Add EH . Add C E . Add EF .
✔
EF = 2
✔
EG = 2
✔
BD = 3
Add EG . Add BD .
GH = 3 AB = 5
PL E
…
At this point the next arc in the list is GH but this would create a cycle E– G– H – E.
✔
EH = 1
✔
C E = 2
Add AB . Although every node is now included the two parts are not connected so a spanning tree is not yet formed.
✔
EF = 2
✔
EG = 2
✔
BD = 3
GH = 3
✔
M
AB = 5
SA
Continue until a spanning tree is formed.
Total weight =
21
Delete all remaining arcs and add the weights of the arcs that have been used to find the total weight of the spanning tree. You should show all the working on one diagram. The ticks and crossing out on the list are evidence that you have used the algorithm correctly.
WORKED EXAMPLE 2.3
Find a different minimum spanning tree for the network in Worked example 2.2.
Use DE instead of BC .
Total weight =
21
Original material © Cambridge University Press 2018