| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Moderate -0.3 This is a straightforward algorithm trace question requiring careful arithmetic with given formulas. Students follow explicit steps with no conceptual insight needed—just substitution and comparison. The golden ratio context is interesting but not exploited mathematically. Slightly easier than average due to mechanical nature, though the multi-step trace and 3dp accuracy requirement prevent it from being trivial. |
| Spec | 7.03a Algorithm definition: input, output, deterministic, finite7.03b Algorithm awareness: uses and practical limitations7.03c Working with algorithms: trace, interpret, adapt7.03f Worst case complexity: T(n) as function of problem size |
| Step 1 | Input A |
| Step 2 | Input B , where \(\mathrm { B } > \mathrm { A }\) |
| Step 3 | Let \(\mathrm { R } = \mathrm { A } + \left( \frac { \sqrt { 5 } - 1 } { 2 } \right) \times ( \mathrm { B } - \mathrm { A } )\) |
| Step 4 | Let \(\mathrm { L } = \mathrm { A } + \mathrm { B } - \mathrm { R }\) |
| Step 5 | Find \(f ( \mathrm {~L} )\) and \(f ( \mathrm { R } )\) |
| Step 6 | If \(\mathrm { f } ( \mathrm { L } ) \leqslant \mathrm { f } ( \mathrm { R } )\) then let \(\mathrm { B } = \mathrm { R }\) and go to Step 8 |
| Step 7 | If \(\mathrm { f } ( \mathrm { L } ) > \mathrm { f } ( \mathrm { R } )\) then let \(\mathrm { A } = \mathrm { L }\) and go to Step 8 |
| Step 8 | If \(\mathrm { B } - \mathrm { A } < 0.1\) then go to step 10 |
| Step 9 | Go to step 3 |
| Step 10 | Print \(\frac { ( \mathrm { A } + \mathrm { B } ) } { 2 }\) and stop |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A=3, L=3.382, R=3.618, B=4, f(L)=2.146, f(R)=1.910 | B1, B1 | R and L; f(R) and f(L). -1 once only for incorrect accuracy, but condone 1.91. Surds OK, but lose accuracy mark. (Q says 3dp.) |
| A=3.382, L=3.618, R=3.764, B=4, f(L)=1.910, f(R)=1.875 | B1, B1, B1 | A; L and R; f(L) and f(R) |
| 3.618 | B1 | A |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Saves a function evaluation | B1 | Has to be a comment about function values. |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. Setting the control on a gas fire to achieve a room temperature of 20°C. Function could be \((\text{temp}-20)^2\). Domain cannot be time based. | B1 | Optimisation with need to sample at discrete intervals. "Deepest point in seabed" example acceptable, assuming depth soundings taken at points, and ignoring the fact that domain is two dimensional rather than one dimensional. |
| [1] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| A=3, L=3.382, R=3.618, B=4, f(L)=2.146, f(R)=1.910 | B1, B1 | R and L; f(R) and f(L). -1 once only for incorrect accuracy, but condone 1.91. Surds OK, but lose accuracy mark. (Q says 3dp.) |
| A=3.382, L=3.618, R=3.764, B=4, f(L)=1.910, f(R)=1.875 | B1, B1, B1 | A; L and R; f(L) and f(R) |
| 3.618 | B1 | A |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Saves a function evaluation | B1 | Has to be a comment about function values. |
| **[1]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. Setting the control on a gas fire to achieve a room temperature of 20°C. Function could be $(\text{temp}-20)^2$. Domain cannot be time based. | B1 | Optimisation with need to sample at discrete intervals. "Deepest point in seabed" example acceptable, assuming depth soundings taken at points, and ignoring the fact that domain is two dimensional rather than one dimensional. |
| **[1]** | | |
---2 This question concerns the following algorithm which operates on a given function, f. The algorithm finds a point between A and B at which the function has a minimum, with a maximum error of 0.05 .
\begin{center}
\begin{tabular}{|l|l|}
\hline
Step 1 & Input A \\
\hline
Step 2 & Input B , where $\mathrm { B } > \mathrm { A }$ \\
\hline
Step 3 & Let $\mathrm { R } = \mathrm { A } + \left( \frac { \sqrt { 5 } - 1 } { 2 } \right) \times ( \mathrm { B } - \mathrm { A } )$ \\
\hline
Step 4 & Let $\mathrm { L } = \mathrm { A } + \mathrm { B } - \mathrm { R }$ \\
\hline
Step 5 & Find $f ( \mathrm {~L} )$ and $f ( \mathrm { R } )$ \\
\hline
Step 6 & If $\mathrm { f } ( \mathrm { L } ) \leqslant \mathrm { f } ( \mathrm { R } )$ then let $\mathrm { B } = \mathrm { R }$ and go to Step 8 \\
\hline
Step 7 & If $\mathrm { f } ( \mathrm { L } ) > \mathrm { f } ( \mathrm { R } )$ then let $\mathrm { A } = \mathrm { L }$ and go to Step 8 \\
\hline
Step 8 & If $\mathrm { B } - \mathrm { A } < 0.1$ then go to step 10 \\
\hline
Step 9 & Go to step 3 \\
\hline
Step 10 & Print $\frac { ( \mathrm { A } + \mathrm { B } ) } { 2 }$ and stop \\
\hline
\end{tabular}
\end{center}
(i) Working correct to three decimal places, perform two iterations of the algorithm for $\mathrm { f } ( x ) = 2 x ^ { 2 } - 15 x + 30$, when $\mathrm { A } = 3$ and $\mathrm { B } = 4$. Start at Step 1 and stop when you reach Step 8 for the second time.\\
(ii) The $\left( \frac { \sqrt { 5 } - 1 } { 2 } \right)$ factor in Step 3 ensures that either the new ' $L$ ' is equal to the old ' $R$ ', or vice versa. Why is this important?\\
(iii) This algorithm is used when the function is not known explicitly, but where its value can be found for any given input. Give a practical example of where it might be used.
\hfill \mbox{\textit{OCR MEI D1 2012 Q2 [8]}}