OCR D2 2007 June — Question 4 16 marks

Exam BoardOCR
ModuleD2 (Decision Mathematics 2)
Year2007
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeDynamic programming minimax route
DifficultyModerate -0.5 This is a standard D2 dynamic programming table completion exercise requiring systematic application of the minimax algorithm. While it involves multiple steps, the procedure is mechanical and follows a well-practiced template with no conceptual challenges or novel problem-solving required.
Spec7.07a Simplex tableau: initial setup in standard format

4 Answer this question on the insert provided. The table shows a partially completed dynamic programming tabulation for solving a minimax problem.
StageStateA ctionWorkingM inimax
\multirow{3}{*}{1}0044
1033
2022
\multirow{9}{*}{2}\multirow{3}{*}{0}0\(\max ( 6,4 ) = 6\)\multirow{3}{*}{3}
1\(\max ( 2,3 ) = 3\)
2\(\max ( 3,2 ) = 3\)
\multirow{3}{*}{1}0\(\max ( 2,4 ) =\)\multirow{3}{*}{}
1\(\max ( 4,3 ) =\)
2\(\max ( 5,2 ) =\)
\multirow{3}{*}{2}0max(2,\multirow{3}{*}{}
1max(3,
2max(4,
\multirow{3}{*}{3}\multirow{3}{*}{0}0max(5,\multirow{3}{*}{}
1max(5,
2max(2,
  1. On the insert, complete the last two columns of the table.
  2. State the minimax value and write down the minimax route.
  3. Complete the diagram on the insert to show the network that is represented by the table.

Question 4:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
Stage 1: state 0, action 0, working 4, minimax 4; state 1, action 0, working 3, minimax 3; state 2, action 0, working 2, minimax 2 Values only credited when seen in table
Stage 2, state 0: \(\max(6,4)=6\), \(\max(2,3)=3\), \(\max(3,2)=3\), minimax 3
Stage 2, state 1: \(\max(2,4)=4\), \(\max(4,3)=4\), \(\max(5,2)=5\), minimax 4M1 For calculating the maxima as 4, 4, 5
A1For calculating the minimax as 4
Stage 2, state 2: \(\max(2,4)=4\), \(\max(3,3)=3\), \(\max(4,2)=4\), minimax 3B1 For completing 4, 3, 2 in the brackets
M1For calculating the maxima as 4, 3, 4 (method)
A1For calculating the minimax as 3 cao
Stage 3, state 0: \(\max(5,3)=5\), \(\max(5,4)=5\), \(\max(2,3)=3\), minimax 3B1 For using their minimax values from stage 2
M1For calculating the maxima for their values
A1For calculating the maxima as 5, 5, 3 cao
A1For calculating the minimax as 3 cao
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
Value = 3M1 For the value from their tabulation
A1For 3 (irrespective of their tabulation) cao
Route: \((0;0) \to (1;1) \to (2;2) \to (3;0)\) (or in reverse)M1 dep For reading route from their tabulation
A1For this route (irrespective of their tabulation) cao
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
Graph structure correct with nodes \((2;0), (1;0), (3;0), (2;1), (1;1), (0;0), (2;2), (1;2)\)B1 For the graph structure correct
Substantially correct attempt at weights (no more than two definite errors or omissions)M1 For a substantially correct attempt at the weights
Weights unambiguously correctA1 For weights unambiguously correct cao
# Question 4:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Stage 1: state 0, action 0, working 4, minimax 4; state 1, action 0, working 3, minimax 3; state 2, action 0, working 2, minimax 2 | | Values only credited when seen in table |
| Stage 2, state 0: $\max(6,4)=6$, $\max(2,3)=3$, $\max(3,2)=3$, minimax 3 | | |
| Stage 2, state 1: $\max(2,4)=4$, $\max(4,3)=4$, $\max(5,2)=5$, minimax 4 | M1 | For calculating the maxima as 4, 4, 5 |
| | A1 | For calculating the minimax as 4 |
| Stage 2, state 2: $\max(2,4)=4$, $\max(3,3)=3$, $\max(4,2)=4$, minimax 3 | B1 | For completing 4, 3, 2 in the brackets |
| | M1 | For calculating the maxima as 4, 3, 4 (method) |
| | A1 | For calculating the minimax as 3 cao |
| Stage 3, state 0: $\max(5,3)=5$, $\max(5,4)=5$, $\max(2,3)=3$, minimax 3 | B1 | For using their minimax values from stage 2 |
| | M1 | For calculating the maxima for their values |
| | A1 | For calculating the maxima as 5, 5, 3 cao |
| | A1 | For calculating the minimax as 3 cao |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Value = 3 | M1 | For the value from their tabulation |
| | A1 | For 3 (irrespective of their tabulation) cao |
| Route: $(0;0) \to (1;1) \to (2;2) \to (3;0)$ (or in reverse) | M1 dep | For reading route from their tabulation |
| | A1 | For this route (irrespective of their tabulation) cao |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph structure correct with nodes $(2;0), (1;0), (3;0), (2;1), (1;1), (0;0), (2;2), (1;2)$ | B1 | For the graph structure correct |
| Substantially correct attempt at weights (no more than two definite errors or omissions) | M1 | For a substantially correct attempt at the weights |
| Weights unambiguously correct | A1 | For weights unambiguously correct cao |

---
4 Answer this question on the insert provided.
The table shows a partially completed dynamic programming tabulation for solving a minimax problem.

\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Stage & State & A ction & Working & M inimax \\
\hline
\multirow{3}{*}{1} & 0 & 0 & 4 & 4 \\
\hline
 & 1 & 0 & 3 & 3 \\
\hline
 & 2 & 0 & 2 & 2 \\
\hline
\multirow{9}{*}{2} & \multirow{3}{*}{0} & 0 & $\max ( 6,4 ) = 6$ & \multirow{3}{*}{3} \\
\hline
 &  & 1 & $\max ( 2,3 ) = 3$ &  \\
\hline
 &  & 2 & $\max ( 3,2 ) = 3$ &  \\
\hline
 & \multirow{3}{*}{1} & 0 & $\max ( 2,4 ) =$ & \multirow{3}{*}{} \\
\hline
 &  & 1 & $\max ( 4,3 ) =$ &  \\
\hline
 &  & 2 & $\max ( 5,2 ) =$ &  \\
\hline
 & \multirow{3}{*}{2} & 0 & max(2, & \multirow{3}{*}{} \\
\hline
 &  & 1 & max(3, &  \\
\hline
 &  & 2 & max(4, &  \\
\hline
\multirow{3}{*}{3} & \multirow{3}{*}{0} & 0 & max(5, & \multirow{3}{*}{} \\
\hline
 &  & 1 & max(5, &  \\
\hline
 &  & 2 & max(2, &  \\
\hline
\end{tabular}
\end{center}

(i) On the insert, complete the last two columns of the table.\\
(ii) State the minimax value and write down the minimax route.\\
(iii) Complete the diagram on the insert to show the network that is represented by the table.

\hfill \mbox{\textit{OCR D2 2007 Q4 [16]}}