| Exam Board | Edexcel |
|---|---|
| Module | FD1 (Further Decision 1) |
| Year | 2023 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Simplex Algorithm |
| Type | Formulate LP from context |
| Difficulty | Standard +0.3 This is a standard LP formulation question from Further Maths Decision module. Part (a) requires translating word problems into constraints (routine for FD1 students), and part (b) involves mechanical setup of a two-stage simplex tableau with clear instructions on which variables to use. While it requires careful attention to detail and understanding of the two-stage method, it follows a well-practiced procedure without requiring novel insight or complex problem-solving. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.07a Simplex tableau: initial setup in standard format |
| b.v. | \(x\) | \(y\) | \(z\) | \(\mathrm { S } _ { 1 }\) | \(S _ { 2 }\) | \(S _ { 3 }\) | \(s _ { 4 }\) | \(a _ { 1 }\) | \(a _ { 2 }\) | Value |
| \(\mathrm { S } _ { 1 }\) | 0 | 0 | 0 | 1 | 1 | 3 | 0 | -1 | -3 | 600 |
| \(z\) | 0 | \(\frac { 4 } { 11 }\) | 1 | 0 | \(- \frac { 1 } { 11 }\) | \(\frac { 1 } { 11 }\) | 0 | \(\frac { 1 } { 11 }\) | \(- \frac { 1 } { 11 }\) | \(\frac { 2000 } { 11 }\) |
| \(x\) | 1 | \(\frac { 7 } { 11 }\) | 0 | 0 | \(\frac { 1 } { 11 }\) | \(- \frac { 12 } { 11 }\) | 0 | \(- \frac { 1 } { 11 }\) | \(\frac { 12 } { 11 }\) | \(\frac { 15600 } { 11 }\) |
| \(s _ { 4 }\) | 0 | \(\frac { 40 } { 11 }\) | 0 | 0 | \(\frac { 1 } { 11 }\) | \(- \frac { 12 } { 11 }\) | 1 | \(- \frac { 1 } { 11 }\) | \(\frac { 12 } { 11 }\) | \(\frac { 15600 } { 11 }\) |
| \(P\) | 0 | \(- \frac { 4 } { 11 }\) | 0 | 0 | \(- \frac { 32 } { 11 }\) | \(- \frac { 56 } { 11 }\) | 0 | \(\frac { 32 } { 11 }\) | \(\frac { 56 } { 11 }\) | \(\frac { 204800 } { 11 }\) |
| I | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 |
| b.v. | \(x\) | \(y\) | \(z\) | \(\mathrm { S } _ { 1 }\) | \(S _ { 2 }\) | \(S _ { 3 }\) | \(s _ { 4 }\) | Value |
| \(s _ { 2 }\) | 0 | 0 | 0 | 1 | 1 | 3 | 0 | 600 |
| \(z\) | 0 | 0 | 1 | \(\frac { 1 } { 10 }\) | 0 | \(\frac { 1 } { 2 }\) | \(- \frac { 1 } { 10 }\) | 100 |
| \(x\) | 1 | 0 | 0 | \(- \frac { 3 } { 40 }\) | 0 | \(- \frac { 9 } { 8 }\) | \(- \frac { 7 } { 40 }\) | 1125 |
| \(y\) | 0 | 1 | 0 | \(- \frac { 1 } { 40 }\) | 0 | \(- \frac { 3 } { 8 }\) | \(\frac { 11 } { 40 }\) | 375 |
| \(P\) | 0 | 0 | 0 | \(\frac { 29 } { 10 }\) | 0 | \(\frac { 7 } { 2 }\) | \(\frac { 1 } { 10 }\) | 20500 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximise \(P = 8x + 20y + 40z\) | B1 | AO3.3 — Correct objective function plus 'maximise'/'max'; '\(P=\)' not required |
| \(4x + 8y + 15z \leqslant 9000\) and \(x + 5y + 12z \geqslant 3600\) | M1 A1 | AO3.3/1.1b — M1 for either one correct inequality; A1 both correct (allow equivalent forms with integer coefficients) |
| \(x + y + z \geqslant 1600\) | B1 | AO3.3 — CAO (allow equivalent, 4 terms only, integer coefficients) |
| \(-x + 3y \leqslant 0\), \((x, y, z \geqslant 0)\) | B1 | AO3.3 — CAO (\(3y \leqslant x\), 2 terms only, integer coefficients) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x+8y+15z+s_1=9000\); \(x+5y+12z-s_2+a_1=3600\); \(x+y+z-s_3+a_2=1600\); \(-x+3y+s_4=0\) | B1 B1 | AO1.1b/2.5 — B1 any two correct equations; B1 all four correct equations |
| \(I=-(a_1+a_2)\) where \(a_1=3600-x-5y-12z+s_2\) and \(a_2=1600-x-y-z+s_3\) | M1 | AO2.1 — Must be exactly two artificial variables; allow slips in forming \(I\) |
| \(I-2x-6y-13z+s_2+s_3=-5200\) and \(P-8x-20y-40z=0\) | A1 | AO2.2a — CAO for both \(I\) and \(P\) rows stated explicitly |
| Initial simplex tableau (all six rows complete, two correct rows) | M1 | AO3.4 |
| CAO all values including b.v. column | A1 | AO2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Pivot (top) row completely correct including change of b.v. (using \(\frac{1}{3}r_1\)) | B1 | AO1.1b |
| All values in one non-pivot row correct, or one of columns \(y, s_1, s_2\), Value correct (must have pivoted on correct value) | M1 | AO2.1 |
| Row operations: \(r_2 - \frac{1}{33}R_1\) | A1 | AO1.1b |
| Row operations: \(r_3 + \frac{12}{33}R_1\) | A1 | AO1.1b |
| Row operations: \(r_4 + \frac{12}{33}R_1\) | B1 | AO2.4 |
| CAO all values including b.v. column; row operations \(\frac{1}{3}r_1,\ r_2-\frac{1}{33}r_1,\ r_3+\frac{12}{33}r_1,\ r_4+\frac{12}{33}r_1,\ r_5+\frac{56}{33}r_1\) | A1 (B1) | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) £20 500 | B1 | AO3.4 — CAO (20 500) |
| (ii) \(1125 + 5(375) + 12(100) = 4200\) minutes, so 70 hours | B1 | AO2.2a — CAO (70 only) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. there is no guarantee that all the books will be sold | B1 | AO3.5b — Must explicitly mention possibility that not all books will be sold; ignore reasons for why, provided they do not relate to the publisher producing less than the optimal number |
# Question 7:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximise $P = 8x + 20y + 40z$ | B1 | AO3.3 — Correct objective function plus 'maximise'/'max'; '$P=$' not required |
| $4x + 8y + 15z \leqslant 9000$ and $x + 5y + 12z \geqslant 3600$ | M1 A1 | AO3.3/1.1b — M1 for either one correct inequality; A1 both correct (allow equivalent forms with integer coefficients) |
| $x + y + z \geqslant 1600$ | B1 | AO3.3 — CAO (allow equivalent, 4 terms only, integer coefficients) |
| $-x + 3y \leqslant 0$, $(x, y, z \geqslant 0)$ | B1 | AO3.3 — CAO ($3y \leqslant x$, 2 terms only, integer coefficients) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x+8y+15z+s_1=9000$; $x+5y+12z-s_2+a_1=3600$; $x+y+z-s_3+a_2=1600$; $-x+3y+s_4=0$ | B1 B1 | AO1.1b/2.5 — B1 any two correct equations; B1 all four correct equations |
| $I=-(a_1+a_2)$ where $a_1=3600-x-5y-12z+s_2$ and $a_2=1600-x-y-z+s_3$ | M1 | AO2.1 — Must be exactly two artificial variables; allow slips in forming $I$ |
| $I-2x-6y-13z+s_2+s_3=-5200$ and $P-8x-20y-40z=0$ | A1 | AO2.2a — CAO for both $I$ and $P$ rows stated explicitly |
| Initial simplex tableau (all six rows complete, two correct rows) | M1 | AO3.4 |
| CAO all values including b.v. column | A1 | AO2.2a |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Pivot (top) row completely correct including change of b.v. (using $\frac{1}{3}r_1$) | B1 | AO1.1b |
| All values in one non-pivot row correct, or one of columns $y, s_1, s_2$, Value correct (must have pivoted on correct value) | M1 | AO2.1 |
| Row operations: $r_2 - \frac{1}{33}R_1$ | A1 | AO1.1b |
| Row operations: $r_3 + \frac{12}{33}R_1$ | A1 | AO1.1b |
| Row operations: $r_4 + \frac{12}{33}R_1$ | B1 | AO2.4 |
| CAO all values including b.v. column; row operations $\frac{1}{3}r_1,\ r_2-\frac{1}{33}r_1,\ r_3+\frac{12}{33}r_1,\ r_4+\frac{12}{33}r_1,\ r_5+\frac{56}{33}r_1$ | A1 (B1) | AO1.1b |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) £20 500 | B1 | AO3.4 — CAO (20 500) |
| (ii) $1125 + 5(375) + 12(100) = 4200$ minutes, so 70 hours | B1 | AO2.2a — CAO (70 only) |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. there is no guarantee that all the books will be sold | B1 | AO3.5b — Must explicitly mention possibility that not all books will be sold; ignore reasons for why, provided they do not relate to the publisher producing less than the optimal number |
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7. A publisher plans to produce three versions of the same book: a paperback, a hardcover, and a deluxe edition.
\begin{itemize}
\item Each paperback takes 4 minutes to print and 1 minute to bind
\item Each hardcover takes 8 minutes to print and 5 minutes to bind
\item Each deluxe edition takes 15 minutes to print and 12 minutes to bind
\end{itemize}
The printing machine is available for at most 150 hours and the binding machine must be used for at least 60 hours.
The publisher decides to produce
\begin{itemize}
\item at least 1600 books in total
\item at least three times as many paperbacks as hardcovers
\end{itemize}
The profit on each paperback sold is $\pounds 8$, the profit on each hardcover sold is $\pounds 20$ and the profit on each deluxe edition sold is $\pounds 40$
Let $x , y$ and $z$ represent the number of paperbacks, hardcovers and deluxe editions produced.
\begin{enumerate}[label=(\alph*)]
\item Formulate this as a linear programming problem, stating the objective and listing the constraints as simplified inequalities with integer coefficients.
The publisher decides to solve this linear programming problem by using the two-stage simplex method.
\item Set up an initial tableau for solving this problem using the two-stage simplex method.
As part of your solution, you must show how
\begin{itemize}
\item the constraints have been made into equations by using slack variables, exactly two surplus variables and exactly two artificial variables
\item the rows for the two objective functions are formed
\end{itemize}
The following tableau is obtained after two iterations of the first stage of the two-stage simplex method.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
b.v. & $x$ & $y$ & $z$ & $\mathrm { S } _ { 1 }$ & $S _ { 2 }$ & $S _ { 3 }$ & $s _ { 4 }$ & $a _ { 1 }$ & $a _ { 2 }$ & Value \\
\hline
$\mathrm { S } _ { 1 }$ & 0 & 0 & 0 & 1 & 1 & 3 & 0 & -1 & -3 & 600 \\
\hline
$z$ & 0 & $\frac { 4 } { 11 }$ & 1 & 0 & $- \frac { 1 } { 11 }$ & $\frac { 1 } { 11 }$ & 0 & $\frac { 1 } { 11 }$ & $- \frac { 1 } { 11 }$ & $\frac { 2000 } { 11 }$ \\
\hline
$x$ & 1 & $\frac { 7 } { 11 }$ & 0 & 0 & $\frac { 1 } { 11 }$ & $- \frac { 12 } { 11 }$ & 0 & $- \frac { 1 } { 11 }$ & $\frac { 12 } { 11 }$ & $\frac { 15600 } { 11 }$ \\
\hline
$s _ { 4 }$ & 0 & $\frac { 40 } { 11 }$ & 0 & 0 & $\frac { 1 } { 11 }$ & $- \frac { 12 } { 11 }$ & 1 & $- \frac { 1 } { 11 }$ & $\frac { 12 } { 11 }$ & $\frac { 15600 } { 11 }$ \\
\hline
$P$ & 0 & $- \frac { 4 } { 11 }$ & 0 & 0 & $- \frac { 32 } { 11 }$ & $- \frac { 56 } { 11 }$ & 0 & $\frac { 32 } { 11 }$ & $\frac { 56 } { 11 }$ & $\frac { 204800 } { 11 }$ \\
\hline
I & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
\item Taking the most negative number in the profit row to indicate the pivot column, perform one complete iteration of the second stage of the two-stage simplex method to obtain a new tableau. Make your method clear by stating the row operations you use.
After three iterations of the second stage of the two-stage simplex method, the following tableau is obtained.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
b.v. & $x$ & $y$ & $z$ & $\mathrm { S } _ { 1 }$ & $S _ { 2 }$ & $S _ { 3 }$ & $s _ { 4 }$ & Value \\
\hline
$s _ { 2 }$ & 0 & 0 & 0 & 1 & 1 & 3 & 0 & 600 \\
\hline
$z$ & 0 & 0 & 1 & $\frac { 1 } { 10 }$ & 0 & $\frac { 1 } { 2 }$ & $- \frac { 1 } { 10 }$ & 100 \\
\hline
$x$ & 1 & 0 & 0 & $- \frac { 3 } { 40 }$ & 0 & $- \frac { 9 } { 8 }$ & $- \frac { 7 } { 40 }$ & 1125 \\
\hline
$y$ & 0 & 1 & 0 & $- \frac { 1 } { 40 }$ & 0 & $- \frac { 3 } { 8 }$ & $\frac { 11 } { 40 }$ & 375 \\
\hline
$P$ & 0 & 0 & 0 & $\frac { 29 } { 10 }$ & 0 & $\frac { 7 } { 2 }$ & $\frac { 1 } { 10 }$ & 20500 \\
\hline
\end{tabular}
\end{center}
Given that the publisher produces the optimal number of each version of the book,
\item \begin{enumerate}[label=(\roman*)]
\item state the maximum profit the publisher can earn,
\item find the number of hours the binding machine will be used.
\end{enumerate}\item Give a reason why the publisher may not earn the profit stated in (d)(i).
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 2023 Q7 [19]}}