Xsanda’s solutions:

1) Matching
i) JF=3-ND=4
1:MA
2:-
3:JF
4:ND
5:JA
6:MJ
ii) SO=5-JA=2
1:MA 2:JA 3:JF 4:ND 5: SO 6:MJ
iii) 3=JF,4=ND,5=SO always
1:MJ 2:MA 3:JF 4:ND 5: SO 6:JA

2) Flow
i) FT = 2, DT = 4, cut = 9
iia) SCBFT +2 (given)
b) SCFBAEDT +1
iii) 8, with cut {SABCF}{DET} of value 8

3) Hungarian
i) Add dummy column X with 100 in each
Reduce rows -86, -92, -88, -98
Reduce cols -0,-2,-0,-2
ii) AS, BT, CL, no match for D, £271 cost
iii) CS, BT; AL costs £83

4) Rock paper scissors
i) scissors with paper: gain of 1, loss of 2: 1 + (-2) ≠ 0
ii) playsafe: scissors
iii) scissors 2-2p-4q, paper 3p+2q-2, rock 2q-p
iv) p=0.5, q=0.25, rock = 0.25

5) critical path analysis
i) F and H also critical
ii) A: -, B: -, C: -, D: A, E: BD, F: BCD, G: E, H: EF, I: GH, J: H
iii) 60 mins, critical ABDEHJ
iv) allocation for 3 people
v) allocation for 2 people
vi) A is critical, but so is B, which happens at the same time, so must reduce both to gain 5 minutes. 200+400=£600. Best is H and J: £500+£500 = £1000

6i) (0;0)-(1;1)-(2;1)-(3;0)-(4;0): weight 7
ii) 1 point
iii) (0;0)-(1;0) 1
(0;0)-(1;1) 2
(1;0)-(2;0) 1
(1;0)-(2;1) 1
(1;1)-(2;0) 2
(1;1)-(2;1) 1
(2;0)-(3;0) 0
(2;0)-(3;1) 0
(2;1)-(3;0) 2
(2;1)-(3;1) 0
(3;0)-(4;0) 1
(3;1)-(4;0) 2
iv) (0;0)-(1;0)-(2;0)-(3;0)-(4;0): 1 point per round maximum

(edited 7 years ago)

Reply 28

Sonnyjimisgod

Original post by marioman

Did anyone on here manage to augment the flow the second time?

Yep man, was just a standard hungarian algo. right?

Reply 29

Original.Name

I'm glad other people think that was a mistake too... I was ****ting myself 😂

Reply 30

Fp3

what mark to people reckon for grade boundaries ?

Reply 31

drandy76

Original post by Fp3

what mark to people reckon for grade boundaries ?

Usually high 50 low 60 for an A and between 6-8 between each grade

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Reply 32

bassala

Original post by marioman

Did anyone on here manage to augment the flow the second time?

That was a mean one but I found a route which you could augment by one, it involved going backwards on one of the arcs - I can't rmemeber which one, but basically, it meant a maximum flow of 8

Reply 33

JoshuaHope

I don't know I think grade boundaries are standard,, the only tricky question was number 6 which was due to the weirdness of it
62 for an A probably

Reply 34

bassala

Original post by xsanda

Xsanda’s solutions:

1) Matching
i) JF=3-ND=4
1:MA
2:-
3:JF
4:ND
5:JA
6:MJ
ii) SO=5-JA=2
1:MA 2:JA 3:JF 4:ND 5: SO 6:MJ
iii) 3=JF,4=ND,5=SO always
1:MJ 2:MA 3:JF 4:ND 5: SO 6:JA

2) Flow
i) FT = 2, DT = 4, cut = 9
iia) SCBFT +2 (given)
b) SCFBAEDT +1
iii) 8, with cut {SABCF}{DET} of value 8

3) Hungarian
i) Add dummy column X with 100 in each
Reduce rows -86, -91, -88, -98
Reduce cols -0,-2,-0,-2
ii) AS, BT, CL, no match for D, £271 cost
iii) CS, BT; AL costs £83

4) Rock paper scissors
i) scissors with paper: gain of 1, loss of 2: 1 + (-2) ≠ 0
ii) playsafe: scissors
iii) scissors 2-2p-4q, paper 3p+2q-2, rock 2q-p
iv) p=0.5, q=0.25, rock = 0.25

5) critical path analysis
i) F and H also critical
ii) A: -, B: -, C: -, D: A, E: BD, F: BCD, G: E, H: EF, I: GH, J: H
iii) 60 mins, critical ABDEHJ
iv) allocation for 3 people
v) allocation for 2 people
vi) A is critical, but so is B, which happens at the same time, so must reduce both to gain 5 minutes. 200+400=£600. Best is H and J: £500+£500 = £1000

6i) (0;0)-(1;1)-(2;1)-(3;0)-(4;0): weight 7
ii) 1 point
iii) (0;0)-(1;0) 1
(0;0)-(1;1) 2
(1;0)-(2;0) 1
(1;0)-(2;1) 1
(1;1)-(2;0) 2
(1;1)-(2;1) 1
(2;0)-(3;0) 0
(2;0)-(3;1) 0
(2;1)-(3;0) 2
(2;1)-(3;1) 0
(3;0)-(4;0) 1
(3;1)-(4;0) 2
iv) (0;0)-(1;0)-(2;0)-(3;0)-(4;0): 1 point per round maximum

Thats what I got :smile:

My explanation for the cost being >200 for reducing activities A and B may have been a bit kerfuffled but other than that, I'm fairly pleased :smile:

Reply 35

JoshuaHope

Original post by bassala

Thats what I got :smile:

My explanation for the cost being >200 for reducing activities A and B may have been a bit kerfuffled but other than that, I'm fairly pleased :smile:

Yeah I am the same, that part was only worth 2 marks though right?

Reply 36

bassala

Original post by JoshuaHope

Yeah I am the same, that part was only worth 2 marks though right?

The explanation part would be one mark and the part where you identify H and J is another mark so yeah :smile:

Reply 37

sarahemilyglass

Found it all fine other than that flow augmentation, I had to say to ignore the one they had added in ii.a , and added two flow augmentation routes (one of flow 2, one of flow 1) to show a flow of 8 through