| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Graphical feasible region identification |
| Difficulty | Moderate -0.8 This is a straightforward algorithmic procedure requiring only arithmetic substitution and basic graph plotting. Part (i) involves mechanical calculation following given steps with no problem-solving, while part (ii) asks for simple recognition that the algorithm plots constraint lines for LP feasible regions—both below average difficulty for A-level. |
| Spec | 7.03a Algorithm definition: input, output, deterministic, finite7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (i) | \(X = \min(25, 8.5) = 8.5\) or equivalent; \(Y = \min(5, 42.5) = 5\) or equivalent; \(X^* = \frac{85-10}{10} = 7.5\) or equivalent; \(Y^* = \frac{25-8.5}{5} = 3.3\) or equivalent | B1, B1, B1, B1 |
| (i) | Graph showing coordinates: \((0,5)\), \((7.5, 5)\), \((8.5, 3.3)\), \((8.5, 0)\) | B1 |
| (ii) | Avoids tiny feasible regions | B1 |
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|-----------|------------|
| (i) | $X = \min(25, 8.5) = 8.5$ or equivalent; $Y = \min(5, 42.5) = 5$ or equivalent; $X^* = \frac{85-10}{10} = 7.5$ or equivalent; $Y^* = \frac{25-8.5}{5} = 3.3$ or equivalent | B1, B1, B1, B1 | OK if only seen once or more on graph; OK if only seen on graph; OK if only seen on graph |
| (i) | Graph showing coordinates: $(0,5)$, $(7.5, 5)$, $(8.5, 3.3)$, $(8.5, 0)$ | B1 | Allow flexibility - sensibly scaled for their X and Y. Disallow if either of the lines in the question could intersect both axes. |
| (ii) | Avoids tiny feasible regions | B1 | Lines can extend beyond segment; condone minor errors in plotting (e.g. 8.5 plotted at 9). Plotting tolerance on axis intersection points within correct small square. |
---2 The algorithm gives a method for drawing two straight lines, if certain conditions are met.
Start with the equations of the two straight lines\\
Line 1 is $a x + b y = c , \quad a , b , c > 0$\\
Line 2 is $d x + e y = f , \quad d , e , f > 0$\\
Let $X =$ minimum of $\frac { c } { a }$ and $\frac { f } { d }$\\
Let $Y =$ minimum of $\frac { c } { b }$ and $\frac { f } { e }$
If $X = \frac { c } { a }$ then $X ^ { * } = \frac { c - b Y } { a }$ and $Y ^ { * } = \frac { f - d X } { e }$
If $X = \frac { f } { d }$ then $X ^ { * } = \frac { f - e Y } { d }$ and $Y ^ { * } = \frac { c - a X } { b }$\\
Draw an $x$-axis labelled from 0 to $X$, and a $y$-axis labelled from 0 to $Y$\\
Join ( $0 , Y$ ) to ( $X , Y ^ { * }$ ) with a straight line\\
Join ( $X ^ { * } , Y$ ) to ( $X , 0$ ) with a straight line\\
(i) Apply the algorithm with $a = 1 , b = 5 , c = 25 , d = 10 , e = 2 , f = 85$.\\
(ii) Why might this algorithm be useful in an LP question?
\hfill \mbox{\textit{OCR MEI D1 2011 Q2 [8]}}