D2 June 2012 Q4

EdexcelOld spec8 marksLinear Programming

4. The tableau below is the initial tableau for a maximising linear programming problem in \(x\), \(y\) and \(z\) which is to be solved.

Basic variable	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	Value
\(r\)	5	\(\tfrac{1}{2}\)	0	1	0	0	5
\(s\)	1	−2	4	0	1	0	3
\(t\)	8	4	6	0	0	1	6
\(P\)	−5	−7	−4	0	0	0	0

(a) Starting by increasing \(y\), perform one complete iteration of the simplex algorithm, to obtain tableau T. State the row operations you use. (5)

(b) Write down the profit equation given by tableau T. (2)

Mark scheme (a)

Scheme

Marks

b.v	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	value	\(\theta\) values
\(r\)	5	\(\tfrac{1}{2}\)	0	1	0	0	5	10
\(s\)	1	−2	4	0	1	0	3	\(-\tfrac{3}{2}\)
\(t\)	8	4	6	0	0	1	6	\(\tfrac{3}{2}\) \(\leftarrow\)
\(P\)	−5	−7	−4	0	0	0	0

b.v.	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	value	Row ops
\(r\)	4	0	\(-\tfrac{3}{4}\)	1	0	\(-\tfrac{1}{8}\)	\(\tfrac{17}{4}\)	\(R1 - \tfrac{1}{2}R3\)
\(s\)	5	0	7	0	1	\(\tfrac{1}{2}\)	6	\(R2 + 2R3\)
\(y\)	2	1	\(\tfrac{3}{2}\)	0	0	\(\tfrac{1}{4}\)	\(\tfrac{3}{2}\)	\(R3 \div 4\)
\(P\)	9	0	\(\tfrac{13}{2}\)	0	0	\(\tfrac{7}{4}\)	\(\tfrac{21}{2}\)	\(R4 + 7R3\)

1M1 1A1
B1
2M1 2A1

(5)

Notes

a1M1 Correct pivot located, attempt to divide row. If choosing negative number as pivot M0B0M0

a1A1 pivot row correct including change of b.v.

a1B1 Row operations CAO – allow if given in terms of old row 3.

a2M1 (ft) Correct row operations used at least once, column \(x\), \(z\), \(t\) or value correct.

a2A1 CAO on the three non-pivot rows.

Scheme	Marks
\(P + 9x + \tfrac{13}{2}z + \tfrac{7}{4}t = \tfrac{21}{2}\)	M1 A1
	(2)

Mark scheme (c)

Scheme	Marks
\(P = \tfrac{21}{2} - 9x - \tfrac{13}{2}z - \tfrac{7}{4}t\), so increasing \(x\) or \(z\) or \(t\) would decrease P	B1
	(1)
	(8 marks)

Notes

c1B1 Explanation, must refer to increasing \(x\), \(z\) and \(t\), condone no ref to \(x = z = t = 0\), must have correct signs in equation in (b). Do not accept ‘no negatives in profit row’ o.e. alone.