OCR Further Discrete 2024 June — Question 6 16 marks

Exam BoardOCR
ModuleFurther Discrete (Further Discrete)
Year2024
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeParametric objective analysis
DifficultyChallenging +1.2 This is a multi-part linear programming question requiring formulation, solving a 2-variable LP graphically, and then analyzing how a parametric change affects optimality. Parts (a)-(c) are standard Further Maths Decision content. Part (d) requires understanding that the optimal solution changes when leftover beads have value, necessitating comparison of corner points with modified objective functions—this adds moderate complexity. Part (e) is a straightforward interpretation question. Overall, this is above-average difficulty due to the parametric analysis in part (d), but the LP itself is routine and the question provides significant scaffolding.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables

6 Sasha is making three identical bead bracelets using amber, brown and red coloured beads. Sasha has 20 amber beads, 12 brown beads and 10 red beads. Each bracelet must use exactly 12 beads.
The profit from selling a bracelet is 6 pence for each amber bead used plus 2 pence for each brown bead used plus 3 pence for each red bead used. Sasha wants to maximise the total profit from selling the three bracelets.
  1. Express Sasha's problem as a linear programming formulation in two variables \(a\) and \(b\), where \(a\) represents the number of amber beads in each bracelet and \(b\) represents the number of brown beads in each bracelet.
  2. Determine how many beads of each colour will be used in each bracelet.
  3. By listing all the feasible solutions, identify an aspect of the optimal solution, other than the profit, that is different from all the other feasible solutions. The beads that are not used in making the bracelets can be sold. The profit from selling each amber bead is \(k\) pence, where \(k\) is an integer, but nothing for each brown or red bead sold. All the previous constraints still apply. Instead of maximising the profit from the bracelets, Sasha wants to maximise the total profit from selling the bracelets and any left over beads. You are given that the optimal solution to the earlier problem does not maximise the total profit from selling the bracelets and any left over beads.
  4. Determine the least possible value of Sasha's maximum total profit.
  5. Why might Sasha not achieve this maximum profit?

Question 6:
AnswerMarks
617
6
AnswerMarks Guidance
65 2
115 7
6(a) 12 – a – b red beads (in each bracelet)
Max P = 3{6a + 2b + 3(12 – a – b)} pence
= 9a – 3b (+ constant)
subject to a ≤ 20/3  a ≤ 6, b ≤ 12/3 = 4
12 – a – b ≤ 10/3  a + b ≥ 9
AnswerMarks
a, b integersB1
M1
A1
B1
B1
AnswerMarks
[5]3.1b
3.4
1.1
3.3
AnswerMarks
3.312 – a – b red beads per bracelet, seen or implied from P expression
Reasonable attempt at objective in terms of these 2 variables only
Or a positive multiple with like terms collected
Correct integer-valued upper bounds for a and b
Or correct with fractions or decimals (3a  20, 3b  12, o.e.) and
statement ‘a, b integers’ o.e. seen
Correct integer-valued lower bound for a + b
Or correct with fractions or decimals (3a + 3b ≥ 26, o.e.) and
statement ‘a+b integer valued’ or ‘a, b integers’ o.e. seen
Ignore inequality a + b  12 (from red  0, but is redundant)
Ignore a ≥ 0, b ≥ 0 if seen (redundant)
AnswerMarks Guidance
6(b) (a, b) = (5, 4) or (6, 3) or (6, 4)
6 amber beads, 3 brown beads, 3 red beadsM1
A1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
3.2aList (at least one) feasible or near feasible (a, b) pair (within + 1 for
each) or sketch constraints
May be implied from final answer
Calculating number of red beads (= 12 – a – b) for their feasible or
near feasible (a, b) pair (within + 1 for each)
with a, b and 12 – a – b non-negative integers
Interpret this as optimal solution for all three colours, using colour
names or initial letters, or variable for red defined, cao.
Not statement 12 – a – b = 3 without ‘red’ or r, o.e.
AnswerMarks Guidance
6(c) (a, b, r) = (5, 4, 3) or (6, 4, 2) or (6, 3, 3)
Optimal solution only uses 3 brown beads in
AnswerMarks
each braceletM1
A1
AnswerMarks
[2]1.1
2.2aGiving all three (and only these) feasible solutions, in any way
Identifying any (known) difference between optimal solution
(6, 3, 3) and both of the others, apart from the profit
AnswerMarks Guidance
6(d) (6, 3, 3) gives profit 153 pence and leaves 2
amber beads (and 3 brown and 1 red)
So total profit is 153 + 2k (pence)
(6, 4, 2) leaves same number of amber beads as
(6, 3, 3) and profit from bracelets is less so total
profit is less
(5, 4, 3) leaves 5 amber beads (and 1 red)
and gives total profit 141 + 5k (pence)
141 + 5k > 153 + 2k  k > 4, min k = 5
When k = 5, total profit is 166 (pence)
Least possible value of Sasha’s maximum total
AnswerMarks
profit is 166 penceM1
A1
A1
M1
A1
AnswerMarks
[5]3.4
3.4
3.4
3.1a
AnswerMarks
2.2aCalculating number of amber beads remaining for any feasible
solution (i.e. 2 amber beads for either a = 6 solution or 5 amber
beads for a = 5 solution, may be implied by sight of 2k or 5k
Explaining why (6, 4, 2) cannot be optimal
(6, 4, 2) leaves 2 amber beads (and 4 red) and gives total profit
150 + 2k < 153 + 2k
May be implied from 153 + 2k as the only total profit for a = 6
Calculating total profit 141 + 5k
Comparing ‘their’ total profits to find a min for k
Or implied from sight of k > 4 or k = 4 or (min) k = 5
166 (pence may be implied) or £1.66
AnswerMarks Guidance
6(e) e.g.
May not sell all the items madeB1
[1]3.5b Any valid reason that does not involve the numerical values of the
profits or the number of beads used or the costs
e.g.
May not have enough time to make all the items
People may not want to purchase beads separately
PMT
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Question 6:
6 | 17
6
6 | 5 | 2 | 7 | 11
11 | 5 | 7 | 7 | 6
6 | (a) | 12 – a – b red beads (in each bracelet)
Max P = 3{6a + 2b + 3(12 – a – b)} pence
= 9a – 3b (+ constant)
subject to a ≤ 20/3  a ≤ 6, b ≤ 12/3 = 4
12 – a – b ≤ 10/3  a + b ≥ 9
a, b integers | B1
M1
A1
B1
B1
[5] | 3.1b
3.4
1.1
3.3
3.3 | 12 – a – b red beads per bracelet, seen or implied from P expression
Reasonable attempt at objective in terms of these 2 variables only
Or a positive multiple with like terms collected
Correct integer-valued upper bounds for a and b
Or correct with fractions or decimals (3a  20, 3b  12, o.e.) and
statement ‘a, b integers’ o.e. seen
Correct integer-valued lower bound for a + b
Or correct with fractions or decimals (3a + 3b ≥ 26, o.e.) and
statement ‘a+b integer valued’ or ‘a, b integers’ o.e. seen
Ignore inequality a + b  12 (from red  0, but is redundant)
Ignore a ≥ 0, b ≥ 0 if seen (redundant)
6 | (b) | (a, b) = (5, 4) or (6, 3) or (6, 4)
6 amber beads, 3 brown beads, 3 red beads | M1
A1
A1
[3] | 1.1
1.1
3.2a | List (at least one) feasible or near feasible (a, b) pair (within + 1 for
each) or sketch constraints
May be implied from final answer
Calculating number of red beads (= 12 – a – b) for their feasible or
near feasible (a, b) pair (within + 1 for each)
with a, b and 12 – a – b non-negative integers
Interpret this as optimal solution for all three colours, using colour
names or initial letters, or variable for red defined, cao.
Not statement 12 – a – b = 3 without ‘red’ or r, o.e.
6 | (c) | (a, b, r) = (5, 4, 3) or (6, 4, 2) or (6, 3, 3)
Optimal solution only uses 3 brown beads in
each bracelet | M1
A1
[2] | 1.1
2.2a | Giving all three (and only these) feasible solutions, in any way
Identifying any (known) difference between optimal solution
(6, 3, 3) and both of the others, apart from the profit
6 | (d) | (6, 3, 3) gives profit 153 pence and leaves 2
amber beads (and 3 brown and 1 red)
So total profit is 153 + 2k (pence)
(6, 4, 2) leaves same number of amber beads as
(6, 3, 3) and profit from bracelets is less so total
profit is less
(5, 4, 3) leaves 5 amber beads (and 1 red)
and gives total profit 141 + 5k (pence)
141 + 5k > 153 + 2k  k > 4, min k = 5
When k = 5, total profit is 166 (pence)
Least possible value of Sasha’s maximum total
profit is 166 pence | M1
A1
A1
M1
A1
[5] | 3.4
3.4
3.4
3.1a
2.2a | Calculating number of amber beads remaining for any feasible
solution (i.e. 2 amber beads for either a = 6 solution or 5 amber
beads for a = 5 solution, may be implied by sight of 2k or 5k
Explaining why (6, 4, 2) cannot be optimal
(6, 4, 2) leaves 2 amber beads (and 4 red) and gives total profit
150 + 2k < 153 + 2k
May be implied from 153 + 2k as the only total profit for a = 6
Calculating total profit 141 + 5k
Comparing ‘their’ total profits to find a min for k
Or implied from sight of k > 4 or k = 4 or (min) k = 5
166 (pence may be implied) or £1.66
6 | (e) | e.g.
May not sell all the items made | B1
[1] | 3.5b | Any valid reason that does not involve the numerical values of the
profits or the number of beads used or the costs
e.g.
May not have enough time to make all the items
People may not want to purchase beads separately
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2024 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
6 Sasha is making three identical bead bracelets using amber, brown and red coloured beads. Sasha has 20 amber beads, 12 brown beads and 10 red beads.

Each bracelet must use exactly 12 beads.\\
The profit from selling a bracelet is 6 pence for each amber bead used plus 2 pence for each brown bead used plus 3 pence for each red bead used.

Sasha wants to maximise the total profit from selling the three bracelets.
\begin{enumerate}[label=(\alph*)]
\item Express Sasha's problem as a linear programming formulation in two variables $a$ and $b$, where $a$ represents the number of amber beads in each bracelet and $b$ represents the number of brown beads in each bracelet.
\item Determine how many beads of each colour will be used in each bracelet.
\item By listing all the feasible solutions, identify an aspect of the optimal solution, other than the profit, that is different from all the other feasible solutions.

The beads that are not used in making the bracelets can be sold.

The profit from selling each amber bead is $k$ pence, where $k$ is an integer, but nothing for each brown or red bead sold.

All the previous constraints still apply.

Instead of maximising the profit from the bracelets, Sasha wants to maximise the total profit from selling the bracelets and any left over beads.

You are given that the optimal solution to the earlier problem does not maximise the total profit from selling the bracelets and any left over beads.
\item Determine the least possible value of Sasha's maximum total profit.
\item Why might Sasha not achieve this maximum profit?
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete 2024 Q6 [16]}}