Mark Scheme 4773
June 2005
Qu. 1
(i)
32*80 = 2560 calories
(ii)
3000/32 = 93.75 kg
(iii) Auxiliary equation is (3x−1)(3x−2) = 0
Solution is un = 13.75(1/3)n − 27.5(2/3)n + 93.75
(iv)
90
90
85.83333
81.66667
78.65741
76.80556
75.78961
75.28807
75.06873
74.98871
74.96962
74.97276
74.98119
(Oscillatory) convergence to 75 kg.
(v)
90
90
82.77778
75.55556
70.33951
67.12963
65.36866
64.49931
64.11913
63.98043
63.94734
63.95278
63.9674
63.98052
M1 A1
M1 A1
M1
M1
A1
M1
A1
B1
A1
particular
93.75 or 3rd eqn
gen homogeneous
correct form
case 1 (u0 = 80)
+ case 2 (u1 = 80)
M1 simultaneous
A1 13.75 and −27.5
B1 final answer
M1
A1
B1
B1
Qu. 2
(i)
E
A
2
2
C
3
2
1 F
2
2
G 2
S
1
1
5
1
1
B
3
T
1
H
D
M1
A1
1
I
(ii)
1
1
B
0
2
B1 1 and 1
B1 0
B1 2
C
(iii) S B D A C G (T)
M1 A1
E
(iv)
A
2
C
3
2
1
1
S
G 1
1
4
3
D
1
1
1
B
2
1 F
H
T
M1
A1
1
1
I
(v)
S B⏐A C D E F G H I T
or S A B D⏐C E F G H I T
M1 a cut, properly
specified
A1 correct cut
(vi)
Max SA + SB + SD
st
SA+CA+DA-AD-AC = 0
SB+CB+DB-BC-BD = 0
AC+BC-CA-CB-CE-CF-CG = 0
SD+BD+AD-DA-DB-DH-DI = 0
SA < 2
SB < 5
SD < 1
AD < 2
DA < 2
BC < 1
CB < 1
AC < 3
BD < 3
CA < 3
CE < 2
CF < 1
CG < 2
DB < 3
DH < 1
DI < 1
end
B1
M1
A1
B1
(vii)
OBJECTIVE FUNCTION VALUE
1)
6.00000
VARIABLE
SA
SB
SD
CA
DA
AD
AC
CB
DB
BC
BD
CE
CF
CG
DH
DI
VALUE
2.000000
3.000000
1.000000
0.000000
1.000000
0.000000
3.000000
0.000000
0.000000
1.000000
2.000000
2.000000
0.000000
2.000000
1.000000
1.000000
REDUCED COST
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
M1
A1
B1
Flows are as listed in the "VALUE" column.
Qu. 3
(i)
Simulating service times (=lookup(rand(),cum.probs,times))
Accumulating (expectation is 207.5 seconds)
B1
B1
(ii)
Repetitions
Mean (not far off 207.5 seconds)
sd (order of magnitude 5 seconds)
(2*1.96*s)2 = (about) 400 repetitions
(assuming a 95% confidence interval half-width of 0.5s)
B1
B1
B1
M1
A1
(iii) Rand()*120
fixed
sorted
M1
A1
B1
(iv) max(arrival time, gate available time)
+ service time
finish time approx as in (i)
B1
B1
B1
(v)
Test barrier free times to see which barrier passenger uses.
Computation of barrier free times, eg:
=if(bar=1, max(arrival t + service t, bar t + service t), bar t)
M1 A1
M1
A1
finish time approx 130s
B1
Qu. 4
(i)
Sched. City
S1
L
S2
L
S3
L
S4
L
S5
B
S6
B
S7
B
S8
B
S9
B
S10
B
S11
M
S12
M
S13
M
S14
P
S15
P
S16
P
S17
P
S18
P
S19
P
S20
P
Flight City
101
B
101
B
101
B
101
B
201
P
201
P
201
P
202
M
202
M
202
M
301
P
302
L
303
B
401
B
401
B
401
B
401
B
402
M
402
M
403
L
Flight City
201
P
201
P
202
M
202
M
402
M
402
M
403
L
301
P
302
L
303
B
403
L
102
B
204
M
201
P
202
M
203
P
202
M
302
L
303
B
102
B
Flight City
402
M
403
L
302
L
301
P
302
L
303
B
102
B
403
L
102
B
102
204
B
M
301
P
303
102
203
203
B
B
P
P
Flight City
302
L
403
102
L
B
102
B
204
M
203
203
P
P
M1
A1
A1
A1
A1
London
Berlin
Milan
Paris
(ii)
Min S1+S2+S3+S4+S5+S6+S7+S8+S9+S10+S11+S12
+S13+S14+S15+S16+S17+S18+S19+S20
st
S1+S2+S3+S4>1
S5+S7+S8+S9+S11+S12+S18+S20>1
S1+S2+S5+S6+S7+S14>1
S3+S4+S8+S9+S10+S15+S17>1
S16+S17+S18+S19+S20>1
S11+S12+S13>1
S4+S8+S11+S15>1
S1+S3+S5+S9+S12+S18>1
S6+S10+S13+S17+S19>1
S14+S15+S16+S17>1
S1+S5+S6+S18+S19>1
S2+S4+S7+S8+S11+S20>1
M1 A1 objective
M1
A3 (−1 each
error/omission)
(iii)
OBJECTIVE FUNCTION VALUE
1)
3.000000
VARIABLE
S1
S2
S3
S4
S5
S6
S7
S8
S9
S10
S11
S12
S13
S14
S15
S16
S17
S18
S19
S20
VALUE
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
REDUCED COST
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
1.000000
0.000000
1.000000
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
3 pilots are used
M1
A1
B1
(iv) Three more runs, with S1=0, S11=0 and S17=0 in turn. All
require 4 pilots
M1 A1 (3 runs)
A1
(4 pilots)
(v)
B1
No account taken of pilot stress
(workload/long day/short changeover)
Mark Scheme 4773
June 2006
4773
Mark Scheme
June 2006
Qu. 1
(i)
(ii)
Variables
ai = amount invested in A in year i, i = 1, 2, 3, 4, 5
bi = amount invested in B in year i, i = 1, 2, 3
ci = amount invested in C in year i, i = 3, 4, 5
M1
A1
A1
A1
Maximise 1.15a5+1.55b3+1.20c5
st
a1+b1 = 50000
a2+b2 = 1.15a1
a3+b3+c3 = 1.15a2
a4+c4 = 1.15a3+1.55b1+1.20c3
a5+c5 = 1.15a4+1.55b2+1.20c4
B1
B1
B1
B1
B1
B1
OBJECTIVE FUNCTION VALUE
1)
114264.0
VARIABLE
VALUE
A5
0.000000
B3
0.000000
C5
95220.000000
A1
50000.000000
B1
0.000000
A2
57500.000000
B2
0.000000
A3
0.000000
C3
66125.000000
A4
0.000000
C4
79350.000000
(iii)
a's
b's
c's
REDUCED COST
0.050000
0.178000
0.000000
0.000000
0.053280
0.000000
0.127200
0.072000
0.000000
0.060000
0.000000
M1
A1
Invest all in A in year 1. Put all into A in year 2
Thence all into C in years 3, 4 and 5.
Gives £114264 at the end of 5 years.
B1
£1.59
M1 A1 (£1.57 to
£1.61)
A1
B1
4773
Mark Scheme
June 2006
Qu. 2
(i)
See below – first two columns of s/sheet
M1 A1 A1
(ii)
x2 – x – 1 = 0
1± 5
x=
2
M1
A1
n
n
⎛ 1+ 5 ⎞
⎛ 1− 5 ⎞
x = A⎜
⎟ + B⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 1+ 5 ⎞
⎛ 1− 5 ⎞
A + B = 1 and A ⎜
⎟ + B⎜
⎟=1
⎝ 2 ⎠
⎝ 2 ⎠
1
giving u n =
5
n
5 + 1⎛ 1+ 5 ⎞
1
⎜
⎟ +
2 ⎝ 2 ⎠
5
1 ⎛ 1+ 5 ⎞
=
⎜
⎟
5⎝ 2 ⎠
n +1
B1
B1 B1
5 −1⎛ 1− 5 ⎞
⎜
⎟
2 ⎝ 2 ⎠
1 ⎛ 1− 5 ⎞
−
⎜
⎟
5⎝ 2 ⎠
n
M1
solving
A1 A1
n +1
B1
(iii) =(1/SQRT(5))*(((1+SQRT(5))/2)^(A2+1)-((1-
M1
A1
SQRT(5))/2)^(A2+1))
plus printout
M1 A1
(iv) See s/sheet below.
B1
Converges to 1.61803…
⎛ 1+ 5 ⎞
⎜
⎟
⎝ 2 ⎠
M1 A1
n F(n) Formula
Ratios
89
89 1.61818
1
11 144
144 1.61798
2
2
12 233
233 1.61806
3
3
1.5
13 377
377 1.61803
4
5
5 1.66667
14 610
610 1.61804
5
8
8
1.6
15 987
987 1.61803
6
13
13
1.625
16 1597 1597 1.61803
7
21
21 1.61538
17 2584 2584 1.61803
8
34
34 1.61905
18 4181 4181 1.61803
9
55
55 1.61765
19 6765 6765 1.61803
0
1
1
1
1
1
2
2
3
10
4773
Mark Scheme
June 2006
Qu. 3
(i)
Min 2W1S1+2W1S2+W1S3+5W1S4+3W2S1+2W2S2
+2W2S3+4W2S4+5W3S1+5W3S2+W3S3+2W3S
B1 variables
M1 objective
A1
4
st
W1S1+W1S2+W1S3+W1S4<20
W2S1+W2S2+W2S3+W2S4<20
W3S1+W3S2+W3S3+W3S4<20
W1S1+W2S1+W3S1>10
W1S2+W2S2+W3S2>15
W1S3+W2S3+W3S3>12
W1S4+W2S4+W3S4>20
M1 w/house
A1 availabilities
M1 shop
A1 requirements
(ii)
OBJECTIVE FUNCTION VALUE
1)
104.0000
VARIABLE VALUE
W1S1
8.000000
W1S2
0.000000
W1S3 12.000000
W1S4
0.000000
W2S1
2.000000
W2S2 15.000000
W2S3
0.000000
W2S4
0.000000
W3S1
0.000000
W3S2
0.000000
W3S3
0.000000
W3S4 20.000000
REDUCED COST
0.000000
1.000000
0.000000
3.000000
0.000000
0.000000
0.000000
1.000000
3.000000
4.000000
0.000000
0.000000
B1
M1
A1
Supply shop 1 with 8 from warehouse 1 and 2 from 2
Supply shop 2 from warehouse 2
Supply shop 3 from warehouse 1
Supply shop 4 from warehouse 3
Cost = £104
B1
4773
Mark Scheme
June 2006
Qu. 3 (cont)
(iii)
Min 2W1S1+2W1S2+W1S3+5W1S4+3W2S1+2W2S2
B1 new variables
+2W2S3+4W2S4+5W3S1+5W3S2+W3S3+2W3S
B1 new objective
4
st
+4S1C1+6S2C1+3S3C1+2S4C1
+S1C2+4S2C2+2S3C2+5S4C2
W1S1+W1S2+W1S3+W1S4<20
W2S1+W2S2+W2S3+W2S4<20
W3S1+W3S2+W3S3+W3S4<20
S1C1+S2C1+S3C1+S4C1=30
S1C2+S2C2+S3C2+S4C2=27
S1C1+S1C2-W1S1-W2S1-W3S1=0
S2C1+S2C2-W1S2-W2S2-W3S2=0
S3C1+S3C2-W1S3-W2S3-W3S3=0
S4C1+S4C2-W1S4-W2S4-W3S4=0
A solution is:
W1 to S3 20
W2 to S3 17
W3 to S4 20
S3 to C1 10
S4 to C1 20
S3 to C2 27
B1 supply
constraints
B1 receipt
constraints
B1 in/out constraints
B1
B1
4773
Mark Scheme
June 2006
Qu. 4
(i)
0.22, 0.2325, 0.5475
(ii)
e.g.
look-up tables
simulation run
M1 A1 A1 A1
wet(1) showery(2)
dry(3)
wet
0
0
0
showery
0.2
0.4
0.15
dry
0.5
0.55
0.4
day
0
1
2
0.14227 0.43734
rand
weather
dry
wet
dry
=IF(B8="wet",LOOKUP(C7,$B$2:$B$4,$A$2:$A$4),(IF(B8="
showery",
LOOKUP(C7,$C$2:$C$4,$A$2:$A$4),
(iii) repeating and tabulating
calculating experimental probabilities
(iv) 20 transitions
handling $s
repeating and tabulating
experimental probabilities (theoretical = 0.22, 0.24 and
0.54)
M1 probability
A1 distributions
M1 selecting dist.
A1 by weather
M1 sampling from
A1 distribution
B1 two days
handled
B1
M1 A1
B1
B1
B1
B1
Mark Scheme 4773
June 2007
106
4773
Mark Scheme
June 2007
1.
(i)
un + 2 = un +1 + pun
M1 A1
(ii)
Auxiliary equation is λ 2 − λ − 0.11 = 0
M1 A1
M1
gen homogeneous
A1
with 1.1 & –0.1
B1
case 1(u0 = 20)
+case 2(u1 = 25)
M1
simultaneous
A1
22.5 and −2.5
B1
final answer
Solution is un = 22.5(1.1)n − 2.5(–0.1)n
(iii)
Rec rel
20.0000
25.0000
27.2000
29.9500
32.9420
36.2365
39.8601
43.8461
48.2307
53.0538
58.3592
64.1951
70.6146
77.6761
85.4437
93.9881
103.3869
113.7256
125.0981
137.6080
151.3687
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Formula
20.0000
25.0000
27.2000
29.9500
32.9420
36.2365
39.8601
43.8461
48.2307
53.0538
58.3592
64.1951
70.6146
77.6761
85.4437
93.9881
103.3869
113.7256
125.0981
137.6080
151.3687
Int RR
20
25
27
30
33
36
40
44
48
53
58
64
70
77
85
93
102
112
123
135
149
Formula: =INT(H3+B$2*H2+0.5)
(iv)
B1
recurrence
relation
B1
checking formula
B1
discretising
B1
vn + 2 = (1 − r ) vn +1 + pvn
M1 A1
107
4773
Mark Scheme
June 2007
1. (cont)
(v)
r=
Pruning
0.026
20
25
27
29
31
33
36
39
42
45
48
52
56
60
65
70
75
81
87
94
101
B1
r = 0.025 to 0.027
B1
108
4773
Mark Scheme
June 2007
2.
(i)
P1
1
P2
2
E
3
M
4
J
5
A
6
M1
A1
7
(ii)
(iii)
e.g. locations 3, 6 and 7 for only two trees, so one must be
rejected. Therefore other 6 locations needed.
P1
1
P2
2
B1
B1
B1
(iv)
E
3
M
4
J
5
A
6
e.g. P2 – 5 – M – 6
M1
A1
P1 P2 E M J A
4 5 1 6 2 3
A1
109
4773
Mark Scheme
June 2007
2 (cont).
(v)
st
Max
P11+P14+P15+P21+P24+P25+E1+E2+M3+M5+M6
+J2+J4+A3+A6+A7
P11+P14+P15<=1
P21+P24+P25<=1
E1+E2<=1
M3+M5+M6<=1
J2+J4<=1
A3+A6+A7<=1
P11+P21+E1<=1
E2+J2<=1
M3+A3<=1
P14+P24+J4<=1
P15+P25+M5<=1
M6+A6<=1
A7<=1
M1
A1
objective
M1
A2
tree constraints
(–1 each error)
M1
A2
location
constraints
(–1 each error)
B1
running
B1
interpretation
End
LP OPTIMUM FOUND AT STEP 13
OBJECTIVE FUNCTION VALUE
1) 6.000000
VARIABLE
P11
P14
P15
P21
P24
P25
E1
E2
M3
M5
M6
J2
J4
A3
A6
A7
VALUE
0.000000
0.000000
1.000000
0.000000
1.000000
0.000000
1.000000
0.000000
0.000000
0.000000
1.000000
1.000000
0.000000
0.000000
0.000000
1.000000
REDUCED COST
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
P1 P2 E M J A
5 4 1 6 2 7
110
4773
Mark Scheme
June 2007
3.
(i)
(ii)
Min
st
e.g. C2 C3 C5 C7 C9 C11
M1 A1
C1+C2+C3+C4+C5+C6+C7+C8+C9+C10+C11+C12
C1+C2+C3+C4>=1
C4+C5+C6>=1
C6+C7+C8+C9+C10>=1
C1+C10+C11>=1
C2>=1
C3+C8+C12>=1
C5+C12>=1
C11>=1
C9>=1
C7>=1
M1
A1
M1
A5
objective
constraints
(–1 each error)
end
(iii)
LP OPTIMUM FOUND AT STEP
OBJECTIVE FUNCTION VALUE
1) 6.000000
VARIABLE
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
VALUE
0.000000
1.000000
0.000000
1.000000
0.000000
0.000000
1.000000
0.000000
1.000000
0.000000
1.000000
1.000000
7
B1
REDUCED COST
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
Use locations 2, 4, 7, 9, 11 and 12.
6 cameras needed
B1
B1
(iv)
New objective:
5C1+2C2+3C3+5C4+4C5+1.5C6+2C7+2C8+5C9
+3C10+4C11+7C12
M1
A1
(v)
Running
Use locations 2, 5, 7, 8, 9 and 11.
Cost = £19000
B1
B1
B1
111
running
4773
Mark Scheme
June 2007
4.
(i)
e.g.
1
2
3
(ii)
0
0.1
0.4
=LOOKUP(RAND(),B1:B3,A1:A3)
B1
B1
B1
rand
probs
outcomes
=LOOKUP(RAND(),$B$3:$B$5,$A$3:$A$5)
+ accumulation
e.g.
M1
A1
B1
formula
repeats
accumulation
M1
A1
first run
B1
repetitions
B1
probabilities
expected no. of
replacements
2
3
2
3
2
3
3
3
3
3
2
2
3
2
2
3
(iii)
2
5
7
10
12
15
18
21
24
27
29
31
34
36
38
41
e.g.
day 14
0
1
0
0
1
0
0
1
0
1
day 15
0
0
0
0
0
1
1
0
1
0
0.4
0.3
day16 no. of replacements
1
5
0
6
1
5
1
5
0
6
0
5
0
6
1
5
0
5
1
6
0.5
B1
5.4
112
4773
Mark Scheme
June 2007
Q4 (cont)
(iv)
e.g.
1
2
(v)
0
0.1
1
1
2
2
2
2
2
2
1
2
2
2
1
2
4
6
8
10
12
14
15
17
19
21
Replacements
day 1 day 2
2
6
0
7
0
7
1
6
1
6
0
7
1
6
0
7
0
7
1
6
0.6
B1
changed
probabilities
B1
repetitions
B1
B1
results
averages
6.5
5.4*(50+25) = 405 versus
0.6*(50+25) + 6.5*(30+25) = 402.5
B1
B1
More repetitions.
B1
113
4773 Decision Mathematics Computation
1.
(i)
XA + XB +XE + XF >= 1
M1 A1 ">" OK
indicator vars
Indicator variables correspond to matrix column B1
<= 5
A (or row A) entries which are less than or equal B1
B1
to 5.
Ensures that at least one such indicator is 1.
B1
(ii) Min XA+XB+XC+XD+XE+XF
st
XA+XB+XE+XF >= 1
M1
XA+XB+XE+XF >= 1
A3
(–1 each error/
XC+XF >= 1
omission)
XD+XE >= 1
allow (correct) reduced
XA+XB+XD+XE+XF >= 1
set of inequalities
XA+XB+XC+XE+XF >= 1
M1 A1 A1
(iii) 2 centres, at F&D or E&C or E&F
M1 A1
(iv) e.g. add XF=0 to force solution E and C
B1
(v) Three solutions are F & D, E & C, E & F.
B1
(vi) Problem is unimodular (or convincing argument).
B1
In the interests of efficiency (and parsimony).
2.
(i)
α=
β=
(ii)
e.g. (candidates should show formulae)
0.01
0.04
10
9
8
8
P(birth)
0.1
0.09
0.08
P(death)
0.4
0.36
0.32
rand
0.4261
0.257
0.8854
rand
0.3537
0.1405
0.8632
birth
0
0
0
death
1
1
0
e.g. 1 1 3 2 0 2 2 3 3 0 – 0.2
(iii) e.g.
β
prob extinction
0.01 0.02 0.03 0.04 0.05 0.06 0.07
0
0
0.1 0.2 0.5 0.7 0.8
(iv) Addition of another rand + another if + extra add-on
(v)
e.g.
β
0.01 0.02 0.03 0.04 0.05 0.06 0.07
prob. extinction 0
0
0
0
0
0.2 0.3
B1
B1
B1
B1
B1
B1
handling
parameters
births
deaths
use of "rand"
use of "if"
updating
population
M1 A1 B1
M1
A1 decent range
A1 reasonable
outcomes
B1
M1 A1
B1
M1
A1
3.
(a)
max AB+AE
st
AB-BC-BD+CB+DB=0
BC+DC-CB-CD-CG=0
BD+ED+CD-DB-DC-DE=0
AE-ED-EF+DE+FE=0
EF-FE-FG=0
AB<8
AE<3
BC<5
CB<5
BD<7
DB<7
CD<1
DC<1
DE<2
ED<2
EF<4
FE<4
CG<8
FG<6
END
B1
objective
M1
A2
flow balance
constraints
M1
A1
capacity
constraints
M1
run
A1
results
B1
interpretation
OBJECTIVE FUNCTION VALUE
1)
VARIABLE
AB
AE
BC
BD
CB
DB
DC
CD
CG
ED
DE
EF
FE
FG
10.00000
VALUE
7.000000
3.000000
5.000000
2.000000
0.000000
0.000000
1.000000
0.000000
6.000000
0.000000
1.000000
4.000000
0.000000
4.000000
REDUCED COST
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
Max flow of 10 with flows of 7 from A to B, … etc.
(b)
min
st
8X12+3X15+8X21+5X23+7X24+5X32+X34+8X37+7X42
+X43+2X45+3X51+2X54+4X56+4X65+6X67+8X73+6X76
X12+X15=10
X21+X23+X24=10
X32+X34+X37=10
X42+X43+X45=10
X51+X54+X56=10
X65+X67=10
X73+X76=10
X21+X51=10
X12+X32+X42=10
X23+X43+X73=10
X24+X34+X54=10
X15+X45+X65=10
X56+X76=10
X37+X67=10
M1
A1
objective
M1
A1
supply
constraints
M1
A1
demand
constraints
M1
run
A1
results
B1
interpretation
END
OBJECTIVE FUNCTION VALUE
1)
VARIABLE
X12
X15
X21
X23
X24
X32
X34
X37
X42
X43
X45
X51
X54
X56
X65
X67
X73
X76
310.0000
VALUE
0.000000
10.000000
0.000000
10.000000
0.000000
0.000000
10.000000
0.000000
10.000000
0.000000
0.000000
10.000000
0.000000
0.000000
0.000000
10.000000
0.000000
10.000000
REDUCED COST
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
6.000000
0.000000
0.000000
0.000000
0.000000
0.000000
6.000000
0.000000
0.000000
0.000000
0.000000
Cost = 310 by sending 10 from W1 to S5, … etc.
4.
(a)
Auxiliary equation:
2λ2 –3λ + 1 = 0
(2λ – 1)(λ – 1) = 0
λ = 1 or ½
M1 A1
M1
A1
un = A + B(½)n
B1 B1
5=A+B
3 = A + ½B
B1
B1
un = 1 + 4(½)n
M1 A1
u2 = 2, u3 = 1.5, u10 = 1.003906
u1000000 ≈ 1
B1 B1
B1
(b)(i) & (ii)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
(iii)
5
3
4.5
8.75
13.625
16.6875
16.40625
12.92188
7.976563
4.042969
3.087891
5.588867
10.29541
14.85425
16.98596
15.62469
11.45108
6.551926
3.376808
3.513287
6.893122
Limited wrt to (very) long-term
M1
A1
A1
3.087891
A1
6.893122
B1
4773
Mark Scheme
June 2009
4773 Decision Mathematics Computation
Question 1.
(i)
Bn+2 = Bn+1 + (0 – Bn)
M1 A1
(ii)
Oscillation: 2, 4, 2, –2, –4, –2, 2, 4, …
M1 A1 B1
(iii)
Bn+2 – Bn+1 +
1
2
Bn = 0
2, 4, 3, 1, –0.5, –1, …, 0.00391, –0.00195
Oscillatory convergence
B1
B1
B1
(iv)
2, 4, 3.5, 2.5, 1.625, 1, …, 0.00022, 0.00012
Faster and uniform convergence
B1
B1 B1
(v)
Auxiliary eqn: x2 – x +
x=
1
4
=0
B1
B1
1
2
Bn = A
( 12 )
2=A
4=1+
1
2
n
+ Bn
( 12 )
B1
n
B1
B1
B giving B = 6
Bn = ( 2 + 6n )
"the same"
( 12 )
n
or (1 + 3n )
( 12 )
n −1
B1
B1
94
4773
Mark Scheme
June 2009
Question 2.
(i)
12
e.g.
2
10
10
10
20
20
18
10
28
10
2
8
18
10
0
10
20
0
12
10
2
8
18
0
18
36
10
0
10
20
10
20
10
0
2
12
10
20
10
2
8
10
18
36
reversal
B1
rest
0
0
20
8
0
M1
A1
10
10
18
2
18
20
0
0
10
10
18
M1
A1
10
10
2
2
8
10
8
12
20
18
10
10
0
18
12
20
20
10
20
10
10
2
2
0
10
10
10
10
0
10
12
10
M1
A1
10
10
95
0
4773
Mark Scheme
June 2009
(ii)
{S, A, C, D, E} / {B, T}
M1 A1
(iii)
e.g.
Max SA + SE
st
SA + BA + CA – AB – AC = 0
AB + CB – BA – BC – BT = 0
AC + DC + BC – CA – CB – CD = 0
SE + DE – ED = 0
CD + ED – DC – DE – DT = 0
SA < 12
SE < 20
AB < 10
BA < 10
AC < 10
CA < 10
BC < 10
CB < 10
CD < 10
DC < 10
ED < 18
DE < 18
BT < 20
DT < 10
end
M1
A1
variables
objective
M1
A1
balancing
M1
A1
A1
capacities
forwards
backwards
B1
running
(iv)
OBJECTIVE FUNCTION VALUE
1)
30.00000
VARIABLE
VALUE
REDUCED COST
SA
12.000000
0.000000
SE
18.000000
0.000000
BA
0.000000
0.000000
CA
0.000000
0.000000
AB
10.000000
0.000000
AC
2.000000
0.000000
CB
10.000000
0.000000
BC
0.000000
0.000000
BT
20.000000
0.000000
DC
8.000000
0.000000
CD
0.000000
0.000000
DE
0.000000
1.000000
ED
18.000000
0.000000
DT
10.000000
0.000000
Solution as per part (i)
B1
96
4773
Mark Scheme
June 2009
Question 3.
(i)
e.g.
-1
1
1
1
1
1
-1
5
4
5
6
7
8
9
8
= if(rand()<0.55,1,–1)
M1
A1
A1
= B1 + A2
"if" or equivalent
accumulation
etc.
(ii)
repeating until a player is ruined
repeating 10 times
estimating the probability (theoretical value is 0.2683)
(iii)
e.g.
-1
1
-1
-1
-1
-1
-1
5
4
5
4
3
2
1
0
M1
A1
M1 A1
= if(rand()<0.45,1,–1)
B1
M1
A1
= B1 + A2
change of parameter
count to ruin
repetitions
M1
A1
etc.
estimating the run length
The theoretical value is 50, so there should be some long
runs seen.
(iv)
e.g.
-1
1
1
-1
1
1
0
-1
0
1
0
1
2
= if(rand()<0.55,1,–1)
= B1 + A2
B1 B1
etc
repetitions + probability estimate
(theoretical answer = 0.599)
(v)
M1
A1
B1
As above
How can one tell when a simulation is not emptying the pot?
97
B1
termination
condition
4773
Mark Scheme
June 2009
Question 4.
(i)
max 4s1m0+7s1m1+8s1m2+9s1m3+11s1m4+3s2m0
+6s2m1+10s2m2+12s2m3+14s2m4+3s3m0
+7s3m1+8s3m2+13s3m3+15s3m4
st
s1m0+s1m1+s1m2+s1m3+s1m4=1
s2m0+s2m1+s2m2+s2m3+s2m4=1
s3m0+s3m1+s3m2+s3m3+s3m4=1
s1m1+2s1m2+3s1m3+4s1m4+s2m1+2s2m2+3s2m3
+4s2m4 +s3m1+2s3m2+3s3m3+4s3m4=4
end
int 15
M1
A1
B1
B1
B1
M1
A1
B1
(ii)
LP OPTIMUM FOUND AT STEP 14
OBJECTIVE VALUE = 24.0000000
NEW INTEGER SOLUTION OF 24.0000000 AT BRANCH 0 PIVOT 14
RE-INSTALLING BEST SOLUTION...
OBJECTIVE FUNCTION VALUE
1)
24.00000
VARIABLE
S1M0
S1M1
S1M2
S1M3
S1M4
S2M0
S2M1
S2M2
S2M3
S2M4
S3M0
S3M1
S3M2
S3M3
S3M4
VALUE
REDUCED COST
0.000000
-4.000000
1.000000
-7.000000
0.000000
-8.000000
0.000000
-9.000000
0.000000
-11.000000
0.000000
-3.000000
0.000000
-6.000000
1.000000
-10.000000
0.000000
-12.000000
0.000000
-14.000000
0.000000
-3.000000
1.000000
-7.000000
0.000000
-8.000000
0.000000
-13.000000
0.000000
-15.000000
ROW SLACK OR SURPLUS
2)
0.000000
3)
0.000000
4)
0.000000
5)
0.000000
M1
A1
DUAL PRICES
0.000000
0.000000
0.000000
0.000000
B3
B1
Invest £1 million at site 1, £2 million at site 2 and £1 million at site 3.
Revenue = £24 million.
(iii)
(a)
(b)
£2 million at site 2 and £1 million at site 3.
Revenue = £21 million.
£1 million at site 1, £3 million at site 2 and £4 million at site
3 (or £1m, £4m and £3m).
Revenue = £34 million.
98
M1
A1
M1
A1
4773
Mark Scheme
June 2010
1.
(i)
(ii)
u n = 1.05u n–1 –60
n
u n = 1000×1.05
M1 A2
1.05
– 60
n

1
M1
A2
A1
0.05
= 1200 – 200×1.05n
or
u n = 1.05n + 
1000 = + 
990 = 1.05+ , etc
(iii) int(log(6)/log(1.05)) = 36 years (or spreadsheet)
M1 A1
(iv)
1000
1025
990.625
1015.391
980.7754
1005.295
M1
A1
A1
etc.
(v)
37 years (+ 6 months OK)
B1
1000
970
989.25
959.25
977.9625
947.9625
M1
A1
A1
A1
cao
(vi)
interest OK
etc.
(vii) 35 years
B1
1
cao
4773
Mark Scheme
June 2010
2.
(i)
(ii)
B1
5 2
(negatives of these OK)
e.g:
min
st
m
p-m<0
-p-m<0
q-m<0
-q-m<0
-p-q-1.414214m<-10
p+q-1.414214m<10
M1
A1
A1
A1
first 2 pairs
first pair
second pair
last pair
end
(iii) Objective value: 2.928932
Variable
M
P
Q
Value
2.928932
2.928932
2.928932
Reduced Cost
0.000000
0.000000
0.000000
All dotted lines are
of length 5 2  2)


B1
M1
A1
A1
drawing
lines
point
B1
equidistant
M1
A1
first pair
A1
second pair
A1
third pair
(iv) e.g:
min
st
m
q-m<0
-q-m<0
p+q-1.414214m<1
-p-q-1.414214m<-1
p-q-1.414214m<-1
-p+q-1.414214m<1
end
(v)
Objective value:
Variable
M
Q
P
0.4142135
Value
0.4142135
0.4142135
0.000000
Reduced Cost
0.000000
0.000000
0.000000
B1
(vi)
All dotted lines are
of length 2  1
2
B1
B1
B1
lines
point
distances
4773
Mark Scheme
June 2010
3.
(i)
(ii)
Min 2x11+3x12+7x13+x21+8x22+4x23
st
x11+x12+x13=10
x21+x22+x23=10
x11+x21<7
x12+x22<7
x13+x23<7
end
Objective value: 55.00000
Variable
Value
X11
3.000000
X12
7.000000
X13
0.000000
X21
4.000000
X22
0.000000
X23
6.000000
Reduced Cost
0.000000
0.000000
3.000000
0.000000
6.000000
0.000000
3 containers from S1 to D1
7 containers from S1 to D2
4 containers from S2 to D1
6 containers from S2 to D3
total cost = 55
supplies
B1
depots
B1
B1
B1
(iii) Min 2y11+3y12+9y13+y14+4y21+7y22+2y23
st
B1
B1
+5y24+y31+5y32+3y33+6y34
y11+y12+y13+y14=7
y21+y22+y23+y24=7
y31+y32+y33+y34=6
y11+y21+y31=7
y12+y22+y32=4
y13+y23+y33=6
y14+y24+y34=3
logistics
cost
B1
B1
depots
B1
demands
end
(iv) Objective value: 37.00000
Variable
Y11
Y12
Y13
Y14
Y21
Y22
Y23
Y24
Y31
Y32
Y33
Y34
Value
0.000000
4.000000
0.000000
3.000000
1.000000
0.000000
6.000000
0.000000
6.000000
0.000000
0.000000
0.000000
Reduced Cost
2.000000
0.000000
11.00000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
1.000000
4.000000
4.000000
4 containers from D1 to C2
3 containers from D1 to C4
1 container from D2 to C1
6 containers from D2 to C3
6 containers from D3 to C1
total cost = 37
B1
B1
3
logistics + cost
4773
(v)
Mark Scheme
min 2x11+3x12+7x13+x21+8x22+4x23+2y11+3y12
+9y13+y14+4y21+7y22+2y23+5y24+y31+5y32
+3y33+6y34
st
x11+x12+x13=10
x21+x22+x23=10
y11+y21+y31=7
y12+y22+y32=4
y13+y23+y33=6
y14+y24+y34=3
x11+x21<7
x12+x22<7
x13+x23<7
y11+y12+y13+y14-x11-x21=0
y21+y22+y23+y24–x12–x22=0
y31+y32+y33+y34-x13-x23=0
end
June 2010
B1
B1
supplies
B1
demands + depots
B1
trans-shipment
(vi) Objective value: 91.00000
Variable
X11
X12
X13
X21
X22
X23
Y11
Y12
Y13
Y14
Y21
Y22
Y23
Y24
Y31
Y32
Y33
Y34
Value
4.000000
6.000000
0.000000
3.000000
0.000000
7.000000
0.000000
4.000000
0.000000
3.000000
0.000000
0.000000
6.000000
0.000000
7.000000
0.000000
0.000000
0.000000
Reduced Cost
0.000000
0.000000
2.000000
0.000000
6.000000
0.000000
0.000000
0.000000
8.000000
0.000000
1.000000
3.000000
0.000000
3.000000
0.000000
3.000000
3.000000
6.000000
4 containers from S1 to D1
6 containers from S1 to D2
3 containers from S2 to D1
7 containers from S2 to D3
4 containers from D1 to C2
3 containers from D1 to C4
6 containers from D2 to C3
7 containers from D3 to C1
total cost = 91
B1
B1
Suboptimising does not give the optimum
4
B1
cao
4773
Mark Scheme
June 2010
4.
(i)
(ii)
e.g. = lookup(rand(),A1:A3,B1:B3) with
A B
1
0
0
2 0.1 1
3 0.6 2
Many approaches possible, but all must allow for 3
applications of part (i)
Offspring from generation 0
Conditional offspring from generation 1(s)
Output
M1 A1
B1
B1
B2
B1
B1 M1A1 M1A1
B1
(iii) Theoretical probabilities (Galton-Watson branching):
0
1
2
3
4
0.154 0.29 0.332 0.16 0.064
M1
A1
(iv) Two independent runs.
Sum the numbers in the two second generations.
(or nested “IF”s)
0, 1, 2, 3, 4, 5, 6, 7, 8
B1
B1
(M1 A1)
B1
5
4773
Mark Scheme
June 2011
1.
(i)
A
m
B
c
C
v
D
s1
E
s2
Number of maximal matchings = 4×4 = 16
(note 2 separate graphs)
(ii)
M1
A1
A1
bipartite
ABD/mv
C E / c s1 s2
B1
cao
eg (note ... 4 solutions)
A
m
B
c
C
v
D
s1
E
s2
(iii)
v
B1
D
m
A
B1
B1
B1
B
D
v
A
m
3
first branch
second branch
third branch
4773
Mark Scheme
June 2011
(iv) e.g.
Max Am+Av+Bm+Bv+Cc+Cs1+Cs2+Dv+Es1+Es2
st
Am+Av<1
Bm+Bv<1
Cc+Cs1+Cs2<1
Dv<1
Es1+Es2<1
Am+Bm<1
Cc<1
Av+Bv+Dv<1
Cs1+Es1<1
Cs2+Es2<1
end
e.g.
LP OPTIMUM FOUND AT STEP 6
OBJECTIVE FUNCTION VALUE
1)
4.000000
VARIABLE VALUE
REDUCED COST
AM
1.000000
0.000000
AV
0.000000
0.000000
BM
0.000000
0.000000
BV
0.000000
0.000000
CC
0.000000
0.000000
CS1
1.000000
0.000000
CS2
0.000000
0.000000
DV
1.000000
0.000000
ES1
0.000000
0.000000
ES2
1.000000
0.000000
Arthur has the macaroni
Bertie goes hungry
Charles has the first spaghetti
David has the vegetable lasagne
Edward has the second spaghetti
(v)
1
1
M1
A1
people constraints
M1
A1
meal constraints
M1
running
A1
A
m
B
c
C
v
D
s1
E
s2
1
S
M1 A1
B1
source and sink
with bipartite
connection
B1
arc capacities out
of S (or into T).
T
1
1
4
4773
Mark Scheme
June 2011
2.
(i)
x1 + x2 ≤ 10000
x3 + x4 ≤ 1.04x1
x5 + x6 ≤ 1.10x2 + 1.04x3
x7
≤ 1.09x4 + 1.04x5
x8
≤ 1.10x6
(Might see extra terms for x4 and x5 feeding to x8.)
(ii)
x3 + x4 –1.04x1 < 0
x5 + x6 - 1.10x2 - 1.04x3 < 0
x7 − 1.09x4 − 1.04x5 < 0
x8 − 1.10x6 < 0
B1 max
B1 B1 (-1 each error)
5
Invest all money in investment 2, thence to investment
number 5, thence to investment number 7.
Gives 12812.80 at end.
(v)
1357
1368
147
257
268
1.04×1.04×1.04×1.12
1.04×1.04×1.10×1.04
1.04×1.09×1.12
1.10×1.04×1.12
1.10×1.10×1.04
x7 ≤
1.09x4
1.04x5
x8 ≤
1.10x6
B1
(iii) max 1.12x7 + 1.04x8
(iv) LP OPTIMUM FOUND AT STEP
OBJECTIVE FUNCTION VALUE
1)
12812.80
VARIABLE VALUE
X7
11440.000000
X8
0.000000
X1
0.000000
X2
10000.000000
X3
0.000000
X4
0.000000
X5
11000.000000
X6
0.000000
B1
B1
B1
B1
B1
 1.26
 1.24
 1.27
 1.28
 1.26
Best is as above
B1
running
M1
A1
B1
words!
strategy
outcome
M1
A1
A1
those starting “1”
those starting “2”
A1
(vi) Formulation + solution would be more easily automated
for larger problems.
5
B1
4773
3.
(i)
(ii)
Mark Scheme
June 2011
u n+1 = 0.896u n + 250
B1
B1
structure
0.896
1  0.89 6n
M1
A1
A1
B1
0.896n
rest OK
limit
M1
A1
A1
shape
497
247
M1
A1
M1
A1
5 hour rules
B1
B1
B1
shape
upper values
lower values
B1
cao
B1
analysis needed
u n  250
Tends to
1  0.89 6
250
1  0.89 6
 497 (iii) 497
247
(iv)
1 2 3 4 5 6 7 8 (v)
250.00 389.60 467.56 511.09 429.06 489.59 523.39 542.26 21
22
23
24
25
26
27
28
496 527
544
277
294
440.70
496.09
527.02
544.29
440.70
496.09
527.02
544.29
441
246
9 hour rule
191
(vi) Level drops to 103mg
No. Converges back to previous levels more quickly
with single dose.
6
4773
Mark Scheme
June 2011
4.
(i)
(ii)
Lookup tables, or equivalent
M1 A1
simulating ...
number of customers
a dish for a customer
B1
B1
B1
simulating the correct number of dishes
M1 A1
collecting results
M1 A1
use of relative and absolute addressing
M1 A1
100 repetitions
B1
means and standard deviations
B1 B1
(iii) Not concerned with the variation of the mean, but with the
variation of an individual day’s demand.
B1 B1
(iv) Analyse the cost of waste against the cost of loss of
goodwill and decide upon an appropriate level of
provision, e.g. mean + 1 standard deviation.
the above would give, arguably, 6, 2, 4 and 8 respectively.
B1
7
B1
4773
Mark Scheme
Question
1
(i)
Answer
Key: ( the forward direction is defined by the order of the letters on the arc
name)
June 2012
Marks
B1
B1 FT
B1 FT
1
(ii)
[3]
B1 B1
e.g. flow-augmenting path ... SACDBT (8)
M1
A1
[4]
5
Guidance
capacities and flows
forward potentials
backward potentials
4773
Mark Scheme
Question
1
(iii)
Answer
e.g. flow-augmenting path ... SBT (15)
June 2012
Marks
B1
Guidance
B1
1
(iv)
[2]
M1
A1
Flow is now 38
But cut SA/BCDT has capacity 38
So flow is maximal
[2]
6
max flow/min cut
cao
4773
Question
1
(v)
Mark Scheme
Answer
Marks
B1
e.g.
max SA + SD + SB
st
SA + CA – AC = 0
AC + DC – CA – CD – CT = 0
SD + CD + BD – DC – DB – DT = 0
SB + DB – BD – BT = 0
SA < 10
SD < 15
SB < 15
AC < 8
CA < 8
CD < 5
DC < 5
BD < 10
DB < 10
CT < 5
DT < 10
BT < 30
end
LP OPTIMUM FOUND AT STEP
2
OBJECTIVE FUNCTION VALUE
1)
38.00000
VARIABLE
SA
SD
SB
CA
AC
DC
CD
CT
BD
DB
DT
BT
June 2012
VALUE
8.000000
15.000000
15.000000
0.000000
8.000000
0.000000
5.000000
3.000000
0.000000
10.000000
10.000000
25.000000
Different solution ... CT down 2
CD, DB, BT up 2 ... with same flow
7
Guidance
objective
M1
A1
4 balancing equations
M1
A1
capacities + back capacities
B1
running
B1
[7]
interpretation
4773
Mark Scheme
Question
2
Answer
June 2012
Marks
Guidance
e.g.
min
st
end
int
int
int
int
int
int
30000xa1 + 33000xa2 + 35000xa3
+ 28000xb1 + 31000xb2 + 33000xb3
+ 27000xc2 + 29000xc3
+ 28000xd2 + 30000xd3
+ 800000c + 1400000d
xa – xa1 – xa2 – xa3 = 0
xb – xb1 – xb2 – xb3 = 0
xc – xc2 – xc3 = 0
Note that indicator
xd – xd2 – xd3 = 0
switching is overxa1 + xb1 > 600
modelled here. Costs of
xa2 + xb2 + xc2 + xd2 > 850
opening a mine do not
xa3 + xb3 + xc3 + xd3 > 750
vary from year 2 to year
xc2 – 1000c2 < 0
3, and are payable only
xc3 – 1000c3 < 0
once. So only one
c2 – c3 < 0
indicator is needed for
xd2 – 1000d2 < 0
each of C and D.
xd3 – 1000d3 < 0
Simpler modelling
d2 – d3 < 0
allowed, see below.
c2 + c3 – 10c < 0
d2 + d3 – 10d < 0
2xa2 + 0.1xb2 – xc2 – 1.9xd2 > 0
2xa3 + 0.1xb3 – xc3 – 1.9xd3 > 0
xa < 650
xb < 950
xc < 625
xd < 840
c2
c3
d2
d3
c
d
8
M1
A1
mining
M1
A1
opening
M1
A1
requirements
M1
A1
A1
indicator switching
for C
for D
M1
A1
quality
M1
A1
availabilties
B1
integer variables
4773
Question
Mark Scheme
Answer
OBJECTIVE FUNCTION VALUE
1)
June 2012
Marks
Guidance
0.6745000E+08
VARIABLE
C2
C3
D2
D3
C
D
XA1
XA2
XA3
XB1
XB2
XB3
XC2
XC3
XD2
XD3
XA
XB
XC
XD
VALUE
1.000000
1.000000
0.000000
0.000000
1.000000
0.000000
0.000000
375.000000
250.000000
600.000000
350.000000
0.000000
125.000000
500.000000
0.000000
0.000000
625.000000
950.000000
625.000000
0.000000
Interpretation needed:
cost (£67.5m)
what is produced where
and when
what is opened
9
B1
submitting
B1
B1
B1
cost (cao) 8
production
location
4773
Mark Scheme
Question
Answer
June 2012
Marks
Simpler LP formulation.
min
30000A1 + 33000A2 + 35000A3 + 28000B1 + 31000B2 + 33000B3
+ 27000C2 + 29000C3 + 28000D2 + 30000D3 + 800000XC +
1400000XD
S.T. A1 + A2 + A3 <= 650
B1 + B2 + B3 <= 950
C2 + C3 <= 625
D2 + D3 <= 840
A1 + B1 >= 600
A2 + B2 + C2 + D2 >= 850
A3 + B3 + C3 + D3 >= 750
2A2 + 0.1B2 – C2 – 1.9D2 >= 0
2A3 + 0.1B3 – C3 – 1.9D3 >= 0
C2 + C3 – 1000XC <= 0
D2 + D3 – 1000XD <= 0
END
INT XC
INT XD
[18]
3
(i)
Month
June
Inflation
3.74
Committee
‒1.24
July
August
September
October
November
2.93
1.69
1.26
2.07
3.31
‒0.43
+0.81
+1.24
+0.43
‒0.81
December
3.74
‒1.24
M1
A1
B1
... thence cycling.
[3]
10
Guidance
4773
Mark Scheme
Question
3
(ii)
Answer
3.74
3.74
2.93
2.93
2.31
2.775
2.095
2.72125
2.19 2.686875
2.3925 2.659219
2.5475 2.635859
2.60125 2.615957
2.5775 2.598975
2.526875
2.58448
2.488125 2.572108
2.474688 2.561548
2.480625 2.552535
(Oscillatory) convergence in the “half” case
Slower, (uniform) convergence on the “one eighth” case
June 2012
Marks
B1
B1
3
(iii)
un+2 = un+1 + 0.125 * (2.5 − un) ... plus reorganisation
3
(iv)
Auxiliary equation ... x2 − x + 0.125 = 0
Solutions ... (2  2)/4
General form ... un = A((2 + 2)/4)n + B((2 − 2)/4)n + 2.5
Simultaneous ... A + B = 1.24 ... A((2 + 2)/4) + B((2 − 2)/4) = 0.43
Solutions ... A  0.3513, B  0.8887
3
(v)
Spreadsheet check
Because ((2 + 2)/4)n  0 and ((2 − 2)/4)n  0 as n increases
3
(vi)
α = 0.25 makes discriminant 0 ... (boundary between oscillatory and uniform
convergence)
spreadsheet
fast (uniform) convergence
11
B1
B1
[4]
B1
[1]
M1
A1 FT
M1
A1 FT
A1
[5]
B1
B1
[2]
B1
B1
B1
[3]
Guidance
4773
Mark Scheme
Question
4
(i)
Answer
Marks
M1 A3
M1 A1
A1
A1
M1 A1
[10]
M1 A1
B1
B1
[4]
B1
B1
[2]
B1 B1
[2]
Lookup tables, or equivalent
simulating par 3
simulating par 4
simulating par 5
modelling matchplay
4
(ii)
repetition of holes, taking account of pars
computing strokeplay result
computing matchplay result
4
(iii)
repetition of rounds
collection and display of results
(iv)
no and no
June 2012
12
Guidance
Allow follow through
4773
Question
1 (i)
Mark Scheme
Guidance
Marks
M1
“lookup” or equivalent
A1
M1
“lookup” or equivalent
A1
M1
A1
B1
[7]
B1
Answer
modelling usage
modelling delivered
modelling end stock
start stock = max(previous end stock, 0)
1
(ii)
outages ... number of days when supplies run out
either max stock or total purchases
1
(iii)
Incorporation of outage count into s/s
Outages very variable and often large ... anything between 0 and 25
Incorporation of other measure into s/s
Max stock variable, but not as variable as outages ... 3 to 7 ish
or
Purchases ... around 50 (Allowed here though not a good measure. Not allowed in (iv).
1
(iv)
June 2013
New modelling for number purchased
B1
[2]
B1
B1
B1
B1
[4]
M1
A1
use of “IF”
correct
Outages ... reduced in variability and number
B1
Max stock consistently 2.25 or thereabouts
B1
Note ... no marks for 50
purchased here
Improvement
B1
[5]
... justified!
5
4773
Question
2 (i)
2 (ii)
2 (iii)
2 (iv)
Mark Scheme
June 2013
Answer
ei = ei-1 + 500 – ei-3
450
370
520
570
410
800
460
930
440
860
530
560
570
130
630
-230
600
-290
530
80
400
810
300 270
1600
en = en-1 + 500 – en-2
450
520
520
460
460
440
440
480
480
540
540
560
2 (v)
en = en-1 + p(500 – en-2)
2 (vi)
Auxiliary equation ... x2 – x + 2/9 = 0
Solutions ... 2/3 and 1/3
General form ... A(2/3)n + B(1/3)n + K
560
520
520
460
460
440
440
480
480
540
540
560
560
Solution ... 165(2/3)n – 480(1/3)n + 500
450.00
502.85
500.07
520.00
501.90
500.05
531.11
501.27
500.03
526.67
500.85
500.02
519.75
500.57
500.01
[2]
B1
[1]
M1
A1
[2]
B1
[1]
B1
B1
B1
B1
B2
Simultaneous equations using 450 and 520
2 (vii)
Marks
B1
[1]
M1
A1
513.83
500.38
500.01
509.44
500.25
500.01
506.36
500.17
NB ... might have day 3 = 460 different if part (v) followed for day 3.
6
504.27
500.11
M1
A2
[9]
M1
A1
[2]
Guidance
cao
third order recurrence
cao for at least 25 days,
including starting days.
cao
second order recurrence
cao for at least 25 days,
including starting days.
or equivalent
both
2/3 and 1/3
+K
Candidates may have 3
simultaneous equations or
may know what K needs
to be.
solving simultaneous eqs
–1 each error, cao
modelling their
recurrence relation and
their recurrence solution
cao
4773
Question
3 (i)
Mark Scheme
min
st
Answer
11.12a1+11.34a2+11.74a3+11.63a4+11.12a5+11.34a6+11.74a7+11.63a8
+12.01b1+12.23b2+11.89b3+12.17b4+12.01b5+12.23b6+11.89b7+12.17b8
+11.24c1+11.09c2+11.56c3+11.65c4+11.24c5+11.09c6+11.56c7+11.65c8
+13.34d1+12.95d2+12.67d3+13.01d4+13.34d5+12.95d6+12.67d7+13.01d8
+12.54e1+12.37e2+12.21e3+12.45e4+12.54e5+12.37e6+12.21e7+12.45e8
+11.87f1+11.74f2+11.35f3+11.21f4+11.87f5+11.74f6+11.35f7+11.21f8
+11.52g1+11.42g2+11.37g3+11.74g4+11.52g5+11.42g6+11.37g7+11.74g8
+12.08h1+12.43h2+12.32h3+12.57h4+12.08h5+12.43h6+12.32h7+12.57h8
a1+a2+a3+a4+a5+a6+a7+a8=1
b1+b2+b3+b4+b5+b6+b7+b8=1
c1+c2+c3+c4+c5+c6+c7+c8=1
d1+d2+d3+d4+d5+d6+d7+d8=1
e1+e2+e3+e4+e5+e6+e7+e8=1
f1+f2+f3+f4+f5+f6+f7+f8=1
g1+g2+g3+g4+g5+g6+g7+g8=1
h1+h2+h3+h4+h5+h6+h7+h8=1
a1+b1+c1+d1+e1+f1+g1+h1=1
a2+b2+c2+d2+e2+f2+g2+h2=1
a3+b3+c3+d3+e3+f3+g3+h3=1
a4+b4+c4+d4+e4+f4+g4+h4=1
a5+b5+c5+d5+e5+f5+g5+h5=1
a6+b6+c6+d6+e6+f6+g6+h6=1
a7+b7+c7+d7+e7+f7+g7+h7=1
a8+b8+c8+d8+e8+f8+g8+h8=1
end
7
June 2013
Guidance
Marks
M1
objective
A1
B1
athlete constraints
B1
position constraints
B1
running
4773
Mark Scheme
Question
Answer
or
min
st
June 2013
Marks
Guidance
11.12a1+11.34a2+11.74a3+11.63a4 +12.01b1+12.23b2+11.89b3+12.17b4
+11.24c1+11.09c2+11.56c3+11.65c4+13.34d1+12.95d2+12.67d3+13.01d4
+12.54e1+12.37e2+12.21e3+12.45e4+11.87f1+11.74f2+11.35f3+11.21f4
+11.52g1+11.42g2+11.37g3+11.74g4+12.08h1+12.43h2+12.32h3+12.57h4
a1+a2+a3+a4 =1
b1+b2+b3+b4 =1
c1+c2+c3+c4 =1
d1+d2+d3+d4 =1
e1+e2+e3+e4 =1
f1+f2+f3+f4 =1
g1+g2+g3+g4 =1
h1+h2+h3+h4 =1
a1+b1+c1+d1+e1+f1+g1+h1=2
a2+b2+c2+d2+e2+f2+g2+h2=2
a3+b3+c3+d3+e3+f3+g3+h3=2
a4+b4+c4+d4+e4+f4+g4+h4=2
end
3
(ii)
e.g.
1
H
2
G
3
B
4
F
46.60
5
A
6
C
7
D
8
E
47.33
93.93 total
Minimises total time for both teams. He will want to minimise the time for the first team, then do what is
best with the rest.
B1
teams cao
B1
[7]
B1
times cao
[1]
8
4773
Question
3 (iii)
Mark Scheme
Guidance
Marks
objective
B1
Answer
min
st
June 2013
11.12a1+11.34a2+11.74a3+11.63a4
+12.01b1+12.23b2+11.89b3+12.17b4
+11.24c1+11.09c2+11.56c3+11.65c4
+13.34d1+12.95d2+12.67d3+13.01d4
+12.54e1+12.37e2+12.21e3+12.45e4
+11.87f1+11.74f2+11.35f3+11.21f4
+11.52g1+11.42g2+11.37g3+11.74g4
+12.08h1+12.43h2+12.32h3+12.57h4
a1+a2+a3+a4<1
b1+b2+b3+b4<1
c1+c2+c3+c4<1
d1+d2+d3+d4<1
e1+e2+e3+e4<1
f1+f2+f3+f4<1
g1+g2+g3+g4<1
h1+h2+h3+h4<1
a1+b1+c1+d1+e1+f1+g1+h1=1
a2+b2+c2+d2+e2+f2+g2+h2=1
a3+b3+c3+d3+e3+f3+g3+h3=1
a4+b4+c4+d4+e4+f4+g4+h4=1
M1
A1
B1
end
1
A
2
C
3
G
4
F
44.79
B1
9
< ... i.e. inequality
4773
Question
3 (iii)
Mark Scheme
Guidance
Marks
M1
selection of remaining
athletes
Answer
min
st
June 2013
12.01b1+12.23b2+11.89b3+12.17b4
+13.34d1+12.95d2+12.67d3+13.01d4
+12.54e1+12.37e2+12.21e3+12.45e4
+12.08h1+12.43h2+12.32h3+12.57h4
b1+b2+b3+b4=1
d1+d2+d3+d4=1
e1+e2+e3+e4=1
h1+h2+h3+h4=1
b1+d1+e1+h1=1
b2+d2+e2+h2=1
b3+d3+e3+h3=1
b4+d4+e4+h4=1
A1
B1
B1
end
5
H
6
E
7
D
8
B
49.29
Note 44.79 + 49.29 = 94.08
10
B1
[10]
4773
Mark Scheme
Question
4 (i)
Answer
Solution is 4 starter/main offers, 4 main/dessert offers and the £4.32 starter, at a cost of £104.32.
4
“s” variables are indicators for starters, “m” variables are indicators for mains and “d” variables are
indicators for desserts.
“sm” indicates the number of starter/main offers.
“md” indicates the number of main/dessert offers.
Use of “number” twice.
“int 17” declares that the first 17 variables are indicator variables. These take the values 0 or 1 and indicate
in the objective whether the individual dishes are paid for or not.
(ii)
4
(iii)
Paying separately for a dessert instead of a starter would mean sm=5 and md=3. So the alternative would
cost an extra £(15.00 – 10.00 + 3.42 – 4.32).
4
(iv)
min
st
8.50s1+7.65s2+4.32s3+5.67s4+5.67s5
+12.42m1+9.85m2+13.36m3+21.25m4+12.42m5+17.85m6+13.63m7+13.63m8
+6.85d1+5.32d2+3.42d3+10.18d4
+15sm
+10md
+17.5smd
m1+m2+m3+m4+m5+m6+m7+m8+sm+md+smd=8
s1+s2+s3+s4+s5+sm+smd=5
d1+d2+d3+d4+md+smd=4
end
int 17
4
(v)
Same solution ... smd=0
4
(vi)
Running at smd cost of £15.15 and at cost of £15.14.
Solution is one 3-course offer, three of each 2-course offer, the £9.85 main and the £4.32 starter (at total
cost of £104.31).
June 2013
Guidance
Marks
B1
coding and running
B1
order ft
B1
cost ft
[3]
B1
M1
M1
A1
M1
E1
[6]
B1
[1]
M1
new variable
A1
in objective correctly
A1
in constraint
A1
ditto
A1
ditto
[5]
B1
[1]
B1
B1
[2]
11