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Decision Maths Networks Kruskals Algorithm

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Wiltshire Networks 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.

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Wiltshire Networks

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Wiltshire Minimum 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.

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Wiltshire Algorithms 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)

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Wiltshire Kruskal`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.

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Wiltshire Example Consider the Network below. It helps to rank the arcs in increasing order.

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Wiltshire Applying the Algorithm 1 – Start by selecting the smallest arc, AB or DE, it makes no difference. Select AB.

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Wiltshire Applying the Algorithm 2 – Now select the next smallest, which is DE.

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Wiltshire Applying The Algorithm 3 – Next we can select CF or ` DF, again it makes no difference. Lets pick DF.

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Wiltshire Applying the Algorithm. Next select CF.

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

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Wiltshire Applying the Algorithm 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.

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Wiltshire Question – Ex 3a pg 66 q1 Find the minimal spanning tree and associated shortest distance for the network below:

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q1

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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Wiltshire Solution – Ex 3a pg 66 q4

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