| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Transportation problem: stepping-stone method |
| Difficulty | Moderate -0.3 This is a standard textbook application of the stepping-stone method with a small 3×3 transportation problem. The question provides the initial feasible solution and asks for one iteration followed by optimality check—both are routine algorithmic procedures taught in D2 with no novel problem-solving required. Slightly easier than average due to the small problem size and explicit guidance on what to show. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables |
| \(P\) | \(Q\) | \(R\) | Supply | |
| \(A\) | 7 | 5 | 7 | 12 |
| \(B\) | 5 | 6 | 5 | 7 |
| \(C\) | 14 | 12 | 9 | 11 |
| Demand | 10 | 9 | 11 |
| \(P\) | \(Q\) | \(R\) | |
| \(A\) | 10 | 2 | |
| \(B\) | 7 | 0 | |
| \(C\) | 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 0\), \(B = 1\), \(C = 5\), \(P = 7\), \(Q = 5\), \(R = 4\) | M1 A1 | |
| \(AR = 7 - 0 - 4 = 3\) | ||
| \(BP = 5 - 1 - 7 = -3 \leftarrow\) | ||
| \(CP = 14 - 5 - 7 = 2\) | ||
| \(CC = 12 - 5 - 5 = 2\) | A1 | |
| Entering cell BP, Existing cell BQ | M1, A1 | \(Q = 7\) |
| Final table: A row: \(P=3\), \(Q=9\); B row: \(P=7\), \(R=0\); C row: \(R=11\) | A1(6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 0\), \(B = -2\), \(C = 2\), \(P = 7\), \(Q = 5\), \(R = 7\) | B1 | |
| \(AR = 7 - 0 - 7 = 0\) | ||
| \(BQ = 6 + 2 - 5 = 3\) | ||
| \(CP = 14 - 2 - 7 = 5\) | ||
| \(CQ = 12 - 2 - 5 = 5\) | B1 | |
| All improvement indices are non-negative, so optimal. | B1(3) |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 0$, $B = 1$, $C = 5$, $P = 7$, $Q = 5$, $R = 4$ | M1 A1 | |
| $AR = 7 - 0 - 4 = 3$ | | |
| $BP = 5 - 1 - 7 = -3 \leftarrow$ | | |
| $CP = 14 - 5 - 7 = 2$ | | |
| $CC = 12 - 5 - 5 = 2$ | A1 | |
| Entering cell BP, Existing cell BQ | M1, A1 | $Q = 7$ |
| Final table: A row: $P=3$, $Q=9$; B row: $P=7$, $R=0$; C row: $R=11$ | A1(6) | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 0$, $B = -2$, $C = 2$, $P = 7$, $Q = 5$, $R = 7$ | B1 | |
| $AR = 7 - 0 - 7 = 0$ | | |
| $BQ = 6 + 2 - 5 = 3$ | | |
| $CP = 14 - 2 - 7 = 5$ | | |
| $CQ = 12 - 2 - 5 = 5$ | B1 | |
| All improvement indices are non-negative, so optimal. | B1(3) | |
---2. The following transportation problem is to be solved.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& $P$ & $Q$ & $R$ & Supply \\
\hline
$A$ & 7 & 5 & 7 & 12 \\
\hline
$B$ & 5 & 6 & 5 & 7 \\
\hline
$C$ & 14 & 12 & 9 & 11 \\
\hline
Demand & 10 & 9 & 11 & \\
\hline
\end{tabular}
\end{center}
A possible north-west corner solution is:
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& $P$ & $Q$ & $R$ \\
\hline
$A$ & 10 & 2 & \\
\hline
$B$ & & 7 & 0 \\
\hline
$C$ & & & 11 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the stepping-stone method once to obtain an improved solution. You must make your shadow costs, improvement indices, entering cell, exiting cell and stepping-stone route clear.
\item Demonstrate that your solution is optimal.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q2 [9]}}