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*Rough*DECISION 1 EDEXCEL JUNE 17TH MARKSCHEME w/Explained Answers

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Original post by Nikhilm
I'm pretty sure you could've done it in 4 bins not 5?


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I ended up with 4 bins
Original post by FugFig
Nah i am pretty sure now its 65. 75 was the second smallest pair, and for the 65 the path was tricky, so i think the others just didnt see it.(at least i hope so)


Yeah the path for 75 was ridiculous, pretty sure I ****ed up that whole question
Also, if I got the objective function wrong (which is obviously -1 mark), are there ecf marks for the next question?
Original post by ediosaurus
In some papers you get penalised 1 mark


My teacher told me that u won't but idk tbh
I've just read a past mark scheme and it says 'condone lack of numbers', just as long as you don't put activities in the nodes...
How many marks were questions 2, 7 and 8?
Original post by Dannyfire
How many marks were questions 2, 7 and 8?


5 11 and 14
Original post by TrueDAN
Getting on this one as fast as possible. I hated this exam, everyone else found it good - maybe decision maths isn't for me but it'll be nice to see how your answers compare.

Question 1:
- A bipartite graph consists of 2 vertices, X and Y. Vertices can only be joined from X to Y, not joined from the same set I.e. X to X.
- It was the most confusing / long winded route I got it as. You were matching P with B.
P-A = ..... M - B and change state.

Question 2:
- Precedence table was with 3 dummies. I did get a nice looking one but I know I made one mistake. C and D were dependent on A. E and F were dependent on B. A and B were dependent on nothing and came from a node labelled O.
Remember to link the all up at the end to create a nice network.

Question 3:
- Quicksort with 10 numbers. I can remember 11 being shifted to the very end after the first pass. Make sure to label pivots clearly and to say that the source is complete.
- First fit decreasing got you 4 bins.
- Lowerbound was 4, the question was asking if it was optimal. Therefore, it was optimal.

Question 4:
- Ouput 135 implying 7 and 25 were multiplied together.
- 122 because using 1/2 would encounter negatives.
- 122 X 1/2 = 61.

Question 5:
Worried about this one - can't remember it well.
- 48 units in length.
- One with J 49 units in length.
- Missed out Prim's question and only noticed at end so rushed like mad - I got 87.
Answer is 82 I believe.
Question 6: Chinese postman
- Weight 459 but I put 460 :frown:
- Look at one excluding F which was EH and 35 in length.
- 419 with a route with 21 nodes, starting at F and finishing at B.

Question 7:
- for answers of w,X,y,z (not in order) 4,11,17 and 22?
- Gantt chart - hate you - you had 4 critical values. I got 6.
- 5 workers as lower bound looking at day 12-14
- Scheduling = ran out of time / didn't bother really.

Question 8:
- 20x + 65y less than or equal to 520.
- Remember that X.y larger than equal to 0. Plot the points to get a region R - one of the equations I struggled with and I think it was something like:
-X + 24y larger than or equal to 24. I brought the X over and got 0,0 and 0,1 as my points. I don't think it interfered with the region - I may have got it wrong please tell me.
- Exact values you had to look at the point of intersection between 7x + 8y = 120 and 20x + 65y = 520.
I did simultaneous equations and got fractions - I checked there decimal points and it did look close to where V was. I got V to be the upper, right hand side point.

9 and 4 I got for the crates
People got 9 and 5
I ink the correct is 11 and 4.
I think I've done horribly on this paper, please help me add on to it to create a nice markscheme! I really wanted an A overall, think I may have not done enough in this module. Anywho, I will update ASAP and try my best to respond quickly. Good luck all :smile:



For Algorithm one, some people say because 1/2 for X will give a never ending result as u'd keep halving the X value ( giving: -/2 -/4 -/8 -/16 ....)

I agree with it even though I didn't get in the exam
Original post by lucabrasi98
They do lol. It says so on the front page

Only an idiot draws diagrams in pen anyway


Umm... I guess I was one ( with the amount of TIME we had to swap pens)
Reply 109
last question: 9x and 5y giving maximum capacity given the budget of 520 as it cost a total of 505
Reply 110
What do you think the grade boundaries will be like? I totally screwed up this paper and I need an A overall, I guess I'm just going to have to ace C3 and C4 lol
For part 2 of the flowchart was the reason not because a fraction can be neither odd or even?
Original post by sadasad
last question: 9x and 5y giving maximum capacity given the budget of 520 as it cost a total of 505



that doesn't fit the constraint of 7x+8y<112
Reply 113
Prims was 69 or 70?
Reply 114
Original post by brokeins1
For Algorithm one, some people say because 1/2 for X will give a never ending result as u'd keep halving the X value ( giving: -/2 -/4 -/8 -/16 ....)

I agree with it even though I didn't get in the exam


Oh yes I suppose - grrr another 2 marks I may have dropped. But wasn't the input of the algorithm:
Is X odd
If yes, X = X-1
So inputting 1/2 would take you to that procedure which encounters negatives? Cannot remember it well enough
Reply 115
Original post by s0phi33
What do you think the grade boundaries will be like? I totally screwed up this paper and I need an A overall, I guess I'm just going to have to ace C3 and C4 lol


I wanted high 60s in that test and have came out with high 40s :frown: Certainly lower than normal, do not think a lot of people will have got over 60 in it due to the time constraints. I do not want to speculate, but I could not see an A being over 60 at all. Maybe even 55, who knows - just smash C3 and C4 and do not worry about D1 :biggrin: It was not a test of how good you are at maths at all - really disappointing exam.
Doesnt anyone have a copy of the paper?!!!!!
Original post by TrueDAN
Oh yes I suppose - grrr another 2 marks I may have dropped. But wasn't the input of the algorithm:
Is X odd
If yes, X = X-1
So inputting 1/2 would take you to that procedure which encounters negatives? Cannot remember it well enough


I guess you have to realate it to the diffenition of an algorithm being finite, and so wouldn't be in thiscase where you have 1/2 =x. OR with this value u'd get a wrong output ( a negative answer, whereas xy= +61

Btw was it not 1 mark?????
So no one had a copy of the paper then?

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