OCR Further Discrete 2022 June — Question 6 15 marks

Exam BoardOCR
ModuleFurther Discrete (Further Discrete)
Year2022
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicThe Simplex Algorithm
TypePerform one Simplex iteration
DifficultyStandard +0.3 This is a standard Further Maths simplex algorithm question requiring routine tableau setup and one iteration, followed by reading values from a given tableau. Part (c)-(d) add mild algebraic manipulation and conceptual understanding, but overall this is a textbook exercise testing procedural competence rather than problem-solving insight. Slightly easier than average A-level due to its mechanical nature.
Spec7.07a Simplex tableau: initial setup in standard format7.07b Simplex iterations: pivot choice and row operations7.07c Interpret simplex: values of variables, slack, and objective7.07d Simplex terminology: basic feasible solution, basic/non-basic variable

6 A linear programming problem is
Maximise \(\mathrm { P } = 2 \mathrm { x } - \mathrm { y }\) subject to $$\begin{aligned} 3 x + y - 4 z & \leqslant 24 \\ 5 x - 3 z & \leqslant 60 \\ - x + 2 y + 3 z & \leqslant 12 \end{aligned}$$ and \(x \geqslant 0 , y \geqslant 0 , z \geqslant 0\)
    1. Represent this problem as an initial simplex tableau.
    2. Carry out one iteration of the simplex algorithm. After two iterations the resulting tableau is
      \(P\)\(x\)\(y\)\(z\)\(s\)\(t\)\(u\)RHS
      10\(\frac { 5 } { 11 }\)0\(- \frac { 6 } { 11 }\)\(\frac { 8 } { 11 }\)0\(30 \frac { 6 } { 11 }\)
      01\(- \frac { 3 } { 11 }\)0\(- \frac { 3 } { 11 }\)\(\frac { 4 } { 11 }\)0\(15 \frac { 3 } { 11 }\)
      00\(- \frac { 5 } { 11 }\)1\(- \frac { 5 } { 11 }\)\(\frac { 3 } { 11 }\)0\(5 \frac { 5 } { 11 }\)
      00\(\frac { 34 } { 11 }\)0\(\frac { 12 } { 11 }\)\(- \frac { 5 } { 11 }\)1\(10 \frac { 10 } { 11 }\)
    1. Write down the basic variables after two iterations.
    2. Write down the exact values of the basic feasible solution for \(x , y\) and \(z\) after two iterations.
    3. State what you can deduce about the optimal value of the objective for the original problem. You are now given that, in addition to the constraints above, \(\mathrm { x } + \mathrm { y } + \mathrm { z } = 9\).
  1. Use the additional constraint to rewrite the original constraints in terms of \(x\) and \(y\) but not \(z\).
  2. Explain why the simplex algorithm cannot be applied to this new problem without some modification.

Question 6:
AnswerMarks Guidance
6(a) (i)
1 -2 1 0 0 0 0 0
0 3 1 -4 1 0 0 24
0 5 0 -3 0 1 0 60
AnswerMarks
0 -1 2 3 0 0 1 12B1
B1
AnswerMarks
[2]1.1
1.1Objective row correct
Three constraint rows correct
AnswerMarks Guidance
6(a) (ii)
P x y z s t u RHS
5 8 2
1 0 − 0 0 16
3 3 3
1 4 1
0 1 − 0 0 8
3 3 3
5 11 5
0 0 − − 1 0 20
3 3 3
7 5 1
0 0 0 1 20
AnswerMarks
3 3 3B1FT
M1FT
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1Positive pivot element for their initial tableau
May be seen in answer to (a)(i)
Or implied e.g. from x column correct (and y, z both non-basis)
Pivot row correct (for their positive pivot value) and four different
basis columns including P and pivot col
cao
AnswerMarks Guidance
6(b) (i)
[1]2.5 Allow P missing
6(b) (ii)
11 11B1
[1]1.1 cao
6(b) (iii)
11
(z value is) negative in the objective row so
solution is not yet optimal, hence P  306
AnswerMarks
max 11M1
A1
AnswerMarks
[2]1.1
2.4306 or 336 or 30.5 (3 s.f.) or better
11 11
‘Negative in top row’ so ‘not optimal’ or ‘at least 306’ or ‘greater
11
than 306’ (since no 0’s in RHS) o.e.
11
AnswerMarks Guidance
6(c) 3x + y – 4(9 – x – y)  24
 7x + 5y  60
5x – 3(9 – x – y)  60  8x + 3y  87
–x + 2y + 3(9 – x – y)  12  4x + y  15
AnswerMarks
x  0, y  0 and 9 – x – y  0  x + y  9M1
A1
A1
B1
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
3.1aSubstitute z = 9 – x – y
Any of the first three constraints correct
(in form ax + by  or  c), allow negative values of c
All three correct (in form ax + by  or  c), allow c < 0
Dealing with non-negativity for z (may imply x  0, y  0)
AnswerMarks Guidance
Px y
6(d) The constraint 4x + y  15 (o.e.)
is not of the form ax + by  c with c  0M1 FT
A1
AnswerMarks
[2]3.2b
2.4Identifying that this constraint is the problem
(FT their constraints provided one is like this)
Explaining that this is not of the required form (… non-negative)
Question 6:
6 | (a) | (i) | P x y z s t u RHS
1 -2 1 0 0 0 0 0
0 3 1 -4 1 0 0 24
0 5 0 -3 0 1 0 60
0 -1 2 3 0 0 1 12 | B1
B1
[2] | 1.1
1.1 | Objective row correct
Three constraint rows correct
6 | (a) | (ii) | Pivot on the 3 in row 2 of the x column
P x y z s t u RHS
5 8 2
1 0 − 0 0 16
3 3 3
1 4 1
0 1 − 0 0 8
3 3 3
5 11 5
0 0 − − 1 0 20
3 3 3
7 5 1
0 0 0 1 20
3 3 3 | B1FT
M1FT
A1
[3] | 1.1
1.1
1.1 | Positive pivot element for their initial tableau
May be seen in answer to (a)(i)
Or implied e.g. from x column correct (and y, z both non-basis)
Pivot row correct (for their positive pivot value) and four different
basis columns including P and pivot col
cao
6 | (b) | (i) | P, x, z, u | B1
[1] | 2.5 | Allow P missing
6 | (b) | (ii) | x = 153, y = 0, z = 55
11 11 | B1
[1] | 1.1 | cao
6 | (b) | (iii) | After two iterations P = 306
11
(z value is) negative in the objective row so
solution is not yet optimal, hence P  306
max 11 | M1
A1
[2] | 1.1
2.4 | 306 or 336 or 30.5 (3 s.f.) or better
11 11
‘Negative in top row’ so ‘not optimal’ or ‘at least 306’ or ‘greater
11
than 306’ (since no 0’s in RHS) o.e.
11
6 | (c) | 3x + y – 4(9 – x – y)  24
 7x + 5y  60
5x – 3(9 – x – y)  60  8x + 3y  87
–x + 2y + 3(9 – x – y)  12  4x + y  15
x  0, y  0 and 9 – x – y  0  x + y  9 | M1
A1
A1
B1
[4] | 3.1a
1.1
1.1
3.1a | Substitute z = 9 – x – y
Any of the first three constraints correct
(in form ax + by  or  c), allow negative values of c
All three correct (in form ax + by  or  c), allow c < 0
Dealing with non-negativity for z (may imply x  0, y  0)
P | x | y | z | s | t | u | RHS
6 | (d) | The constraint 4x + y  15 (o.e.)
is not of the form ax + by  c with c  0 | M1 FT
A1
[2] | 3.2b
2.4 | Identifying that this constraint is the problem
(FT their constraints provided one is like this)
Explaining that this is not of the required form (… non-negative)
6 A linear programming problem is\\
Maximise $\mathrm { P } = 2 \mathrm { x } - \mathrm { y }$\\
subject to

$$\begin{aligned}
3 x + y - 4 z & \leqslant 24 \\
5 x - 3 z & \leqslant 60 \\
- x + 2 y + 3 z & \leqslant 12
\end{aligned}$$

and $x \geqslant 0 , y \geqslant 0 , z \geqslant 0$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Represent this problem as an initial simplex tableau.
\item Carry out one iteration of the simplex algorithm.

After two iterations the resulting tableau is

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
$P$ & $x$ & $y$ & $z$ & $s$ & $t$ & $u$ & RHS \\
\hline
1 & 0 & $\frac { 5 } { 11 }$ & 0 & $- \frac { 6 } { 11 }$ & $\frac { 8 } { 11 }$ & 0 & $30 \frac { 6 } { 11 }$ \\
\hline
0 & 1 & $- \frac { 3 } { 11 }$ & 0 & $- \frac { 3 } { 11 }$ & $\frac { 4 } { 11 }$ & 0 & $15 \frac { 3 } { 11 }$ \\
\hline
0 & 0 & $- \frac { 5 } { 11 }$ & 1 & $- \frac { 5 } { 11 }$ & $\frac { 3 } { 11 }$ & 0 & $5 \frac { 5 } { 11 }$ \\
\hline
0 & 0 & $\frac { 34 } { 11 }$ & 0 & $\frac { 12 } { 11 }$ & $- \frac { 5 } { 11 }$ & 1 & $10 \frac { 10 } { 11 }$ \\
\hline
\end{tabular}
\end{center}
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down the basic variables after two iterations.
\item Write down the exact values of the basic feasible solution for $x , y$ and $z$ after two iterations.
\item State what you can deduce about the optimal value of the objective for the original problem.

You are now given that, in addition to the constraints above, $\mathrm { x } + \mathrm { y } + \mathrm { z } = 9$.
\end{enumerate}\item Use the additional constraint to rewrite the original constraints in terms of $x$ and $y$ but not $z$.
\item Explain why the simplex algorithm cannot be applied to this new problem without some modification.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete 2022 Q6 [15]}}