The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

D1 (Decision 1) 17 May 2013 Official Thread

Scroll to see replies

Original post by tiny hobbit
I teach my students not to put them in at all. Having just checked up on several students who did D1 last summer, they all got full marks on the Dijkstra question, so not putting the larger values in is fine.


quick question when drawing the activity network, where you have 2 activties sharing the same start vertex and end vertex because of uniquely represented, you use a dummy but does it matter which one out of those 2 activities to put the dummies? :smile:
Reply 641
Avatar for tobywalsh
tobywalsh
Original post by otrivine
quick question when drawing the activity network, where you have 2 activties sharing the same start vertex and end vertex because of uniquely represented, you use a dummy but does it matter which one out of those 2 activities to put the dummies?


Do you mean if you have activity A and B both starting off from nothing. Then Activity C follows, dependant on both A and B being completed? If so; No :smile:
Original post by Westeros
M can't do activity 1?



http://www.examsolutions.net/a-level-maths-papers/Edexcel/Decision-Maths/Decision-Maths-D1/2010-June/questions/q8.gif

Here is the question, it specifically asks for the lower bound and the mark scheme says 4 :mad:

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/june2010-ms/6689_01_msc_20100618.pdf

I guess in that case I stand correct on the term lower bound :redface: They just basically wanted you to do a bit of scheduling in your head and figure out how many workers are actually needed.

But I don't know why they used the term lower bound however :confused: That is bloody confusing
Original post by tobywalsh
Do you mean if you have activity A and B both starting off from nothing. Then Activity C follows, dependant on both A and B being completed? If so; No :smile:


No what I mean is, when you have you know the 2 activities can not start and end at same place/node right, so you have to use a dummy as they are uniquely represented , if you have A and B starting and finishing at same node , does it matter if you putty the dummy in A and not in B or vice versa?
Original post by posthumus
I guess in that case I stand correct on the term lower bound :redface: They just basically wanted you to do a bit of scheduling in your head and figure out how many workers are actually needed.

But I don't know why they used the term lower bound however :confused: That is bloody confusing

so does this prove that to get the minimum number of workers, there is no relation between lower bound and using ur gantt because you can different answers , lower bound gets 3 workers and using gantt gets 4
Reply 645
Avatar for lizz-ie
lizz-ie
18
Please could someone help me with Critical Path Analysis? I can't do any of it :frown:
Reply 646
Avatar for tobywalsh
tobywalsh
@ otrivine do you mean this? Either one works :smile:
D1 Explanatin.png
Original post by otrivine
quick question when drawing the activity network, where you have 2 activties sharing the same start vertex and end vertex because of uniquely represented, you use a dummy but does it matter which one out of those 2 activities to put the dummies? :smile:


Doesn't matter which
Original post by otrivine
No what I mean is, when you have you know the 2 activities can not start and end at same place/node right, so you have to use a dummy as they are uniquely represented , if you have A and B starting and finishing at same node , does it matter if you putty the dummy in A and not in B or vice versa?


ImageUploadedByStudent Room1368717720.997983.jpg
You mean like this? the dummy between activity 8 and 9 is to represent uniqueness, so that K and L don't share the same start and end event number. It doesn't matter which dummy you put it on, because if you look at this example, both K and L are dependent on I H and G :smile:


Posted from TSR Mobile
Original post by tiny hobbit
Doesn't matter which


OK good, and it would be great if you could check for me question 26) part d) page 172? i dont get it
Original post by Westeros
ImageUploadedByStudent Room1368717720.997983.jpg
You mean like this? the dummy between activity 8 and 9 is to represent uniqueness, so that K and L don't share the same start and end event number. It doesn't matter which dummy you put it on, because if you look at this example, both K and L are dependent on I H and G :smile:


Posted from TSR Mobile


yes thats the one so for 8-9 even if I put the dummy on L or K, I will still get the marks right
Reply 651
Avatar for Arsey
Arsey
10
Original post by otrivine
A quck question sir, why is it sometimes that the lower bound does not tell you how many workers cause sometimes when you use the chart you get 4 and lower bound gets 3 why does it not work all the time


the lower bound is the smallest amount of workers in an ideal world, it is not necessarily possible.

when you add up the activities and divide by the critical time this leads to an a lower bound (but this may not be possible).

If a Gannt chart is used properly this will lead to a better lower bound which will definitely be possible solution if done correctly.

You can get harder situations when working out a lower bound from a Gannt.

Say for example on day 10, 3 activities must be happening and a fourth may be happening. The LB is 3.

On day 14, a different 3 activities must be occurring, again indicating an LB of 3.

However, what if the fourth activity on day 10 only escaped by pushing it into day 14; then the LB becomes 4!
Original post by tobywalsh
@ otrivine do you mean this? Either one works :smile:
D1 Explanatin.png


Yes thats the one :smile: Brilliant thanks!
Reply 653
Avatar for Arsey
Arsey
10
Original post by otrivine
oh so we round up everytime the book is correct? even if we have 5.02 do we round to 6


yep, always round up no matter how close to the integer below
Original post by Arsey
the lower bound is the smallest amount of workers in an ideal world, it is not necessarily possible.

when you add up the activities and divide by the critical time this leads to an a lower bound (but this may not be possible).

If a Gannt chart is used properly this will lead to a better lower bound which will definitely be possible solution if done correctly.

You can get harder situations when working out a lower bound from a Gannt.

Say for example on day 10, 3 activities must be happening and a fourth may be happening. The LB is 3.

On day 14, a different 3 activities must be occurring, again indicating an LB of 3.

However, what if the fourth activity on day 10 only escaped by pushing it into day 14; then the LB becomes 4!




Thank you so much! in the exam could they ask why when calculating lower bound is not as accurate as using a gantt chart to get minimium number of workers?
Original post by otrivine
so does this prove that to get the minimum number of workers, there is no relation between lower bound and using ur gantt because you can different answers , lower bound gets 3 workers and using gantt gets 4


Original post by posthumus
I guess in that case I stand correct on the term lower bound :redface: They just basically wanted you to do a bit of scheduling in your head and figure out how many workers are actually needed.

But I don't know why they used the term lower bound however :confused: That is bloody confusing


I think what I've found from past papers, the Critical Path Analysis questions which ask for a lower bound which don't involve a gantt/cascade chart, you find the lower bound by doing:
total weight of network/critical time
But the question which involve a gantt chart, they have selected a time and stated how activities must be happening during that specific time! Luckily they're only 2 mark questions :biggrin:
Original post by Westeros
I think what I've found from past papers, the Critical Path Analysis questions which ask for a lower bound which don't involve a gantt/cascade chart, you find the lower bound by doing:
total weight of network/critical time
But the question which involve a gantt chart, they have selected a time and stated how activities must be happening during that specific time! Luckily they're only 2 mark questions :biggrin:


It's so overwhelming because I was told you never had to use your brain for D1 :tongue: And yh I think then you can select a time to give an example of what you mean :smile: I hate these type of questions which are hard to explain!

Thanks for sharing, I'd never come across such questions !
Original post by Arsey
yep, always round up no matter how close to the integer below


Will you be posting answers for the D1 paper?
Also a quick question :colondollar:
Can you explain regarding linear programming - when it asks to find the optimal solution/amount, how do I know to use integer values? For example, in one question I used point testing and found (11.2,6) to be the optimal solution, but the mark scheme used (11,7), and my graph was definitely correct? :biggrin:
Original post by lizz-ie
Please could someone help me with Critical Path Analysis? I can't do any of it :frown:


Is there anything specifically though that is troubling you ?

Would be easier also if you could post questions :smile:

If you don't mind me asking... are you attempting to learn all the algorithms/ techniques 1 / 2 days before the exam ? :redface:
Reply 659
Avatar for P!=NP
P!=NP
How many past papers has everyone done and how many more do they plan to do? Anybody done every paper after 2009? If so, which one was the hardest?

Edit: Apart from maths247, any other last minute revision sources?
(edited 10 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you