| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Transportation problem: stepping-stone method |
| Difficulty | Moderate -0.5 This is a standard algorithmic question on the stepping-stone method for transportation problems. Students follow a well-defined procedure: calculate shadow costs using given values, compute improvement indices from a formula, identify the most negative index, and trace a stepping-stone path. While it requires careful arithmetic and understanding of the method, it involves no problem-solving insight—just mechanical application of a taught algorithm. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06c Working with constraints: algebra and ad hoc methods7.06d Graphical solution: feasible region, two variables |
| 1 | 2 | 3 | 4 | Supply | |
| A | 22 | 36 | 19 | 37 | 35 |
| B | 29 | 35 | 30 | 36 | 15 |
| C | 24 | 32 | 25 | 41 | 20 |
| D | 23 | 30 | 23 | 38 | 30 |
| Demand | 30 | 20 | 30 | 20 |
| 1 | 2 | 3 | 4 | |
| A | 30 | 5 | ||
| B | 15 | 0 | ||
| C | 20 | |||
| D | 10 | 20 |
| 1 | 2 | 3 | 4 | |
| A | x | x | ||
| B | x | x | ||
| C | 8 | 2 | x | 1 |
| D | 9 | 2 | x | x |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | The solution would otherwise be degenerate | B1 |
| (1) | ||
| Part (b) | M1, A1 | |
| \(\begin{array}{c\ | cccc} & 22 & 36 & 31 & 46 \\ \hline & 1 & 2 & 3 & 4 \\ 0 & A & x & x & -12 & -9 \\ -1 & B & 8 & x & x & -9 \\ -6 & C & (8) & (2) & x & (1) \\ -8 & D & (9) & (2) & x & x \end{array}\) | M1, A1 |
| (4) | ||
| Part (c) | Route is e.g. \(A3 - B3 - B2 - A2\) entering cell A3, exiting cell B3 | M1, A1, A1 |
| (3) | ||
| 8 marks |
| **Part (a)** | The solution would otherwise be **degenerate** | B1 | a1B1: CAO |
|---|---|---|---|
| | | **(1)** | |
| **Part (b)** | | M1, A1 | |
| | $\begin{array}{c\|cccc} & 22 & 36 & 31 & 46 \\ \hline & 1 & 2 & 3 & 4 \\ 0 & A & x & x & -12 & -9 \\ -1 & B & 8 & x & x & -9 \\ -6 & C & (8) & (2) & x & (1) \\ -8 & D & (9) & (2) & x & x \end{array}$ | M1, A1 | |
| | | **(4)** | |
| **Part (c)** | Route is e.g. $A3 - B3 - B2 - A2$ entering cell A3, exiting cell B3 | M1, A1, A1 | c1M1: A valid route (possibly drawn), their most negative II chosen, only one empty square used, $\theta$'s balance; c1A1: CAO – stepping stone route **stated** or clearly shown on **separate diagrams**; c2A1: CAO for entering and exiting cells |
| | | **(3)** | |
| | | **8 marks** | |
**Alternative shadow costs:**
- $1(0) \quad 2(14) \quad 3(9) \quad 4(24)$
- $A(22) \quad B(21) \quad C(16) \quad D(14)$
---
3. Table 1 below shows the cost, in pounds, of transporting one unit of stock from each of four supply points, $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and D , to four demand points $1,2,3$ and 4 . It also shows the stock held at each supply point and the stock required at each demand point. A minimum cost solution is required.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
& 1 & 2 & 3 & 4 & Supply \\
\hline
A & 22 & 36 & 19 & 37 & 35 \\
\hline
B & 29 & 35 & 30 & 36 & 15 \\
\hline
C & 24 & 32 & 25 & 41 & 20 \\
\hline
D & 23 & 30 & 23 & 38 & 30 \\
\hline
Demand & 30 & 20 & 30 & 20 & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
Table 2 shows an initial solution given by the north-west corner method.\\
Table 3 shows some of the improvement indices for this solution.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& 1 & 2 & 3 & 4 \\
\hline
A & 30 & 5 & & \\
\hline
B & & 15 & 0 & \\
\hline
C & & & 20 & \\
\hline
D & & & 10 & 20 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& 1 & 2 & 3 & 4 \\
\hline
A & x & x & & \\
\hline
B & & x & x & \\
\hline
C & 8 & 2 & x & 1 \\
\hline
D & 9 & 2 & x & x \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 3}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Explain why a zero has been placed in cell B3 in Table 2.\\
(1)
\item Calculate the shadow costs and the missing improvement indices and enter them into Table 3 in your answer book.
\item Taking the most negative improvement index to indicate the entering cell, state the steppingstone route that should be used to obtain the next solution. You must state your entering cell and exiting cell.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2013 Q3 [8]}}