OCR D2 2008 January — Question 3 12 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Year2008
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeDynamic programming minimax route
DifficultyModerate -0.5 This is a standard dynamic programming minimax problem requiring systematic backward induction through a given table structure. While it involves multiple stages and careful bookkeeping, the algorithm is mechanical and follows a well-defined procedure taught in D2. The table structure is provided, reducing the problem to routine calculation rather than problem formulation, making it easier than average A-level questions that require more independent reasoning.
Spec7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation
3
  1. StageStateActionWorkingMinimax
    \multirow{3}{*}{1}001
    103
    202
    \multirow{6}{*}{2}\multirow{2}{*}{0}0(4,\multirow{2}{*}{}
    1(2,
    \multirow{2}{*}{1}1(3,\multirow{2}{*}{}
    2(5,
    \multirow{2}{*}{2}0(2,\multirow{2}{*}{}
    2(4,
    \multirow{3}{*}{3}\multirow{3}{*}{0}0(5,\multirow{3}{*}{}
    1(3,
    2(1,
  2. Minimax value = \(\_\_\_\_\) Minimax route = \(\_\_\_\_\)
  3. \includegraphics[max width=\textwidth, alt={}, center]{95fbb09b-0301-4fc1-b694-838b8d0b64a6-10_958_1527_1539_351}
Question 3:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Minimax column for stage 1 shows 1, 3, 2B1 Identified in some way
1, 3, 2 transferred to working column for stage 2 correctlyM1
Calculating maximum values in working column for stage 2M1
Minimax column for stage 2 shows 3, 3, 2A1 Identified in some way (cao)
Calculating maximum values in working column for stage 3, correct methodM1
Minimax column for stage 3 shows 2A1 Identified in some way (cao)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Minimax value \(= 2\)B1 2, cao
Minimax route \(= (3;0)-(2;2)-(1;0)-(0;0)\) (or in reverse)M1 Tracing their route (whatever problem solved)
This route from correct workingA1 Using network \(\Rightarrow\) M0
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
All vertices labelled correctlyB1
Arcs correct, need not be directed; condone stage boundaries shownM1
Arc weights correctA1 Be generous in interpretation of which weight is attached to which arc 
# Question 3:

## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Minimax column for stage 1 shows 1, 3, 2 | B1 | Identified in some way |
| 1, 3, 2 transferred to working column for stage 2 correctly | M1 | |
| Calculating maximum values in working column for stage 2 | M1 | |
| Minimax column for stage 2 shows 3, 3, 2 | A1 | Identified in some way (cao) |
| Calculating maximum values in working column for stage 3, correct method | M1 | |
| Minimax column for stage 3 shows 2 | A1 | Identified in some way (cao) |

## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Minimax value $= 2$ | B1 | 2, cao |
| Minimax route $= (3;0)-(2;2)-(1;0)-(0;0)$ (or in reverse) | M1 | Tracing their route (whatever problem solved) |
| This route from correct working | A1 | Using network $\Rightarrow$ M0 |

## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| All vertices labelled correctly | B1 | |
| Arcs correct, need not be directed; condone stage boundaries shown | M1 | |
| Arc weights correct | A1 | Be generous in interpretation of which weight is attached to which arc |

---
3 (i)

\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Stage & State & Action & Working & Minimax \\
\hline
\multirow{3}{*}{1} & 0 & 0 & 1 &  \\
\hline
 & 1 & 0 & 3 &  \\
\hline
 & 2 & 0 & 2 &  \\
\hline
\multirow{6}{*}{2} & \multirow{2}{*}{0} & 0 & (4, & \multirow{2}{*}{} \\
\hline
 &  & 1 & (2, &  \\
\hline
 & \multirow{2}{*}{1} & 1 & (3, & \multirow{2}{*}{} \\
\hline
 &  & 2 & (5, &  \\
\hline
 & \multirow{2}{*}{2} & 0 & (2, & \multirow{2}{*}{} \\
\hline
 &  & 2 & (4, &  \\
\hline
\multirow{3}{*}{3} & \multirow{3}{*}{0} & 0 & (5, & \multirow{3}{*}{} \\
\hline
 &  & 1 & (3, &  \\
\hline
 &  & 2 & (1, &  \\
\hline
\end{tabular}
\end{center}

(ii) Minimax value = $\_\_\_\_$

Minimax route = $\_\_\_\_$\\
(iii)\\
\includegraphics[max width=\textwidth, alt={}, center]{95fbb09b-0301-4fc1-b694-838b8d0b64a6-10_958_1527_1539_351}

\hfill \mbox{\textit{OCR D2 2008 Q3 [12]}}
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Q1 14 Q2 17 Q3 12 Q4 14 Q5 15