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# D1 (Decision 1) 17 May 2013 Official Thread Watch

1. (Original post by posthumus)
Erm a minimum spanning tree is where nodes have no more than 1 arc between them and there are no cycles. The weight of this tree will be the minimum ?

Define : Bipartite graph & Also maximum matching
Not quite, minimum spanning tree, is where it includes all the vertices of G and ensures that the total length of arc is as small as possible.

Bipartitie graph is where it consists of 2 sets of distinct vertices X and vertices Y in which the arcs can join from vertices in X to vertices in Y, but not within the set of vertices.

Maximum matching , is the (1-1) pairing of all the elements of X to the elements in Y
2. Does anyone know what year they changed the spec?? Also really having trouble with scheduling diagrams !!!!!

This was posted from The Student Room's iPhone/iPad Ap
3. (Original post by Flounder1)
Does anyone know what year they changed the spec?? Also really having trouble with scheduling diagrams !!!!!

This was posted from The Student Room's iPhone/iPad Ap
I think the Paper's started getting updated from Jun 2009.

Scheduling Diagrams are not always Unique, so you might find your Answer is slightly different to the Markscheme, not too different though.

Check out Maths24/7 on youtube. He put up some good videos which really helped me.
4. (Original post by m4ths/maths247)
Nice post.
I like D1. It could be argued that its not as mathematically challenging as some units but its still intrinsically rewarding.

I have just finished all the D1 videos and managed to get scheduling up this weekend as that seemed to be the topic most people dislike!
Hi Sir,

Hope you are well.

I had a question regarding your video on scheduling which is fantastic. But I am not sure when you calculated the lower bound , it came to 2.161--> so this is close to 2 , why do you round it to 3 and not 2?
5. (Original post by otrivine)
Hi Sir,

Hope you are well.

I had a question regarding your video on scheduling which is fantastic. But I am not sure when you calculated the lower bound , it came to 2.161--> so this is close to 2 , why do you round it to 3 and not 2?
You can't have 2.161 workers So you'll need 3 workers !
6. (Original post by posthumus)
You can't have 2.161 workers So you'll need 3 workers !
No what i meant was , it should be 2 workers not 3 /
7. (Original post by otrivine)
No what i meant was , it should be 2 workers not 3 /
Yh but you can't ever round down because... what about that 0.161 work left ? Who's going to do that ?

A third worker... the only difference is - that this third worker will have a lot of free time on his/her hands
8. A little help here guys please ...

I thought here there would be 2 routes (or paths) to the end node...

BDJ
BDHL

I realize that they said to state the critical activities and not the "critical path" ... maybe this is why I got it wrong.

9. I'm still getting djikstra wrong...
My teacher said it seems as though I'm getting worse lol

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10. (Original post by otrivine)
Hi Sir,

Hope you are well.

I had a question regarding your video on scheduling which is fantastic. But I am not sure when you calculated the lower bound , it came to 2.161--> so this is close to 2 , why do you round it to 3 and not 2?
Hi
Sorry I have not posted much this week.
Ok as the poster has put above:

Let's say all the activities take 50 hours and you need to get a project done in 20 hours (thats the critical time) without delay.
The LEAST number of workers must be 3 as with 2 they have 25 hours of work to complete in 20 which, clearly, can't be done.
Thereofre 50/20 give 2.5 but they need at LEAST one more worker. hence rounding up.
The lower bound doesn't always tell you how many are required, it just tells you the least amount by calculation.
Say you have 4 activities starting at the same time, 3 will not be enough in this case even if it is the lower bound.
11. (Original post by m4ths/maths247)
Hi
Sorry I have not posted much this week.
Ok as the poster has put above:

Let's say all the activities take 50 hours and you need to get a project done in 20 hours (thats the critical time) without delay.
The LEAST number of workers must be 3 as with 2 they have 25 hours of work to complete in 20 which, clearly, can't be done.
Thereofre 50/20 give 2.5 but they need at LEAST one more worker. hence rounding up.
The lower bound doesn't always tell you how many are required, it just tells you the least amount by calculation.
Say you have 4 activities starting at the same time, 3 will not be enough in this case even if it is the lower bound.
thanks sir, so even if we get the lowest value such as 2.03 we round to 3 workers? correct
12. (Original post by otrivine)
thanks sir, so even if we get the lowest value such as 2.03 we round to 3 workers? correct
Yes because it needs more than 2.

If bags of sugar were only sold in 200g bags and you needed 410g to make a cake how many would you have to buy?
13. (Original post by m4ths/maths247)
Yes because it needs more than 2.

If bags of sugar were only sold in 200g bags and you needed 410g to make a cake how many would you have to buy?
I see your point now, thank you so much
14. (Original post by otrivine)
I see your point now, thank you so much
You are welcome. Good luck with the exam.
I may do some more D1 stuff in the week if I find any nice questions that take my fancy!!
15. In case anyone forgot, I just looked at some Posts which helped me earlier.

Here are some of Tiny Hobbit's Wisdom (Teacher Extraordinaire). I will probably refer to this again:

(Original post by tiny hobbit)
For a list of 7 numbers, the maximum number of comparisons in a bubble sort would be 6+5+4+3+2+1, which is an arithmetic series. Sn = 0.5 x 6(1 + 6) i.e. 0.5 n (first + last)
(Original post by tiny hobbit)
There are 2 sorts of scheduling questions:

1. If you have got to get it done in the critical time, regardless of how many workers it takes, then yes, give a path of critical activities to one worker. It can then be useful to work out the latest start time for each activity, i.e late time at finish - duration, to see which activities are most urgent. You must keep checking on the network as to whether an activity can start yet.

2. Fixed number of workers, the project will have to finish late. In this case, do NOT give the critical activities to one worker. Start using the latest start times for the activities straight away, doing those with the smallest latest start times first.

With either version, don't forget to answer the question, i.e. state either the minimum number of workers or the time.
16. (Original post by tiny hobbit)
For a list of 7 numbers, the maximum number of comparisons in a bubble sort would be 6+5+4+3+2+1, which is an arithmetic series. Sn = 0.5 x 6(1 + 6) i.e. 0.5 n (first + last)
Hi,
When trying to work out the number of workers needed from a schudling diagram - how do you do it? Any tips on where to look to see which activites must be happening therefore that amount of workers?

Thanks
17. When doing a linear programming question in the exam do you have to use the ruler method to find the optimal point? Or can you just test the points on the edges?
18. (Original post by Smko)
Hi,
When trying to work out the number of workers needed from a schudling diagram - how do you do it? Any tips on where to look to see which activites must be happening therefore that amount of workers?

Thanks
do you mean from a cascade chart? If so, find the time (usually something + 0.5) t which the greatest number of activities must be happening.

If you mean from a scheduling diagram, how many rows did you have to use so that all of the activities happened on time and no activity started before its preceding ones had finished.
19. (Original post by IWantSomeMushu)
When doing a linear programming question in the exam do you have to use the ruler method to find the optimal point? Or can you just test the points on the edges?
If the question doesn't say, either will do, but make sure if you are using point testing that you try all of the vertices of R and show your working.

There has recently been one question that told you to use an objective line and one question that told you to use point testing.
20. (Original post by tiny hobbit)
do you mean from a cascade chart? If so, find the time (usually something + 0.5) t which the greatest number of activities must be happening.

If you mean from a scheduling diagram, how many rows did you have to use so that all of the activities happened on time and no activity started before its preceding ones had finished.
Sorry meant cascade chart! how do you know where the time is? DO you just have to look?

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