Tuesday 24 June 2014 – Morning
A2 GCE MATHEMATICS (MEI)
4773/01
Decision Mathematics Computation
Candidates answer on the Answer Booklet.
* 3 1 4 5 2 9 4 3 8 0 *
OCR supplied materials:
•
12 page Answer Booklet (OCR12)
(sent with general stationery)
•
Graph paper
•
MEI Examination Formulae and Tables (MF2)
Other materials required:
•
Scientific or graphical calculator
•
Computer with appropriate software and
printing facilities
Duration: 2 hours 30 minutes
*
4
7
7
3
0
1
*
INSTRUCTIONS TO CANDIDATES
•
•
•
•
•
•
•
•
Write your name, centre number and candidate number in the spaces provided on the
Answer Booklet. Please write clearly and in capital letters.
Use black ink. HB pencil may be used for graphs and diagrams only.
Answer all the questions.
Read each question carefully. Make sure you know what you have to do before starting
your answer.
You are permitted to use a graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.
Additional sheets, including computer print-outs, should be fastened securely to the
Answer Booklet.
Do not write in the bar codes.
INFORMATION FOR CANDIDATES
•
•
•
•
•
•
The number of marks is given in brackets [ ] at the end of each question or part
question.
In each of the questions you are required to write spreadsheet or other routines to carry
out various processes.
For each question you attempt, you should submit print-outs showing the routine you
have written and the output it generates.
You are not expected to print out and submit everything your routine produces, but
you are required to submit sufficient evidence to convince the examiner that a correct
procedure has been used.
The total number of marks for this paper is 72.
This document consists of 8 pages. Any blank pages are indicated.
COMPUTING RESOURCES
•
Candidates will require access to a computer with a spreadsheet program, a linear
programming package and suitable printing facilities throughout the examination.
© OCR 2014 [Y/102/2662]
DC (NF/SW) 81136/3
OCR is an exempt Charity
Turn over
2
1
The owner of a grocery store aims to keep 150 cartons of UHT milk in stock. At the end of each working day
the store keeper counts the stock and places an order. If the stock is less than 150 he orders the difference
between 150 and the stock. If the stock is 150 or more then he does not order any. Orders take one working
day to be processed and are then delivered at the start of the next working day after that. The store is closed
on Sundays.
At the end of Friday the store keeper ordered 35 cartons of UHT milk. These arrived on Monday morning.
On Saturday he received cartons from Thursday’s order and sold 50 cartons. He ended the day with 130
cartons in stock and ordered 20.
On Monday he received 35 from Friday’s order and again sold 50. So at the close of trading he had 115 in
stock, and he ordered 35.
(i) Build a spreadsheet to model u n , the number of cartons in stock at the beginning of day n before
receiving that day’s delivery, starting with Monday as day 1. You will need to model the possibility of
him ordering no cartons, and you need to show your formulae.
Assuming that demand remains constant at 50 per day, report on how stock levels behave in the long
run.
[5]
(ii) For this scenario the owner does not ever order 0 cartons. So, ignoring the possibility that the stock
might exceed 150 at the end of a day, construct a recurrence relation to model the scenario (ie demand
constant at 50 per day).
[2]
(iii) Find the auxiliary equation for your recurrence relation, and show that it has no real roots.
Explain how you could have seen that the auxiliary equation would have no real roots by inspecting
your spreadsheet.
[3]
The owner’s daughter suggests that her father tries ordering 50 cartons plus a fraction, a , of the difference
between 150 and the closing stock, 0 1 a 1 1, for closing stocks less than or equal to 150. (You are not
required to model the possibility of closing stocks greater than 150.) So on Saturday he would have ended
the day with 130 cartons in stock and have ordered 50 + 20a . On Monday he would have had 115 in stock at
the close of trading, and would have ordered 50 + 35a .
(iv) Construct the recurrence relation for u n , the number of cartons in stock at the beginning of day n,
with Monday being day 1 and assuming that demand remains constant at 50 per day. Show that for
[4]
a = 0.25 , the recurrence relation’s auxiliary equation has one real root.
(v) Add a model of this scenario to your spreadsheet. Use your spreadsheet to demonstrate
that for a = 0.25 and for demand constant at 50, the solution to the recurrence relation is
n
n
[2]
u n = 60 ^0.5h - 100n ^0.5h + 150 .
(vi) The recurrence relation given in part (v) is a mathematical approximation to a realistic ordering policy
since it allows non-integer values. Produce a spreadsheet model with a = 0.25 , but in which order
quantities are to the nearest whole number.
[1]
(vii) Use your spreadsheet from part (vi) to investigate what increased level of demand leads to stock
running out.
[1]
© OCR 2014
4773/01 Jun14
3
2
A cable car installation at a ski resort has two cabins connected by the cable. Skiers board the lower cabin
until it is full. It then travels up whilst the empty cabin travels down from the top station.
The cabins can each hold 160 skiers. The journey time from bottom to top is 4 minutes. So the system can
move 2400 skiers per hour, or an average rate of 2 skiers every 3 seconds.
Skiers arrive at an average rate of 1 skier every 2 seconds at intervals given by the following distribution.
time interval (secs)
probability
1
2
3
4
0.4
0.3
0.2
0.1
(i) Build a spreadsheet simulation for loading and transporting 160 skiers, assuming that an empty cabin
is waiting for the first skier. For each skier, record the time between his or her arrival and the time at
which the cabin completes its journey, and find the mean of these times. This is an estimate of the
“mean time in system”.
[6]
(ii) Explain why the assumption that an empty cabin is waiting for the first skier will also model the
situation in which the first skier arrives just after a cabin has departed, assuming that skiers are arriving
at a rate of 1 skier every 2 seconds.
[1]
(iii) Run your simulation 10 times, recording the mean time in system for each of your 10 runs.
[2]
(iv) Criticise the modelling assumptions.
[1]
In a modernisation the cabins are replaced by cabins which each hold 120 skiers, but which can complete
the journey in 3 minutes, so that the capacity is unchanged at 2400 skiers per hour.
(v) Simulate this system for 120 skiers, assuming that an empty cabin is waiting for the first skier. Repeat
your simulation 10 times and compare this mean time in system to that in the old system.
[3]
In fact, the new cabins are programmed to depart when full or at 4-minute intervals, whichever happens
first.
(vi) Simulate this system, assuming that an empty cabin is waiting for the first skier. Explain how you dealt
with the limit of no more than 4 minutes between departures.
Repeat your simulation 10 times, each time assuming that an empty cabin is waiting for the first skier.
Record how many skiers were loaded in each simulation. Compare the mean time in system to that
which you observed in part (iii).
[4]
(vii) When would it become necessary to modify the modelling assumption that an empty cabin is waiting
for the first skier?
[1]
© OCR 2014
4773/01 Jun14
Turnover
4
3
(i) Put the following LP into LINDO format, run it, and explain what it does.
max M
st
M < 65
M < 37
M < 19
M < 54
M < 23
end
[4]
The table gives the complete set of shortest distances between the vertices of a network.
A
B
C
D
E
A
0
23
42
35
52
B
23
0
37
29
43
C
42
37
0
18
50
D
35
29
18
0
32
E
52
43
50
32
0
The following LP finds the minimum of the row maxima. The “R” variables give the row maxima. “M”
gives the minimum of these, using the approach of part (i). Maximising Y forces the R variables to be as
small as possible and M to be as large as possible at the same time.
maximise
st
end
© OCR 2014
Y
Y=M−2R1−2R2−2R3−2R4−2R5
M < R1
M < R2
M < R3
M < R4
M < R5
R1 > 23
R1 > 42
R1 > 35
R1 > 52
R2 > 23
R2 > 37
R2 > 29
R2 > 43
R3 > 42
R3 > 37
R3 > 18
R3 > 50
R4 > 35
R4 > 29
R4 > 18
R4 > 32
R5 > 52
R5 > 43
R5 > 50
R5 > 32
4773/01 Jun14
5
(ii) Put the LP into LINDO format, and run it. You will need to put the instruction “Free Y” on a line after
your “End” statement, since Y will take a negative value.
Interpret the output. You should explain what gives the minimum of the row maxima, and how you can
find which row contains this.
[6]
John is designing a hot water system for his new house. The vertices on the network represent positions
for taps. The arcs represent possible connections between taps. The weights represent the lengths in metres
of pipes that would be needed to connect taps. John wants to position the hot water cylinder so that the
maximum distance from it to a tap is as small as possible.
B
5
A
2
3
9
3
F
E
C
6
4
7
5
6
D
(iii) Explain the steps that you would go through to use LP to find the best vertex for the location of John’s
hot water cylinder. You do not need to produce or run an LP.
[5]
(iv) Use inspection to find the four best vertices at which to position the cylinder, each with a maximum
distance of 8 metres from other vertices.
[1]
(v) Why might LP be used to solve real-world problems of this nature when they seem easy to do by
inspection?
[1]
(vi) Show that, in fact, the best position for the cylinder is not at a vertex.
[1]
Question4beginsonpage6
© OCR 2014
4773/01 Jun14
Turnover
6
4
KPQ Logistics is a distribution company. It has two distribution centres, one in the north of the country
(N) and one in the south (S). It serves producers of goods, taking deliveries into its distribution centres and
delivering from the centres on to shops.
The company maintains a large database detailing the costs of delivery from producers to the company’s
distribution centres, and costs of delivery from the centres to customers.
The company can move up to 200 tonnes a week between its two centres at a cost of £5 per tonne.
HJS Manufacturers produce a chemical in its two factories, F1 and F2. It is sold by the tonne, mainly to
10 customers located in different parts of the country. HJS employs KPQ to distribute this chemical.
Next week’s production schedule and orders are detailed below, as are the transport costs from factories to
centres and from centres to customers.
Next week’s production (tonnes)
Customer
Next week’s orders (tonnes)
F1
F2
1000
750
A
B
C
D
E
F
G
H
I
J
170
70
400
150
80
120
50
175
200
300
Transport costs (£ per tonne)
F1
F2
N
12
7
S
5
15
A
B
C
D
E
F
G
H
I
J
N
13
15
7
7
20
21
14
11
8
9
S
4
12
14
17
11
10
7
8
14
15
KPQ wishes to construct a transport plan for next week which will minimise its costs.
(i) Formulate this as an LP problem.
[10]
(ii) Solve your LP, interpret the solution, and report the results for KPQ.
(iii) Suggest two ways in which costs might be reduced further without incurring any capital expense. [2]
ENDOFQUESTIONPAPER
© OCR 2014
4773/01 Jun14
[6]
GCE
Mathematics (MEI)
Unit 4773: Decision Mathematics Computation
Advanced GCE
Mark Scheme for June 2014
Oxford Cambridge and RSA Examinations
4773
1.
Mark Scheme
Annotations and abbreviations
Annotation in scoris
and 
BOD
FT
ISW
M0, M1
A0, A1
B0, B1
SC
^
MR
Highlighting
Other abbreviations
in mark scheme
E1
U1
G1
M1 dep*
cao
oe
rot
soi
www
Meaning
Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or
unstructured) and on each page of an additional object where there is no candidate response.
Benefit of doubt
Follow through
Ignore subsequent working
Method mark awarded 0, 1
Accuracy mark awarded 0, 1
Independent mark awarded 0, 1
Special case
Omission sign
Misread
Meaning
Mark for explaining
Mark for correct units
Mark for a correct feature on a graph
Method mark dependent on a previous mark, indicated by *
Correct answer only
Or equivalent
Rounded or truncated
Seen or implied
Without wrong working
1
June 2014
4773
2.
a
Mark Scheme
June 2014
Subject-specific Marking Instructions for GCE Mathematics (MEI) Decision strand
Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full
marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded.
For subsequent marking you must make it clear how you have arrived at the mark you have awarded.
b
An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to
assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on
the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always
be looked at and anything unfamiliar must be investigated thoroughly.
Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work
must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according
to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should
contact your Team Leader.
c
The following types of marks are available.
M
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are
not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an
intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by
substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be
specified.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the
associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded.
B
Mark for a correct result or statement independent of Method marks.
2
4773
Mark Scheme
June 2014
E
A given result is to be established or a result has to be explained. This usually requires more working or explanation than the
establishment of an unknown result.
Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is
ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a
candidate passes through the correct answer as part of a wrong argument.
d
When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says
otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is
dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone
wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when
two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
e
The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results.
Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not
given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there
may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme
rationale. If this is not the case please consult your Team Leader.
Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow
through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within
the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.
f
Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates
are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small
variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally
be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation
regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme
rationale. If in doubt, contact your Team Leader.
g
Rules for replaced work
If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do
as the candidate requests.
3
4773
Mark Scheme
June 2014
If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last
(complete) attempt and ignore the others.
NB Follow these maths-specific instructions rather than those in the assessor handbook.
h
For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark
according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate,
though this may differ for some units. This is achieved by withholding one A mark in the question.
Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
4
4773
Mark Scheme
Question
1 (i)
Answer
130
115
85
70
85
115
130
115
85
70
85
115
130
115
85
70
85
115
130
June 2014
Marks
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
Settles to cycle ...
85
70
85
115
130
115
Guidance
B1
130 and 115 used
correctly
B1
un = un-1 – 50 + x
B1
x = 150 – un-2
B1
dealing with -ve orders
B1
cycling
[5]
1
(ii)
un+2 = un+1 ‒ 50 + (150 ‒ un), i.e. un+2 ‒ un+1 + un = 100
M1
A1
[2]
1
linear 2nd order RR for
un+2
either expression
4773
Question
1 (iii)
Mark Scheme
Answer
2
–
=
Marks
+1=0
1
1 4
2
M1
1
3
A1
2
“oscillations” or “cycles”
1
(iv)
June 2014
Guidance
2
–
+ 1 ft
ft if discriminant < 0
B1
un+2 = un+1 ‒ 50 + 50 + (150 ‒ un) (i.e. un+2 ‒ un+1 + un = 150 )
auxiliary equation 2 ‒ + = 0
discriminant of auxillary = 1 ‒ 4 = 0 for = 0.25
OR
[3]
B1
B1
B1 B1
OR
for getting ( ‒ 0.5)2 = 0 when
B1B1
= 0.25, so only one solution
discriminant + soln
factorisation +
comment
[4]
1
(v)
130
115
120
128.75
136.25
141.5625
145
147.1094
148.3594
149.082
149.4922
149.7217
149.8486
149.9182
149.9561
B1
by RR
B1
by formula
[2]
2
4773
Question
1 (vi)
Mark Scheme
Answer
130
115
120
129
137
142
145
147
148
149
150
150
June 2014
Marks
Guidance
B1
rounding
[1]
1
(vii)
OK for demand
87
B1
[1]
3
4773
Question
2 (i)
Mark Scheme
June 2014
Answer
arr rate
0.5
arrival int arrival time number
1
1
1
2
3
2
2
5
3
2
7
4
1
8
5
1
9
6
capacity
run time
serv time
550
548
546
544
543
542
160
240
dep time
arr time
311
551
mean serv
i.e.
395.56 secs
6 mins
36
secs
Marks
Guidance
M1
arrival interval table +
A1
lookup
B1
arrival times
M1
subtraction of arrival
times
A1
finding service times
B1
mean service time
[6]
2 (ii)
2 (iii)
At an arrival rate of 1 skier every 2 seconds the queue length will not exceed the cabin capacity, so skiers will
simply be waiting on the platform rather than in the cabin – each for the same time.
e.g.
6m45s
6m38s
6m49s
6m40s
6m24s
6m33s
6m27s
6m54s
6m21s 6m34s
B1
[1]
M1A1
[2]
2 (iv)
Skiers rarely arrive singly
B1
[1]
4
4773
Question
2 (v)
Mark Scheme
June 2014
Answer
arr rate
0.5
arrival int arrival time number
2
2
1
3
1
4
3
7
2
9
2
11
e.g. 5.02 4.52 4.46 5.25 5.13
capacity
run time
Marks
120
180
dep time
serv time
1
419
arr time
2
418
3
417
4
414
mean serv
5
412
i.e.
6
410
4.47 4.53 5.05 5.01 4.43
Down from about 6.5 mins to 5 mins.
241
B1
421
new fill criterion +
new run time
296.60 secs
4 mins
57
secs
B1
B1
[3]
5
Guidance
4773
Question
2 (vi)
Mark Scheme
June 2014
Answer
arr rate
arrival
int
0.5
arrival
time
3
1
2
1
3
3
232
233
235
236
239
242
capacity
run time
120
180
loading
number serv time flag
115
188
1
116
187
1
117
185
1
118
184
1
119
181
1
120
178
0
Marks
dep time
loaded
188
187
185
184
181
0
240
119
arr time
mean serv
i.e.
B1
B1
420
297.92 secs
4 mins
Mean time should be reduced slightly.
58
secs
B1
B1
[4]
2 (vii)
If the arrival rate is such that more than 120 skiers could arrive during the journey time of 3 minutes.
B1
(A queue would then have to be modelled, with some skiers being left behind to impinge on the next trip.)
[1]
6
Guidance
recording 10
dep times and
loadings
departure time
= min of full
time & 4 mins
adjusting
mean service
time
correctlydifficultneedn’t be
automated
4773
Question
3 (i)
Mark Scheme
Answer
max
st
June 2014
Marks
M
M<65
M<37
M<19
M<54
M<23
end
Gives M = 19 ... minimum
B1
entering
B1
running
B1
B1
M = 19
minimum
[4]
7
Guidance
4773
Question
3 (ii)
Mark Scheme
Answer
max
st
Y
M-R1<0
M-R2<0
M-R3<0
M-R4<0
M-R5<0
Y-M+2R1+2R2+2R3+2R4+2R5=0
R1>23
R1>42
R1>35
R1>52
R2>23
R2>37
R2>29
R2>43
R3>42
R3>37
R3>18
R3>50
R4>35
R4>29
R4>18
R4>32
R5>52
R5>43
R5>50
R5>32
end
free Y
June 2014
Marks
Guidance
B1
B1
B1
Y constraint
M constraints
rest
B1
running
M gives the row minimax.
B1
The row is given by the subscript on the Ri which matches M.
B1
[6]
3
(iii)
Need the minimax of the shortest distances from each vertex
Find the matrix of shortest distances. Need the minimax row (or column).
Solve using LP as per part (ii) (or by inspection for this small problem).
B1
minimax
B1
shortest distances
B1B1
B1
[5]
8
4773
Mark Scheme
Question
3 (iv)
Answer
Best vertices are A, B, D and F (all with a minimax of 8).
3
Problem size ... big
(v)
June 2014
Marks
B1
[1]
B1
[1]
3
(vi)
e.g.
Point midway between A and B has minimax of 6.5.
(Or 6 if 0.6 of way from A to B.)
B1
[1]
9
Guidance
4773
Question
4 (i)
Mark Scheme
Answer
min
st
12F1N+5F1S+7F2N+15F2S+5NS+5SN+13NA+4SA+15NB+12SB+7NC+14SC+7ND+17SD+20NE
+11SE+21NF+10SF+14NG+7SG+11NH+8SH+8NI+14SI+9NJ+15SJ
F1N+F1S<1000
F2N+F2S<750
SN<200
NS<200
NA+SA=170
NB+SB=70
NC+SC=400
ND+SD=150
NE+SE=80
NF+SF=120
NG+SG=50
NH+SH=175
NI+SI=200
NJ+SJ=300
NA+NB+NC+ND+NE+NF+NG+NH+NI+NJ+NS-F1N-F2N-SN=0
SA+SB+SC+SD+SE+SF+SG+SH+SI+SJ+SN-F1S-F2S-NS=0
June 2014
Marks
M1
A2
B1
B1
B1
M1
A1
B1
B1
end
[10]
10
Guidance
–1 each error/omission
production constraints
internal movement
constraints (both)
supply requirements
linking equations
4773
4
Question
(ii)
Mark Scheme
Answer
June 2014
Marks
Variable Value
F1N
0.000000
F1S
965.0000
F2N
750.0000
F2S
0.000000
NS
0.000000
SN
200.0000
NA
0.000000
SA
170.0000
NB
0.000000
SB
70.00000
NC
400.0000
SC
0.000000
ND
150.0000
SD
0.000000
NE
0.000000
SE
80.00000
NF
0.000000
SF
120.0000
NG
0.000000
SG
50.00000
NH
0.000000
SH
175.0000
NI
200.0000
SI
0.000000
NJ
200.0000
SJ
100.0000
B1
11
Guidance
running
4773
Mark Scheme
Question
June 2014
Answer
Marks
To centres:
N
S
F1
0
965
F2
750
0
N
0
0
S
200
0
D
150
0
E
0
80
M1
A1
From centres:
N
S
A
0
170
B
0
70
C
400
0
F
0
120
G
0
50
H
0
175
Total cost = £25175
I
200
0
J
200
100
M1
A1
B1
[6]
4
(iii)
Deliver direct from factories to some shops.
B1
Relax/remove constraint on tonnage moved between centres.
B1
[2]
12
Guidance
For a description of how UMS marks are calculated see:
www.ocr.org.uk/learners/ums_results.html
Unit level raw mark and UMS grade boundaries June 2014 series
AS GCE / Advanced GCE / AS GCE Double Award / Advanced GCE Double Award
GCE Mathematics (MEI)
4751/01 (C1) MEI Introduction to Advanced Mathematics
4752/01 (C2) MEI Concepts for Advanced Mathematics
4753/01
4753/02
4753/82
4753
4754/01
(C3) MEI Methods for Advanced Mathematics with Coursework: Written Paper
(C3) MEI Methods for Advanced Mathematics with Coursework: Coursework
(C3) MEI Methods for Advanced Mathematics with Coursework: Carried Forward Coursework Mark
(C3) MEI Methods for Advanced Mathematics with Coursework
(C4) MEI Applications of Advanced Mathematics
4755/01 (FP1) MEI Further Concepts for Advanced Mathematics
4756/01 (FP2) MEI Further Methods for Advanced Mathematics
4757/01 (FP3) MEI Further Applications of Advanced Mathematics
4758/01
4758/02
4758/82
4758
4761/01
(DE) MEI Differential Equations with Coursework: Written Paper
(DE) MEI Differential Equations with Coursework: Coursework
(DE) MEI Differential Equations with Coursework: Carried Forward Coursework Mark
(DE) MEI Differential Equations with Coursework
(M1) MEI Mechanics 1
4762/01 (M2) MEI Mechanics 2
4763/01 (M3) MEI Mechanics 3
4764/01 (M4) MEI Mechanics 4
4766/01 (S1) MEI Statistics 1
4767/01 (S2) MEI Statistics 2
4768/01 (S3) MEI Statistics 3
4769/01 (S4) MEI Statistics 4
4771/01 (D1) MEI Decision Mathematics 1
4772/01 (D2) MEI Decision Mathematics 2
4773/01 (DC) MEI Decision Mathematics Computation
4776/01
4776/02
4776/82
4776
4777/01
(NM) MEI Numerical Methods with Coursework: Written Paper
(NM) MEI Numerical Methods with Coursework: Coursework
(NM) MEI Numerical Methods with Coursework: Carried Forward Coursework Mark
(NM) MEI Numerical Methods with Coursework
(NC) MEI Numerical Computation
4798/01 (FPT) Further Pure Mathematics with Technology
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Raw
UMS
Max Mark
72
100
72
100
72
18
18
100
90
100
72
100
72
100
72
100
72
18
18
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
18
18
100
72
100
72
100
a
61
80
57
80
58
15
15
80
68
80
63
80
60
80
57
80
63
15
15
80
57
80
57
80
55
80
48
80
61
80
60
80
61
80
56
80
51
80
46
80
46
80
54
14
14
80
55
80
57
80
b
56
70
51
70
52
13
13
70
61
70
57
70
54
70
51
70
56
13
13
70
49
70
49
70
48
70
41
70
53
70
53
70
54
70
49
70
46
70
41
70
40
70
48
12
12
70
47
70
49
70
c
51
60
45
60
47
11
11
60
54
60
51
60
48
60
45
60
50
11
11
60
41
60
41
60
42
60
34
60
46
60
46
60
47
60
42
60
41
60
36
60
34
60
43
10
10
60
39
60
41
60
d
46
50
39
50
42
9
9
50
47
50
45
50
42
50
39
50
44
9
9
50
34
50
34
50
36
50
28
50
39
50
40
50
41
50
35
50
36
50
31
50
29
50
38
8
8
50
32
50
33
50
e
42
40
33
40
36
8
8
40
41
40
40
40
36
40
34
40
37
8
8
40
27
40
27
40
30
40
22
40
32
40
34
40
35
40
28
40
31
40
26
40
24
40
32
7
7
40
25
40
26
40
u
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Raw
UMS
Raw
UMS
Raw
UMS
Max Mark
72
100
72
100
72
100
a
61
80
55
80
56
80
b
53
70
48
70
48
70
c
46
60
41
60
41
60
d
39
50
34
50
34
50
e
32
40
27
40
27
40
u
0
0
0
0
0
0
GCE Statistics (MEI)
G241/01 (Z1) Statistics 1
G242/01 (Z2) Statistics 2
G243/01 (Z3) Statistics 3
Unit level raw mark and UMS grade boundaries June 2013 series: GCE
1