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# Decision Maths Help !!! :/ Watch

1. I'm stuck on a few of the questions on D1:

The diagram shows the time, in minutes, for a traffic warden to walk along a network of roads, where x>0

a) Explain why a section of roads from A to E has to be repeated

b) The route ACE is the second shortest route connecting A to E. Find the range of possible values of x.

c) Find, in terms of x, an expression for the minimum distance that the traffic warden must walk and write down a possible route that he could take.

d) Starting at A, the traffic warden wants to get to F as quickly as possible. On Figure 1 use Dijkstra's algorithm to find, in terms of x, the minimum time for this journey, stating the route that he should take. (I know how to use Dijkstra's algorithm but not with x values)
2. *bump*
3. for 1) what properties of graphs do you know?

for 2) consider other paths that take you from A to E
4. Also I'm pretty sure you haven't wrote the full question out, since part a) makes no sense without reading the later parts
5. Not sure if these are right, but I hope it helps...

a) The section of roads must be repeated to make the graph traversable keeping to the minimum distance starting and ending at A as possible.

b) ACE is 20+2x, its the second shortest route and I assume the shortest route would be ABE (14+12=26), therefore solve 20x+2>26

c) Just do the route inspection algorithm, add up all the terms and it should be in terms of x.

d) I think all you have to do is use Dijkstra's algorithm and from part (b) you should know the value of x, not too sure but I'll see if I can do it.

btw, Im sort of confused why EF is 16 and the repeat of EF is 13+x
6. aaah sorry forgot a little bit at the top....
7. (Original post by ForGreatJustice)
for 1) what properties of graphs do you know?

for 2) consider other paths that take you from A to E
x > 0
8. (Original post by Sai4)
Not sure if these are right, but I hope it helps...

a) The section of roads must be repeated to make the graph traversable keeping to the minimum distance starting and ending at A as possible.

b) ACE is 20+2x, its the second shortest route and I assume the shortest route would be ABE (14+12=26), therefore solve 20x+2< or = 26

c) Just do the route inspection algorithm, add up all the terms and it should be in terms of x.

d) I think all you have to do is use Dijkstra's algorithm and from part (b) you should know the value of x, not too sure but I'll see if I can do it.

btw, Im sort of confused why EF is 16 and the repeat of EF is 13+x

for b) shortest is A to D to E which is 25.5, so i got x can range from 2.75 to 3. Which one is route inspection algorithm? is that Kruskal's one?
9. (Original post by EternalDoom)
for b) shortest is A to D to E which is 25.5, so i got x can range from 2.75 to 3. Which one is route inspection algorithm? is that Kruskal's one?
No, another name for the Route inspection algorithm is Chinese Postman algorithm, the one where you find the shortest route starting and ending at the same vertex.
10. (Original post by Sai4)
No, another name for the Route inspection algorithm is Chinese Postman algorithm, the one where you find the shortest route starting and ending at the same vertex.
Haven't done that in class yet XD

So I'll try and do it with Kruskal's or Prim's
11. Oh, if you haven't done that yet, you won't be able to answer this question
12. (Original post by Sai4)
Oh, if you haven't done that yet, you won't be able to answer this question
Awesome, gotta love teachers!
13. Yeah, The route inspection algorithm is the longest algorithm in D1 I think, really easy to make mistakes.
14. (Original post by Sai4)
Oh, if you haven't done that yet, you won't be able to answer this question
I just did it with Kruskal O.o

10 + 12 + 12.5 + 13 + 13 + x
= 60.5x

One possible route is A - D - E - F - E - C - B - C - E - D - A

Is this wrong?
15. hi has anyone got the june 2010 paper for ocr decision maths (D1)
thanks

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