Decision Mathematics D2 (6690)
Practice paper B mark scheme
Question
number
(a)
1.
Scheme
Marks
85
B1 (1)
C1 = 140
C2 = 104
B1 B1 (2)
For example,
S B D F H J T
–4
S B D F G T
–1
S B D F C H I T
–2
A1
S B D F C H J T
–2
A1
S B D E G T
– 10
M1 A1
A1 (5)
Max flow – min cut theorem, flow is 104, mincut is C2
M1 A1 (2)
(10 marks)
2.
(a)
7x + 10y + 10z + r = 3600
6x + 9y + 12z + s = 3600
2x + 3y + 4z + t = 2400
P – 35x – 55y – 60z = 0
(b)
bv
x
y
z
r
s
t
value
Row ops
r
2
5
2
0
1
– 56
0
600 R1 – 10R2
z
1
2
3
4
1
0
1
12
0
300 R2  12
T
0
0
0
0
– 13
1
1200 R3 – 4R2
P
–5
–10
0
0
5
0
18000 R4 + 60R2
bv
x
y
z
r
s
t
value
r
4
5
1
0
2
5
– 13
0
240 R1 
z
– 101
0
1
– 103
1
3
0
120 R2 –
T
0
0
0
0
– 13
1
1200 R3 stet
P
3
0
0
4
5
3
0
Row ops
5
2
3
4
R1
20400 R4 + 10R1
Question
number
(c) P = 20400
Scheme
Marks
x=0
y = 240
z = 120
r=0
s=0
t = 1200
M1
A2ft, 1, 0
(3)
(16 marks)
3.
(a) Adds 32 to AB + BA (ACB)
B1
47 to AE + EA (ACDE)
B1
32 to CE + EC
(CDE)
B1
53 to DG + GD (DCG)
B1 (4)
(b) A C B D E
F G
A
M1 A1
15 + 17 + 38 + 11 + 31 + 30 + 23 = 165 miles
A1 (3)
(c) For example, BC, CD, DE, EF, FG
B
C
17
D
21
E
11
F
31
G
30
Weight of RSMT = 110 miles
A1
Lower band = 110 + 15 + 23
M1
= 148 miles
A1 ft (4)
(11 marks)
4.
Let xij be the number of units transported from i to j, in
1000 litres where i  {F, G, H} and j  {S, T, U }.
B2, 1, 0 (2)
Minimise C = 23xfs + 31xft + 46xfu +
35xgs + 38xgt + 51xgu +
41xhs + 50xht + 63xhu
Subject to xfs + xft + xfu
B1
unbalanced
B1 (2)
 540
xgs + xgt + xgu  789
M1
xhs + xht + xhu  673
xfs + xgs + xhs
 257
xft + xgt + xht
 348
A1
xfu + xgu + xhu  412
A1 (3)
xij  0
B1 (1)
Accept introduction of a dummy demand methods.
(8 marks)
Question
number
(a)
5.
Scheme
Marks
Row 1 dominates row 2, so A will never choose R2
Column 1 dominates column 3, so B will never choose C3
M1 A1
Thus row 2 and column 3 may be deleted.
A1 (3)
(b) Let A play row 3 with probability p and hence row 3 with
probability (1 – p)
If B plays 1, A’s expected gain 3p + 6(1 – p) = 6 – 3p
If B plays 2, A’s expected gain is 5p + 3(1 – p) = 2p + 3
M1 A1
Optimal when 6-3p = 2p+3
5p = 3
p=
3
5
A1
Hence A should play row 1 with probability
3
5
and row 2 with probability
2
5
A1 ft (4)
Similarly, let B play column 1 with probability q
3q + 5(1 – q) = 6q + 3(1 – q)
M1 A1
5q = 2
q=
2
5
A1 ft
So B should play column 1 with probability
2
5
and column 2 with probability
3
5
Value of game is 4 15 to A
B1 (4)
(11 marks)
Question
number
6.
Scheme
Marks
Let xij = 1 if worker does the task, xij = 0 otherwise, where xij
indicates the arc from node i to node j, i  P,Q,R and j  1,2,3
xP1 + xP2 + xP3 = 1
xQ1 + xQ2 + xQ3 = 1
and
xR1 + xR2 + xR3 = 1
B1 B1 (2)
xP1 + xQ1 + xR1 = 1
M1
xP1 + xQ2 + xR2 = 1
A1
xP1 + xQ2 + xP3 = 1
A1 (3)
minimising,
C = 8  P1 + 7  P2 + 3  P3 + 9  P1 + 5  P2 + 6  P3 +
10  P1 + 4  P2 + 4  P3, where C is in hundreds of pounds.
7.
Stage
1
(Sept)
2
(Aug)
3
(July)
4
(June)
5
(May)
State Action Dest
2
1
0
2
Value
0
2
3
4
5
4
3
5
4
5
0
0
0
2
1
0
1
0
0
200+200 = 400
200+100 = 300
200
= 200
200+200+500+400 = 1300
200+200
+300 = 700
200+200
+200 = 600
200+100
+300 = 1100
200+100
+200 = 500
200
+500+200 = 900
2
5
0
200+200+500+900 = 1800
2
1
0
0
3
4
5
5
4
3
2
2
2
2
1
0
200+200
+1800 = 2200
200+100
+1800 = 2100
200
+500+1800 = 2500
200
+500+2200 = 2900
200
+2100 = 2300
200
+2500 = 2700
1
month
production schedule
Cost: £2300
May June July Aug Sept
4
B1 B1 (2)
(7 marks)
4
5
5
4
M1 A1ft
A1ft (3)
(12 marks)