in the given network as shown below, giving the SP from vertex 1 to every other vertex.

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The shortest distance tree:

Question: is this a minimum weight spanning tree?

No, in MST, (6,5) replaces (3,6)

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You answer should consist

Sequence of arcs. (according to the order the arcs are moved to P)

Tree (Draw it!!). Total weight.

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Theorem 3.1 Dijkstra’s algorithm finds the SD from vertex 1

to every other vertex i (i = 2, …, n). Proof. We prove the theorem by induction

on the cardinality of P. We will show that for each P generated in the algorithm,

(1) for every i P, L(i) is the SD from 1 to i. (2) for every j T, L’(j) is the length of an SP

from 1 to j under the restriction that every intermediate vertex is in P.

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First, when |P| = 1, i.e. P = {1}, T = {2, 3, …, n}, the two conclusions hold obviously. We now show that if conclusions (i)

and (ii) are true when |P| = k-1, then they also hold if |P| = k.

shore shore

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Without loss of generality, assume P = {1, 2, …, k-1}, T = {k, …, n}. By the assumption, (i) for i P, L(i) is the SD from 1 to i; (ii) for j T, L’(j) is the SD from 1 to j under the restriction that every intermediate vertex is in P.

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Also assume that in the current iteration, vertex k moves to P, i.e.

So, L(k) = L’(k), and

(△)

(*) (j)}{L' min (k)L'T j

, {k}TT {k}, PP

Tj j)},a(k, (k)L'(j),{L'min)j('L

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We need to show that (i) L(k) is the SD from 1 to k. If it is not true, let d be the SD from 1 to

k. So, d < L(k) = L’(k). As L’(k) is the SD from 1 to k provided that every intermediate vertex of the SP is in P, it means that along any SP from 1 to k, there must be at least one vertex in T. Let v be the first vertex in T along the SP from 1 to k.

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Let the SD from 1 to v be d’. Then d’ = L’(v) and d’ d < L’(k).

So, L’(v) < L’(k), which contradicts (*). Therefore, L(k) must be the SD from 1

to k.

(*) (j)}{L' min (k)L'T j

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(ii) We need to show that for each j = {k+1, …, n}, ’(j) is the SD from 1 to j under the restriction that all intermediate vertices are in .

Let be the SD from 1 to j when all intermediate vertices are in . The corresponding SP may have two possibilities:

TL

P

dP

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(a) The SP does not go through vertex k.

In this case, is the SD from 1 to j under the restriction that every intermediate vertex is in P = {1, …, k-1}. So, = L’(j).

d

d

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(b) The SP includes vertex k.

In this case, k must be the last vertex before j in the SP. If not, the SP arrives at k, then reaches a q P, and at last comes to j.

It means the shortest path from 1 to q must pass through k, which is impossible because q enters P before k does, i.e. the SP from 1 to q need not to pass k.

Now since k is the last vertex in the SP before reaching j, the SD from 1 to j must be the SD from 1 to k plus a(k, j): L’(k) + a(k, j).


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Combining the two cases (a) and (b), = min {L’(j), L’(k) +a(k, j)}. So, by the formula (△), ’(j) = , i.e.

’(j) is the SD from 1 to j under the restriction that every intermediate vertex of SP is in . So, the proof is completed.

dL d L

P

Tj j)},a(k, (k)L'(j),{L'min)j('L (△)

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3.2 All shortest path algorithm

Let G = (V, A) be a directed network. V = {1, 2, …, n}.

Let auv be the weight of the arc (u, v). (in the method we allow some negative auv)

We want to calculate the SD from every vertex u to every other vertex v.

We will use The Floyd - Warshall Method

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The main idea used Suppose v5 is the second vertex in

the shortest path from v1 to v9. That is, the shortest path from v1 to

v9 is (v1, v5, ???, v9). Besides, suppose (v5, v6, v7, v8, v9)

is the shortest path from v5 to v9. Then, we can assure the shortest

path from v1 to v9 is (v1, v5, v6, v7, v8, v9).

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Let A = [auv] be the n n weight matrix;

P = [Puv] be the n n matrix with Puv = v, i.e.

The F&W method is an iterative method which needs to have n iterations. In the j-th iteration, we will have two n n matrices A(j) = [auv(j)] and P(j)=[Puv(j)] (j = 1, …, n). A(0)=A and P(0)=P.

n...21

.........

n...21

n...21

P

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In A(j), auv (j) shall be the SD from u to v with intermediate vertices in the vertex set {1, 2, …, j}, and the corresponding element puv(j) in P(j) gives the vertex immediately after u in the path to attain the above SD auv(j).

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The algorithm is as follows. Step 0 Let j = 1, A(0) = A and P(0) =

P. Step 1 for all (u,v), If auv (j-1) < auj (j-1)

+ ajv (j-1), then auv (j) = auv (j-1) and puv(j) = puv (j-1), otherwise

auv (j) = auj (j-1) + ajv (j-1) and

puv (j) = puj (j-1).

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Step 2 If in matrix A(j), one diagonal element is negative, stop, the problem has a negative cycle, and some SD from a vertex to another vertex are unbounded.

If j = n, go to step 3; otherwise let j ← j+1 and return to Step 1.

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Step 3 Each element auv(n) of the matrix A(n) gives the SD from vertex u to vertex v.

To find the SP from u to v, if puv (n) = j1, then the first arc of the SP is (u, j1). If j1 v, we continue by checking pj1v(n). If pj1v(n)=j2, then the second arc is (j1, j2). Repeat the procedure until we reach vertex v. Then the SP is u → j1 → j2 → …. → v.

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We use an example to explain the algorithm.

Example 3.2 Obtain the SD matrix and the SP

matrix in the directed network as shown below.

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3.2 All shortest path algorithm We begin with the following

matrices:

4321

4321

4321

4321

P(0) P and

0665

3010

703

340

A(0) A M

M

M

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4121

4321

4321

4321

P(1) and

0265

3010

703

340

A(1)M

M

M

3.2 All shortest path algorithm Iteration 1, based on vertex 1 (j

= 1) at the end of the first iteration we

have the following matrices:

4321

4321

4321

4321

P(0) P and

0665

3010

703

340

A(0) A M

M

M

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Matrix A(1) is obtained from A(0) by the following “triangle” operation:

Draw two lines on row k and column k respectively (here k=1, and when we calculate A(2), k=2, ……).

In A(0), for each element which is neither in row k, nor in column k, find its two projections on the two lines. For example for element , its two projections are and respectively.

Compare the sum of the two values at the projections with the value at the element. (e.g., compare with ). Then we use the rule in the F-W method to obtain A(1) and P(1).)0(43a

)0(43a)0(13a)0(41a

)0()0( 1341 aa

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Iteration 2, based on vertex 2 (j=2)

It begins with A(1) and P(1). The triangle operations are carried out as in the previous operation (but now k=2) – at this stage the auv(2) = min{auv(1), au2(1) + a2v(1)}. At the end of this iteration we have the following matrices:

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Iteration 2, based on vertex 2 (j=2)

4222

4322

4321

4321

P(2) and

0163

30107

703

3-40

A(2)M

M

4121

4321

4321

4321

P(1) and

0265

3010

703

340

A(1)M

M

M

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Iteration 3, based at vertex 3 (j=3)

It begins with A(2) and P(2). Applying the triangle operation (let k=3), we obtain the following matrices:

4222

4322

3321

3321

P(3) and

0163

30107

4703

0340

A(3)

4222

4322

4321

4321

P(2) and

0163

30107

703

3-40

A(2)M

M

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3.2 All shortest path algorithm Iteration 4, based on vertex 4

(j=4) It begins with A(3) and P(3).

Applying the triangle operation (let k=4) , we get the following matrices:

4222

4344

3321

3321

P(4) and

0163

3096

47-03-

03-40

A(4)

4222

4322

3321

3321

P(3) and

0163

30107

4703

0340

A(3)

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At this stage we have the SD from every vertex to every other vertex in the network, which can be obtained readily from A(4).

The various shortest paths are obtained as follows:

From 1 to 2: p12(4)=2. So the SP is 1→2, i.e., the arc (1, 2).

From 1 to 3: p13(4) = 3. So the SP from 1 to 3 is the arc (1, 3). 1 2 3 3

1 2 3 3P(4)

4 4 3 4

2 2 2 4

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From 1 to 4: p14(4) = 3. So take the arc (1, 3) and join it to the SP from 3 to 4. We have p34 (4) = 4. So the SP from 3 to 4 is the arc (3, 4). Thus the SP is 1→3→4.

From 2 to 1: the SP is the arc (2, 1). From 2 to 3: the SP is the arc (2, 3). From 2 to 4: the SP is 2→3→4. From 3 to 1: the SP is 3→4→2→1. From 3 to 2: the SP is 3→4→2. From 3 to 4: the SP is the arc (3, 4). From 4 to 1: the SP is 4→2→1. From 4 to 2: the SP is the arc (4, 2). From 4 to 3: the SP is 4→2→3.

1 2 3 3

1 2 3 3P(4)

4 4 3 4

2 2 2 4

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Locating negative cycles Consider a network for which the

weight matrix is as follows:

0342

0

3102

10

A MMM

MM

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At the end of the second iteration we have the following matrices:

0 1 1 2 3 4

2 0 1 3 1 2 3 4A(2) and P(2)

0 1 2 3 4

2 4 3 1 2 2 2 2

M M

M M M

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In A(2), the diagonal element a44(2)= -1 <0, indicating the presence of a negative cycle in the network. The negative cycle can be located by using P(2) matrix, which is 4 → 2 → 4.

So, the F&W algorithm will stop calculation after obtaining A(2) and P(2).

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Theorem 3.2 If the network contains no negative cycle, then in using the Floyd-Warshall algorithm, auv(n), the

(u, v)th element in A(n), is equal to the SD from u to v.

Proof Suppose the algorithm works until obtaining A(n) and P(n), and there is no negative cycle.

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We use the induction method to show that for j =1, …, n, auv(j) is the SD from u to v with each intermediate vertex w j.

For j=1, as auv(1)= ,

obviously auv(1) is the SD from u to v if only vertex 1 can be a possible intermediate vertex.

otherwise aa

aaa if a

1u1

1vu1uvuv

v

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Suppose auv(j-1) is the SD from u to v with each intermediate vertex w j-1. Let Quv(j) be the SP from u to v with every intermediate vertex w j.

It has two possibilities: (1) Quv(j) does not take j as an intermediate

vertex. Then Quv(j) = Quv(j-1)

and hence auv(j) = auv(j-1). (a)

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3.2 All shortest path algorithm(2) Quv(j) has vertex j as an intermediate

vertex.

Then the part of Quv(j) from u to j must

be the SP from u to j with every intermediate vertex w j-1, i.e., it is Quj(j-1), and the part of Quv(j) from j to v must be Qjv(j-1). So,

auv(j) = auj(j-1) + ajv(j-1). (b)

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Combining (a) and (b), we know that the SD from u to v with each intermediate vertex w j is

auv(j) = min{auv(j-1), auj(j-1) + ajv(j-1)}, which is just the formula in step 1. Therefore we proved that the

conclusion is true for j, and the induction proof is completed.

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An exercise for F-W method. For the network shown on the next page, when we use F-W method to obtain all pairs of shortest distances, the final matrices A(6) and P(6) are:

023345

101423

430512

134034

563401

652620

)6(A

655355

654644

554222

666322

311321

644221

)6(P

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Graph of the exercise

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a. Using matrix P(6), show the shortest paths from vertex 3 to all other vertices and the shortest paths from vertex 4 to all other vertices.

b. Using matrix A(6) to verify your answer for question a.

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3.3 Medians and centers

Location Problem In a network, we need to find a vertex

such that the distances to other vertices meet some optimality criterion.

Suppose G=(V,E), V = {1, 2, …n}. The following two criteria (Minsum Problems & Minmax Problems) are often used.

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1. Minsum Problems Find a vertex j such that the sum

of the distances from j to all other vertices is as small as possible.

Let d(j, k) be the shortest distance from j to k ( d( j, j )=0 ), then the Minsum problem is

n

1kn1,2,..., j

k)d(j,min

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This problem is also called l1 norm location problem as for the non-negative vector

dj = (d (j,1), d (j,2), …, d (j,n)), the l1 norm ||dj||1 = , (if all d(j,k) are non-negative) and the location problem is to find

n

1k

k)d(j,

1jn1,2,..., j

||d|| min

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3.3 Medians and centers Once we obtain the n n shortest

distance matrix D = [d(i, k)] (for example by the F-W method), it is easy to solve the minsum location problem:

let then the best location is the vertex j such

that . The vertex j is also called a median

(vertex).

n

1ki k)d(i,s

in1,2,..., i

j smins

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3.3 Medians and centers For example:

S1=21

S2=19

S3=15

S4=15

S5=11

S6=17

median

vertex

023345

101423

430512

134034

563401

652620

)6(A

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2. Minmax problems Find a vertex j such that the

distance from j to the farthest vertex is minimized, i.e. to solve the problem:

k)d(j,Max Minn1,...,kn1,.., j

The vertex j is also called a center (vertex).

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This is also called the norm location problem, because

= Max (d (j, 1), d (j, 2), …, d (j,

n))(if all d(j,k) are non-negative) and we want to find

L

||d|| j

||d|| Min j

n1,.., j

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The Minmax location problem can also be solved easily once we obtain the all pair SD matrix D=[d(i, k)].

For each row i, let Then the best location for the

minmax problem is vertex j such that

k)d(i, Max n1,.., k

ie

in1,.., i

e Min

je

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3.3 Medians and centers For example:

e1=6

e2=6

e3=4

e4=5

e5=4

e6=5

center vertex

023345

101423

430512

134034

563401

652620

)6(A

centervertex

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It depends on different purposes to choose Minsum or Minmax criterion.

For example, a delivery company which transports goods to all vertices may use the Minsum criterion.

A fire station may use the Minmax criterion to decide its place.

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Use NETSOLVE to calculate shortest paths Specify directed or undirected

graph by typing D or U to answer the question by NETSOLVE, depending on the particular problem.

No node data required. Need to enter edge (arc) data.

Enter the two node names and the weight of the edge (arc).

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Use NETSOLVE to calculate shortest paths

NETSOLVE can calculate 4 types of shortest paths.

Type 1: from a given source vertex (say 2) to a given destination vertex (say 5).

Command: SPATH 2 5Type 2: from a given source vertex (say

1) to all other vertices.Command: SPATH 1 *

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Type 3: from all vertices to a given destination vertex (say 3).

Command: SPATH * 3Type 2: from any vertex to all other

vertices.Command: SPATH * *

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Example. For the network shown below, calculate all pairs of shortest paths

Network Optimization

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