| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2013 |
| Session | January |
| 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 algorithmic trace question requiring careful bookkeeping but minimal mathematical insight. Part (i) involves mechanically following given steps with basic arithmetic, while part (ii) requires recognizing a pattern from part (i). The conceptual demand is low—no proof, problem-solving, or deep understanding of why the algorithm works is needed, just accurate execution of instructions. |
| Spec | 7.03a Algorithm definition: input, output, deterministic, finite7.03c Working with algorithms: trace, interpret, adapt |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Step 1: \(x = 0.44\) | B1 | cao |
| Step 2: \(oldr = 1\); Steps 3,4,5: \(i=1, j=0.5, k=0.5\); Step 6: \(change = 0.22\); Step 7: \(newr = 1.22\); Step 9: \(oldr = 1.22\) | B1 | set-up (i.e. as far as 1.22) |
| Steps 10,11,12: \(i=2, j=-0.5, k=-0.125\); Step 13: \(change = -0.0242\); Step 14: \(newr = 1.1958\); Step 15: \( | change | = 0.0242\); Step 9: \(oldr = 1.1958\) |
| Steps 10,11,12: \(i=3, j=-1.5, k=0.0625\); Step 13: \(change = 0.005324\); Step 14: \(newr = 1.201124\); Step 15: \( | change | = 0.005324\); Step 9: \(oldr = 1.201124\) |
| Steps 10,11,12: \(i=4, j=-2.5, k=-0.03906\); Step 13: \(change = -0.0014641\); Step 14: \(newr = 1.1996599\); Step 15: \( | change | = 0.0014641\); Step 17: \(1.1996599\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.22 - 0.0242 - 0.005324 - 0.0014641 = 0.7490119\) | M1 A1 | use of \(-0.44\); as shown; SC1 (cao) for algorithm repeated or answer only |
| [2] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Step 1: $x = 0.44$ | B1 | cao |
| Step 2: $oldr = 1$; Steps 3,4,5: $i=1, j=0.5, k=0.5$; Step 6: $change = 0.22$; Step 7: $newr = 1.22$; Step 9: $oldr = 1.22$ | B1 | set-up (i.e. as far as 1.22) |
| Steps 10,11,12: $i=2, j=-0.5, k=-0.125$; Step 13: $change = -0.0242$; Step 14: $newr = 1.1958$; Step 15: $|change| = 0.0242$; Step 9: $oldr = 1.1958$ | B1 | 3 steps correct |
| Steps 10,11,12: $i=3, j=-1.5, k=0.0625$; Step 13: $change = 0.005324$; Step 14: $newr = 1.201124$; Step 15: $|change| = 0.005324$; Step 9: $oldr = 1.201124$ | B1 | new estimate (1.1958) |
| Steps 10,11,12: $i=4, j=-2.5, k=-0.03906$; Step 13: $change = -0.0014641$; Step 14: $newr = 1.1996599$; Step 15: $|change| = 0.0014641$; Step 17: $1.1996599$ | B1 B1 | iteration (to 1.201124); iteration and end |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.22 - 0.0242 - 0.005324 - 0.0014641 = 0.7490119$ | M1 A1 | use of $-0.44$; as shown; SC1 (cao) for algorithm repeated or answer only |
| **[2]** | | |
---3 The following algorithm computes an estimate of the square root of a number which is between 0 and 2.\\
Step 1 Subtract 1 from the number and call the result $x$\\
Step 2 Set oldr = 1\\
Step 3 Set $i = 1$\\
Step 4 Set $j = 0.5$\\
Step 5 Set $k = 0.5$\\
Step 6 Set change $= x ^ { i } \times k$\\
Step 7 Set newr $=$ oldr + change\\
Step 8 If $- 0.005 <$ change < 0.005 then go to Step 17\\
Step 9 Set oldr = newr\\
Step 10 Set $i = i + 1$\\
Step 11 Set $j = j - 1$\\
Step 12 Set $k = k \times j \div i$\\
Step 13 Set change $= x ^ { i } \times k$\\
Step 14 Set newr $=$ oldr + change\\
Step 15 If $- 0.005 <$ change < 0.005 then go to Step 17\\
Step 16 Go to Step 9\\
Step 17 Print out newr\\
(i) Use the algorithm to find an estimate of the square root of 1.44 , showing all of the steps.\\
(ii) Consider what happens if the algorithm is applied to 0.56 , and then use your four values of change from part (i) to calculate an estimate of the square root of 0.56 .
\hfill \mbox{\textit{OCR MEI D1 2013 Q3 [8]}}