Decision 1 Worked Solutions June 2005

1
The Further Mathematics Network
Worked Solutions to MEI Decision Maths 1
June 2005
2
1
C
D
F
(i)
E
B
A
Total number of connections 6 ? 2 12
Mains
Any connected tree will do
3
1
C
D
F
(ii)
E
B
A
Total number of connections 12 2 14
Mains
Junction box creates two extra connections
He might be able to save cable. He might avoid
overloading.
(iii)
4
1
C
D
F
(iv)
E
B
A
Mains
Minimum spanning tree will have 7 1 6
arcs Each arc results in 2 connections Total
number of connections 6 ? 2 12, as before
5
2
(i)
Janet
John
(ii)
Yes - Janets route traces south and west
walls attachments
Johns route traces north and east walls
attachments
6
2
(iii)
Janet
John
X
X
(iv)
Yes All avenues covered by forward and backward
passes
i.e. Johns original route Janets route
7
3
(i)
8
3
3
(ii)
2
0
5
Forward pass finding earliest event times
Minimum completion time 5 hours
9
3
3
3
(ii)
2
4
0
5
5
0
Minimum completion time 5 hours
Backward pass finding latest event times
10
3
3
3
(ii)
3
4
0
5
5
0
Critical activities
A 3 0 3
C 5 0 5
D 5 3 2
11
3
3
3
(iii)
3
4
0
5
5
0
B 4 0 2 2
Total Float
E 5 3 1 1
B 3 0 2 1
Independent Float
E 5 4 1 0
12
49
8
49
4
31
6
31
(i)
45
7
45
25
5
26
25
12
2
12
14
3
15
4
14
15
0
1
Perm label
Order
SHORTEST PATH
Temp labels
13
49
8
49
4
31
6
31
(i)
45
7
45
25
5
26 25
12
2
12
14
3
15
4
14
15
0
1
Shortest Distances from C
SHORTEST PATH
14
49
8
49
4
31
6
31
(i)
45
7
45
25
5
26 25
12
2
12
14
3
15
4
14
15
0
1
Shortest Route from P to C
P ? T
? S
? C
SHORTEST PATH
15
49
8
49
4
31
6
31
(i)
45
7
45
25
5
26 25
12
2
12
14
3
15
4
14
15
0
1
Shortest Route from V to C
V ? U
? S
? C
SHORTEST PATH
16
MINIMUM CONNECTOR
4
(ii)
Total length 8
17
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 16
18
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 10 26
19
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 10 10 36
20
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 10 10 12 48
21
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 10 10 12 14 62
22
MINIMUM CONNECTOR
4
(ii)
Total length 8 8 10 10 12 14 18
80
23
MINIMUM CONNECTOR
Bridge between P and Q
4
(iv)
12
Total length 8 8 10 10 12 14 18
80
24
MINIMUM CONNECTOR
Bridge between P and Q
4
(iv)
12
Total length 8 8 10 10 12 14 18
80
Total length 8 8 10 10 12 14 12
74
25
5
Laptop Loan Requests
(i)
Random numbers
Simulated Laptop Loan Requests
(ii)
1
0
1
0
0
2
3
3
3
4
26
5
Laptop Loan Returns
(iii)
Random numbers
Simulated Laptop Loan Returns
(iv)
1
0
1
2
2
1
0
1
3
3
27
5
Laptop Returns
Initially 7 laptops out on loan, 3 laptops in
stock
(v)
28
5
Laptop Loan Returns if less than 3 in stock
(vi)
Random numbers
Simulated Laptop Loan Returns
29
5
Laptop Returns
Initially 7 laptops out on loan, 3 laptops in
stock
(vi)
Only 1 disappointed under new policy, against 4
under old policy
30
6
(i)
Let f represent the number of litres of
Flowerbase produced
Let g represent the number of litres of Growmuch
produced
31
6
(i)
Fibre constraint 0.75f 0.5g ? 12000
? 3f 2g ? 48000
Nutrient mix constraint f 2g ? 25000
Objective function Maximise 9f 20g
pence or (0.09f 0.2g)
32
g
6
(ii)
Evaluate profit at vertices of feasible region
3f 2g 48000
A
B
f 2g 25000
C
f
Max profit of 2500 by producing 12500 litres of
Growmuch
33
g
6
(iii)
(A)
Evaluate profit at vertices of feasible region
3f 2g 40000
A
B
f 2g 25000
C
f
Max profit of 2500 by producing 12500 litres of
Growmuch
34
g
6
(ii)
(B)
If price of fibre is increased, profit on
Flowerbase will be reduced by more than that on
Growmuch since it uses more fibre. Even less
attractive to produce any Flowerbase, so no
change.
3f 2g 48000
A
B
f 2g 25000
C
f
Max profit of 2500 by producing 12500 litres of
Growmuch
35
g
6
(iii)
(C)
Evaluate profit at vertices of feasible region
3f 2g 48000
A
B
f 2g 30000
C
f
Max profit of 3000 by producing 15000 litres of
Growmuch