OCR Further Discrete AS 2024 June — Question 7 13 marks

Exam BoardOCR
ModuleFurther Discrete AS (Further Discrete AS)
Year2024
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeBin packing problems
DifficultyStandard +0.3 This is a straightforward bin packing question testing standard algorithms and basic reasoning. Parts (a)-(c) are routine applications of packing methods, (d) is bookwork, (e) is simple counting, and (f)-(g) require modest logical thinking but no sophisticated insight. The multi-part structure inflates marks but each component is below average difficulty for A-level.
Spec7.03l Bin packing: next-fit, first-fit, first-fit decreasing, full bin7.03m Packing extensions: 2D/3D packing and knapsack problems

7 Six items need to be packed into bins. Each bin has size \(k\), where \(k\) is an integer. The sizes of the items are \(\begin{array} { l l l l l l } 8 & 5 & 4 & 7 & 3 & 3 \end{array}\)
  1. Show a way to pack the items into two bins of size \(k = 18\).
  2. If exactly three bins are needed, determine the possible values of \(k\).
  3. Use the first-fit method to pack the items into bins of size 12.
  4. Explain why, in general, the first-fit algorithm does not necessarily find the most efficient solution to a packing problem. A possible way to calculate the run-time of the first-fit method is to count for each item the number of bins, excluding full bins, that are considered before the bin where each item is packed. The total count for all packed items is then used to calculate the run-time of the method.
  5. Determine the total count for the packing used in part (c). A student says that when \(k = 13\) the total count is 0 because the first bin fills up before the second bin is considered.
  6. Explain why the student's argument is not correct.
  7. Determine the least possible value of \(k\) for which the total count is 0 . \section*{END OF QUESTION PAPER}

Question 7:
AnswerMarks Guidance
7(a) e.g.
Bin 1 8 5 4
AnswerMarks Guidance
Bin 2 7 3 3B1
[1]1.1 Any valid packing of 8 5 4 7 3 3 into two bins of size k = 18
e.g. Bin 1 8 7 3 e.g. Bin 1 8 7
Bin 2 5 4 3 Bin 2 5 4 3 3
AnswerMarks Guidance
7(b) Total weight = 30
Need three bins, two are not enough so k < 15
(but 14 is possible)
Three bins are sufficient so k 10
k = 10 would mean no waste but bin with 8
would not be full, hence k  11
k = 11 is achievable, e.g. 8+3, 7+4, 5+3
AnswerMarks
So 11  k  14B1
M1
A1
AnswerMarks
[3]2.1
3.1b
AnswerMarks
2.2aUB < 15 or UB = 14
Stated or by showing packing into two bins each with exactly 15
Sensible attempt to find LB as 10 or 11
k  {11, 12, 13, 14}
Given that k is an integer so 11 k < 15, 10 < k  14 or 10 < k < 15
are also correct
Or given as separate statements
AnswerMarks Guidance
7(c) 8 5 4 7 3 3
Bin 1 8 4
Bin 2 5 7
AnswerMarks
Bin 3 3 3M1
A1
AnswerMarks
[2]1.1
1.1First-fit with k = 12, evidenced by 8, 5, 4 correctly placed
All correct
AnswerMarks Guidance
7(d) (Solution is not necessarily optimal because)
there may be a packing that uses fewer binsB1
[1]2.2b First-fit will find a good packing but may not find the packing that
uses the fewest bins
AnswerMarks Guidance
7(e) 8  0 (first item), 5  1 (check bin 1)
4  0 (fits in bin 1), bin 1 full
7  0 (fits in first bin that is not full), bin 2 full
3  0, 3  0 (fits in first bin that is not full)
AnswerMarks
0 + 1 + 0 + 0 + 0 + 0 = 1M1
A1
AnswerMarks
[2]1.1
1.1Evidence that 5  1 and 4  0
e.g. 0 + 1 + 0 + …
1 from valid working seen
AnswerMarks Guidance
7(f) When k = 13, 8 and 5 will go in bin 1 and it
will then be full and no longer considered
4 and 7 then go in bin 2 which is not full
But bin 2 does not have enough room for a 3
So need to check bin 2 before putting 3 (or 3’s)
AnswerMarks
into bin 3M1
A1
AnswerMarks
[2]2.4
2.3Bin 2 will not be full
This may be done by showing packing
Bin 1: 8 5 Bin 2: 4 7 Bin 3: 3 3
May be implied from correct A mark
Explain why argument is wrong
First 3 adds 1 to the count
Each 3 adds 1 to the count

Total count will be 0+0+0+0+1+1 = 2

AnswerMarks Guidance
7(g) Need full bins until last
8+5+4
AnswerMarks
= 17M1
A1
AnswerMarks
[2]2.2b
1.1May show packing with 8, 5, 4 in Bin 1 and 7, 3, 3 in Bin 2
(packed in this order)
17
SC B1 17 with no appropriate reasoning
PMT
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Question 7:
7 | (a) | e.g.
Bin 1 8 5 4
Bin 2 7 3 3 | B1
[1] | 1.1 | Any valid packing of 8 5 4 7 3 3 into two bins of size k = 18
e.g. Bin 1 8 7 3 e.g. Bin 1 8 7
Bin 2 5 4 3 Bin 2 5 4 3 3
7 | (b) | Total weight = 30
Need three bins, two are not enough so k < 15
(but 14 is possible)
Three bins are sufficient so k 10
k = 10 would mean no waste but bin with 8
would not be full, hence k  11
k = 11 is achievable, e.g. 8+3, 7+4, 5+3
So 11  k  14 | B1
M1
A1
[3] | 2.1
3.1b
2.2a | UB < 15 or UB = 14
Stated or by showing packing into two bins each with exactly 15
Sensible attempt to find LB as 10 or 11
k  {11, 12, 13, 14}
Given that k is an integer so 11 k < 15, 10 < k  14 or 10 < k < 15
are also correct
Or given as separate statements
7 | (c) | 8 5 4 7 3 3
Bin 1 8 4
Bin 2 5 7
Bin 3 3 3 | M1
A1
[2] | 1.1
1.1 | First-fit with k = 12, evidenced by 8, 5, 4 correctly placed
All correct
7 | (d) | (Solution is not necessarily optimal because)
there may be a packing that uses fewer bins | B1
[1] | 2.2b | First-fit will find a good packing but may not find the packing that
uses the fewest bins
7 | (e) | 8  0 (first item), 5  1 (check bin 1)
4  0 (fits in bin 1), bin 1 full
7  0 (fits in first bin that is not full), bin 2 full
3  0, 3  0 (fits in first bin that is not full)
0 + 1 + 0 + 0 + 0 + 0 = 1 | M1
A1
[2] | 1.1
1.1 | Evidence that 5  1 and 4  0
e.g. 0 + 1 + 0 + …
1 from valid working seen
7 | (f) | When k = 13, 8 and 5 will go in bin 1 and it
will then be full and no longer considered
4 and 7 then go in bin 2 which is not full
But bin 2 does not have enough room for a 3
So need to check bin 2 before putting 3 (or 3’s)
into bin 3 | M1
A1
[2] | 2.4
2.3 | Bin 2 will not be full
This may be done by showing packing
Bin 1: 8 5 Bin 2: 4 7 Bin 3: 3 3
May be implied from correct A mark
Explain why argument is wrong
First 3 adds 1 to the count
Each 3 adds 1 to the count
Total count will be 0+0+0+0+1+1 = 2
7 | (g) | Need full bins until last
8+5+4
= 17 | M1
A1
[2] | 2.2b
1.1 | May show packing with 8, 5, 4 in Bin 1 and 7, 3, 3 in Bin 2
(packed in this order)
17
SC B1 17 with no appropriate reasoning
PMT
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For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
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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.
7 Six items need to be packed into bins.

Each bin has size $k$, where $k$ is an integer.

The sizes of the items are\\
$\begin{array} { l l l l l l } 8 & 5 & 4 & 7 & 3 & 3 \end{array}$
\begin{enumerate}[label=(\alph*)]
\item Show a way to pack the items into two bins of size $k = 18$.
\item If exactly three bins are needed, determine the possible values of $k$.
\item Use the first-fit method to pack the items into bins of size 12.
\item Explain why, in general, the first-fit algorithm does not necessarily find the most efficient solution to a packing problem.

A possible way to calculate the run-time of the first-fit method is to count for each item the number of bins, excluding full bins, that are considered before the bin where each item is packed. The total count for all packed items is then used to calculate the run-time of the method.
\item Determine the total count for the packing used in part (c).

A student says that when $k = 13$ the total count is 0 because the first bin fills up before the second bin is considered.
\item Explain why the student's argument is not correct.
\item Determine the least possible value of $k$ for which the total count is 0 .

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete AS 2024 Q7 [13]}}