Decision Maths Networks Kruskals Algorithm Wiltshire Networks A Network is a weighted graph, which...

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Decision Maths

NetworksKruskal’s Algorithm

WiltshireNetworks

A Network is a weighted graph, which just means there is a number associated with each edge.

The numbers can represent distances, costs, times in real world applications.

Obvious examples include maps and similar geographical networks.

WiltshireNetworks

WiltshireMinimum Connector Problem

Basically you need to travel to every node using the least total length.

Consider 4 houses in a Network shown in the diagram below. The weight on each arc represents the distance between each house.

An Electricity company wants to supply every house by using as little cable as possible.

Clearly the shortest possible route is to go from A to B to C and then to D.

So 4 + 3 + 3 = 10, there is no shorter way of supplying every house.

WiltshireAlgorithms

The previous example was a simple one and the solution was very easy to spot.

For more complicated examples you will need to use an algorithm.

An Algorithm is simply a list of instructions that solve a particular problem.

(You will cover Algorithms in more depth later on in the course)

WiltshireKruskal`s Algorithm

There are 3 steps to follow in Kruskal`s Algorithm. Step 1 – Select the shortest arc in the

network. Step 2 – Select the shortest arc from those

which are remaining. Ensure that you do not create a cycle.

If you do ignore and move on to the next shortest arc.

Step 3 – If all the vertices are connected then stop. If not return to step 2.

WiltshireExample

Consider the Network below.

It helps to rank the arcs in increasing order.

WiltshireApplying the Algorithm

1 – Start by selecting the smallest arc,

AB or DE, it makes no difference.

Select AB.

2 – Now select the next smallest, which is

WiltshireApplying The Algorithm

3 – Next we can select CF or ` DF, again it makes no

difference. Lets pick DF.

WiltshireApplying the Algorithm.

Next select CF.

The next smallest length is EF. However there is already a route from E to F, so this arc is not required.

Adding CD will again create a loop so the last arc to add is AF. All vertices are now joined so the problem is complete.

WiltshireQuestion – Ex 3a pg 66 q1

Find the minimal spanning tree and associated shortest distance for the network below:

WiltshireSolution – Ex 3a pg 66 q1

WiltshireSolution – Ex 3a pg 66 q4

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