| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Network Flows |
| Type | Transportation problem: stepping-stone method |
| Difficulty | Standard +0.3 This is a standard transportation problem using the stepping-stone method, a routine algorithmic procedure taught in D2. While it requires careful bookkeeping across multiple iterations (parts a-d) and formulating an LP (part e), it involves no novel insight—just methodical application of a learned algorithm. The multi-part structure and computational length place it slightly above average difficulty, but it remains a textbook exercise. |
| Spec | 7.03l Bin packing: next-fit, first-fit, first-fit decreasing, full bin |
| P | Q | R | Supply | |
| A | 20 | 5 | 13 | 74 |
| B | 7 | 15 | 8 | 58 |
| C | 9 | 14 | 21 | 63 |
| D | 22 | 16 | 10 | 85 |
| Demand | 145 | 57 | 78 |
| P | Q | R | Supply | |
| A | 74 | 74 | ||
| B | 58 | 58 | ||
| C | 13 | 50 | 63 | |
| D | 7 | 78 | 85 | |
| Demand | 145 | 57 | 78 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Valid loop using \(\theta\): e.g. \(A \to P = 74-\theta\), \(A \to Q = \theta\), \(C \to P = 13+\theta\), \(C \to Q = 50-\theta\) giving solution \(A:24, B:58, C:63, D\) with \(Q=50, R=7, R=78\) | M1 A1 | (2) a1M1: Valid route, only one empty square, AQ used, \(\theta\)'s balance. a1A1: Correct route, up to improved solution (six numbers, no zeros) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shadow costs: \(0, -13, -11, 11\) for rows \(A,B,C,D\); \(20, 5, -1\) for cols \(P,Q,R\); Improvement indices: \(P:X, Q:X, R:14\) for \(A\); \(P:X, Q:23, R:22\) for \(B\); \(P:X, Q:20, R:33\) for \(C\); \(P:-9, Q:X, R:X\) for \(D\) | M1 A1 | b1M1: Finding 7 shadow costs and 6 improvement indices. b1A1: Shadow costs correct |
| New solution: \(A:P=24-\theta\), \(A:Q=50+\theta\), \(D:\theta\), \(D:Q=7-\theta\) giving \(A:17, B:58, C:63, D:7\) with \(Q=57, R=78\) | M1 A1 | (4) b2M1: Valid route, most negative II chosen, only one empty square, \(\theta\)'s balance. Entering cell DP, exiting cell DQ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shadow costs: \(0, -13, -11, 2\) for rows; \(20, 5, 8\) for cols; Improvement indices: \(A:P=X, Q=X, R=5\); \(B:P=X, Q=23, R=13\); \(C:P=X, Q=20, R=24\); \(D:P=X, Q=9, R=X\) | M1 A1 A1 | (3) c1M1: Finding 7 shadow costs and all 6 IIs or at least 1 negative II found. c1A1: Shadow costs correct. c2A1: CSO + reason + optimal |
| Optimal since no negative improvement indices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\pounds)\ 2532\) | B1 | (1) d1B1: CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(x_{ij}\) be the number of units transported from \(i\) to \(j\) | B1 | e1B1: \(x_{ij}\) defined correctly (must include 'number of' and 'from \(i\) to \(j\)') |
| where \(i \in \{A,B,C,D\}\), \(j \in \{P,Q,R\}\) and \(x_{ij} \geq 0\) | B1 | e2B1: Defining set of values for \(i\) and \(j\) including non-negativity constraint |
| Minimise \(C = 20x_{AP} + 5x_{AQ} + 13x_{AR} + 7x_{BP} + 15x_{BQ} + 8x_{BR} + 9x_{CP} + 14x_{CQ} + 21x_{CR} + 22x_{DP} + 16x_{DQ} + 10x_{DR}\) | M1 A1 | e1M1: Objective function (allow one error in coefficient or variable). e1A1: CAO and minimise |
| \(x_{AP} + x_{AQ} + x_{AR} \leq 74\) | ||
| \(x_{BP} + x_{BQ} + x_{BR} \leq 58\) | ||
| \(x_{CP} + x_{CQ} + x_{CR} \leq 63\) | M1 | e2M1: At least 3 constraints with unit coefficients, rhs values correct |
| \(x_{DP} + x_{DQ} + x_{DR} \leq 85\) | A1 | e2A1: At least 5 correct constraints |
| \(x_{AP} + x_{BP} + x_{CP} + x_{DP} \leq 145\) | A1 | (7) e3A1: All 7 constraints correct |
| \(x_{AQ} + x_{BQ} + x_{CQ} + x_{DQ} \leq 57\) | ||
| \(x_{AR} + x_{BR} + x_{CR} + x_{DR} \leq 78\) |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Valid loop using $\theta$: e.g. $A \to P = 74-\theta$, $A \to Q = \theta$, $C \to P = 13+\theta$, $C \to Q = 50-\theta$ giving solution $A:24, B:58, C:63, D$ with $Q=50, R=7, R=78$ | M1 A1 | **(2)** a1M1: Valid route, only one empty square, AQ used, $\theta$'s balance. a1A1: Correct route, up to improved solution (six numbers, no zeros) |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shadow costs: $0, -13, -11, 11$ for rows $A,B,C,D$; $20, 5, -1$ for cols $P,Q,R$; Improvement indices: $P:X, Q:X, R:14$ for $A$; $P:X, Q:23, R:22$ for $B$; $P:X, Q:20, R:33$ for $C$; $P:-9, Q:X, R:X$ for $D$ | M1 A1 | b1M1: Finding 7 shadow costs and 6 improvement indices. b1A1: Shadow costs correct |
| New solution: $A:P=24-\theta$, $A:Q=50+\theta$, $D:\theta$, $D:Q=7-\theta$ giving $A:17, B:58, C:63, D:7$ with $Q=57, R=78$ | M1 A1 | **(4)** b2M1: Valid route, most negative II chosen, only one empty square, $\theta$'s balance. Entering cell DP, exiting cell DQ |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shadow costs: $0, -13, -11, 2$ for rows; $20, 5, 8$ for cols; Improvement indices: $A:P=X, Q=X, R=5$; $B:P=X, Q=23, R=13$; $C:P=X, Q=20, R=24$; $D:P=X, Q=9, R=X$ | M1 A1 A1 | **(3)** c1M1: Finding 7 shadow costs **and** all 6 IIs **or** at least 1 negative II found. c1A1: Shadow costs correct. c2A1: CSO + reason + optimal |
| Optimal since no negative improvement indices | | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\pounds)\ 2532$ | B1 | **(1)** d1B1: CAO |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $x_{ij}$ be the number of units transported from $i$ to $j$ | B1 | e1B1: $x_{ij}$ defined correctly (must include 'number of' and 'from $i$ to $j$') |
| where $i \in \{A,B,C,D\}$, $j \in \{P,Q,R\}$ and $x_{ij} \geq 0$ | B1 | e2B1: Defining set of values for $i$ and $j$ **including** non-negativity constraint |
| Minimise $C = 20x_{AP} + 5x_{AQ} + 13x_{AR} + 7x_{BP} + 15x_{BQ} + 8x_{BR} + 9x_{CP} + 14x_{CQ} + 21x_{CR} + 22x_{DP} + 16x_{DQ} + 10x_{DR}$ | M1 A1 | e1M1: Objective function (allow one error in coefficient **or** variable). e1A1: CAO and minimise |
| $x_{AP} + x_{AQ} + x_{AR} \leq 74$ | | |
| $x_{BP} + x_{BQ} + x_{BR} \leq 58$ | | |
| $x_{CP} + x_{CQ} + x_{CR} \leq 63$ | M1 | e2M1: At least 3 constraints with unit coefficients, rhs values correct |
| $x_{DP} + x_{DQ} + x_{DR} \leq 85$ | A1 | e2A1: At least 5 correct constraints |
| $x_{AP} + x_{BP} + x_{CP} + x_{DP} \leq 145$ | A1 | **(7)** e3A1: All 7 constraints correct |
| $x_{AQ} + x_{BQ} + x_{CQ} + x_{DQ} \leq 57$ | | |
| $x_{AR} + x_{BR} + x_{CR} + x_{DR} \leq 78$ | | |
**Total: 17 marks**
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5. The table 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 each of three sales points, $\mathrm { P } , \mathrm { Q }$ and R . It also shows the stock held at each supply point and the amount required at each sales point. A minimum cost solution is required.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& P & Q & R & Supply \\
\hline
A & 20 & 5 & 13 & 74 \\
\hline
B & 7 & 15 & 8 & 58 \\
\hline
C & 9 & 14 & 21 & 63 \\
\hline
D & 22 & 16 & 10 & 85 \\
\hline
Demand & 145 & 57 & 78 & \\
\hline
\end{tabular}
\end{center}
The north-west corner method gives the following initial solution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
& P & Q & R & Supply \\
\hline
A & 74 & & & 74 \\
\hline
B & 58 & & & 58 \\
\hline
C & 13 & 50 & & 63 \\
\hline
D & & 7 & 78 & 85 \\
\hline
Demand & 145 & 57 & 78 & \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Taking AQ as the entering cell, use the stepping stone method to find an improved solution. Make your route clear.
\item Perform one further iteration of the stepping stone method to obtain an improved solution. You must make your method clear by stating your shadow costs, improvement indices, route, entering cell and exiting cell.
\item Determine whether your current solution is optimal. Justify your answer.
\item State the cost of the solution you found in (b).
\item Formulate this problem as a linear programming problem. You must define your decision variables and make the objective function and constraints clear.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2015 Q5 [17]}}