OCR (NOT MEI) - D1 Exam - January 2012

I got something like 998 for the upper bound, and looking at the previous post that's wrong. poop.

Did people x=6 for the simplex?

Part a my outputs were 8, 2, 1 and 0
Part b algorithm never terminated

yep got x =6 , y=0 and z = 0

I got something like 998 for the upper bound, and looking at the previous post that's wrong. poop.

Did people x=6 for the simplex?

I did this quickly after the exam, but I haven't been through them to check them, but my answers are below. I might have some of the numbering wrong, and by the sounds of it, I've got one wrong, as I got 922 not 967. I normally get some basic arithmetic wrong on D1, so take these with a pinch of salt. Let me know what to correct.

1. Shuttle sort with 8 passes, 26 comparisons and 21 swaps.
2. i) 5 arcs – e.g. all outer arcs except one
ii) 15 arcs – complete graph - all pairings connected exactly once
iii) All vertices must have an even order, so we need to reduce each vertex order by 1. If we pair the 6 vertices into 3 pairs, we can remove the arc joining each pair and all vertices will be of order 4. So 12 arcs.
iv) It is semi-Eulerian as there are exactly two nodes of odd order: F and D. E.g. FABECAD
3. i) ACBDF weight 13
ii) 38+6=44
iii) Duplicate AC and DF 38+7=45
4. i) x represents the number of red bags
ii) 3x+7y+6z represents the total number of sweets need, where x,y,z represent the number of red, yellow and blue bags respectively
$5x+4y+6z\leq40$
$5x+2y+3z\leq30$
$x,y,z\geq0$ and x,y,z are integers
iv) Assume that she has enough party bags of each colour.
v)
P x y Z s t u RHS
1 -1 -1 -1 0 0 0 0
0 3 7 6 1 0 0 80
0 5 4 6 0 1 0 40
0 5 2 3 0 0 1 30
vi) Ratios: sweets 26.7, balloons 8, toys 6
P x y z s t u RHS
1 0 -0.6 -0.4 0 0 0.2 6
0 0 5.8 4.2 1 0 -0.6 62
0 0 2 3 0 1 -1 10
0 1 0.4 0.6 0 0 0.2 6
x=6, y=0, z=0, so 6 red bags and no others
vii) 10 yellow bags
5. i) completed network
ii) total weight 563
iii) 563+50+59=672
iv) 970
v) FAD = 435, CBE = 278
total weight 922, route such as FADEBCF
6. i)
A B C D E F Output
10 128 12.8 12 120 8 8
10 12 1.2 1 10 2 2
10 1 0.1 0 0 1 1
10 0
ii)
A B C D E F Output
10 -13 -1.3 -2 -20 7 7
10 -2 -0.2 -1 -10 8 8
10 -1 -0.1 -1 -10 9 9
10 -1 -0.1 -1 -10 9 9
The algorithm will continue to repeat.

For the assumption, I think it's we assume she sells them all. But your way works too :P

I did this quickly after the exam, but I haven't been through them to check them, but my answers are below. I might have some of the numbering wrong, and by the sounds of it, I've got one wrong, as I got 922 not 967. I normally get some basic arithmetic wrong on D1, so take these with a pinch of salt. Let me know what to correct.

1. Shuttle sort with 8 passes, 26 comparisons and 21 swaps.
2. i) 5 arcs – e.g. all outer arcs except one
ii) 15 arcs – complete graph - all pairings connected exactly once
iii) All vertices must have an even order, so we need to reduce each vertex order by 1. If we pair the 6 vertices into 3 pairs, we can remove the arc joining each pair and all vertices will be of order 4. So 12 arcs.
iv) It is semi-Eulerian as there are exactly two nodes of odd order: F and D. E.g. FABECAD
3. i) ACBDF weight 13
ii) 38+6=44
iii) Duplicate AC and DF 38+7=45
4. i) x represents the number of red bags
ii) 3x+7y+6z represents the total number of sweets need, where x,y,z represent the number of red, yellow and blue bags respectively
$5x+4y+6z\leq40$
$5x+2y+3z\leq30$
$x,y,z\geq0$ and x,y,z are integers
iv) Assume that she has enough party bags of each colour.
v)
P x y Z s t u RHS
1 -1 -1 -1 0 0 0 0
0 3 7 6 1 0 0 80
0 5 4 6 0 1 0 40
0 5 2 3 0 0 1 30
vi) Ratios: sweets 26.7, balloons 8, toys 6
P x y z s t u RHS
1 0 -0.6 -0.4 0 0 0.2 6
0 0 5.8 4.2 1 0 -0.6 62
0 0 2 3 0 1 -1 10
0 1 0.4 0.6 0 0 0.2 6
x=6, y=0, z=0, so 6 red bags and no others
vii) 10 yellow bags
5. i) completed network
ii) total weight 563
iii) 563+50+59=672
iv) 970
v) FAD = 435, CBE = 278
total weight 922, route such as FADEBCF
6. i)
A B C D E F Output
10 128 12.8 12 120 8 8
10 12 1.2 1 10 2 2
10 1 0.1 0 0 1 1
10 0
ii)
A B C D E F Output
10 -13 -1.3 -2 -20 7 7
10 -2 -0.2 -1 -10 8 8
10 -1 -0.1 -1 -10 9 9
10 -1 -0.1 -1 -10 9 9
The algorithm will continue to repeat.

I did this quickly after the exam, but I haven't been through them to check them, but my answers are below. I might have some of the numbering wrong, and by the sounds of it, I've got one wrong, as I got 922 not 967. I normally get some basic arithmetic wrong on D1, so take these with a pinch of salt. Let me know what to correct.

1. Shuttle sort with 8 passes, 26 comparisons and 21 swaps.
2. i) 5 arcs – e.g. all outer arcs except one
ii) 15 arcs – complete graph - all pairings connected exactly once
iii) All vertices must have an even order, so we need to reduce each vertex order by 1. If we pair the 6 vertices into 3 pairs, we can remove the arc joining each pair and all vertices will be of order 4. So 12 arcs.
iv) It is semi-Eulerian as there are exactly two nodes of odd order: F and D. E.g. FABECAD
3. i) ACBDF weight 13
ii) 38+6=44
iii) Duplicate AC and DF 38+7=45
4. i) x represents the number of red bags
ii) 3x+7y+6z represents the total number of sweets need, where x,y,z represent the number of red, yellow and blue bags respectively
$5x+4y+6z\leq40$
$5x+2y+3z\leq30$
$x,y,z\geq0$ and x,y,z are integers
iv) Assume that she has enough party bags of each colour.
v)
P x y Z s t u RHS
1 -1 -1 -1 0 0 0 0
0 3 7 6 1 0 0 80
0 5 4 6 0 1 0 40
0 5 2 3 0 0 1 30
vi) Ratios: sweets 26.7, balloons 8, toys 6
P x y z s t u RHS
1 0 -0.6 -0.4 0 0 0.2 6
0 0 5.8 4.2 1 0 -0.6 62
0 0 2 3 0 1 -1 10
0 1 0.4 0.6 0 0 0.2 6
x=6, y=0, z=0, so 6 red bags and no others
vii) 10 yellow bags
5. i) completed network
ii) total weight 563
iii) 563+50+59=672
iv) 970
v) FAD = 435, CBE = 278
total weight 922, route such as FADEBCF
6. i)
A B C D E F Output
10 128 12.8 12 120 8 8
10 12 1.2 1 10 2 2
10 1 0.1 0 0 1 1
10 0
ii)
A B C D E F Output
10 -13 -1.3 -2 -20 7 7
10 -2 -0.2 -1 -10 8 8
10 -1 -0.1 -1 -10 9 9
10 -1 -0.1 -1 -10 9 9
The algorithm will continue to repeat.

Ah, that's a much better answer. I didn't like my one at all. Thanks.

I put she makes a profit of some kind? Will i get the mark or nah?

I can only see a few things that look different to what I had but I may have remembered some stuff wrong.
In question 1 I seem to remember 36 comparisons, not 26
Also for 5v I also got 967
And for the assumption I put that they are all sold but there was probably other answers for that question.

Cant see any other answers where I didnt get the same but not sure.

I can only see a few things that look different to what I had but I may have remembered some stuff wrong.
In question 1 I seem to remember 36 comparisons, not 26
Also for 5v I also got 967
And for the assumption I put that they are all sold but there was probably other answers for that question.

Cant see any other answers where I didnt get the same but not sure.

I got 36 comparisons too, but I'm absolutely rubbish at counting comparisons and swaps so I could easily be wrong...

A couple of things with this.

2ii) All of the 6 nodes cannot have an order of four. Someone showed me this and I said damn. I got this wrong too.

4iv) Apparantly this is she sells all the bags she makes?

4vii) I thought it was 10 blue bags? The z column can be changed if you use the solution from the first iteration? Are there two answers to this too? Or am I just wrong. LOL

5v) What two arcs did you use to join these two? As I got BD and BF. Which was 104 or something near it and 50? These were the two shortest I thought? And it asked for a closed route for the nearest neighbour?

I would have thought that with a shuttle sort and 9 items it would be 1+2+3+4+5+6+7+8=36?

am I doing something wrong?

Yeahh this is not always true because if you're sorting say the first 4 numbers for example, you know that from the previous passes that the first 3 numbers are already in order and so if you compare the 4th number with the 3rd and it does not swap then no more comparisons need to be made for this pass hope this helps and it's not too confusing