D2 June 2006 Q8

EdexcelOld spec16 marksLinear Programming

8. The tableau below is the initial tableau for a maximising linear programming problem.

Basic variable	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	Value
\(r\)	7	10	10	1	0	0	3600
\(s\)	6	9	12	0	1	0	3600
\(t\)	2	3	4	0	0	1	2400
\(P\)	−35	−55	−60	0	0	0	0

(a) Write down the four equations represented in the initial tableau above. (4)

(b) Taking the most negative number in the profit row to indicate the pivot column at each stage, solve this linear programming problem. State the row operations that you use. (9)

Scheme	Marks
\(7x + 10y + 10z + r = 3600\) \(6x + 9y + 12z + s = 3600\) \(2x + 3y + 4z + t = 2400\)	B2, 1, 0
\(\text{P} - 35x - 55y - 60z = 0\)	B2, 0
	(4)

Mark scheme (b)

Scheme

Marks

(i)

b.v.	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	value	Row ops
\(r\)	\(2\)	\(\boldsymbol{\tfrac{5}{2}}\)	\(0\)	\(1\)	\(-\tfrac{5}{6}\)	\(0\)	\(600\)	\(\text{R}_1 - 10\text{R}_2\)
\(z\)	\(\tfrac{1}{2}\)	\(\tfrac{3}{4}\)	\(1\)	\(0\)	\(\tfrac{1}{12}\)	\(0\)	\(300\)	\(\text{R}_2 \div 12\)
\(t\)	\(0\)	\(0\)	\(0\)	\(0\)	\(-\tfrac{1}{3}\)	\(1\)	\(1200\)	\(\text{R}_3 - 4\text{R}_2\)
\(P\)	\(-5\)	\(-10\)	\(0\)	\(0\)	\(5\)	\(0\)	\(18000\)	\(\text{R}_4 + 60\text{R}_2\)

A1
M1
A1ft
B1 (5)

(ii)

b.v.	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	value	Row ops
\(y\)	\(\tfrac{4}{5}\)	\(1\)	\(0\)	\(\tfrac{2}{5}\)	\(-\tfrac{1}{3}\)	\(0\)	\(240\)	\(\text{R}_1 \div \tfrac{5}{2}\)
\(z\)	\(-\tfrac{1}{10}\)	\(0\)	\(1\)	\(-\tfrac{3}{10}\)	\(\tfrac{1}{3}\)	\(0\)	\(120\)	\(\text{R}_2 - \tfrac{3}{4}\text{R}_1\)
\(t\)	\(0\)	\(0\)	\(0\)	\(0\)	\(-\tfrac{1}{3}\)	\(1\)	\(1200\)	\(\text{R}_3\ \text{stet}\)
\(P\)	\(3\)	\(0\)	\(0\)	\(4\)	\(\tfrac{5}{3}\)	\(0\)	\(20400\)	\(\text{R}_4 + 10\text{R}_1\)

M1
M1
A1ft
M1
A1 (4)

(9)

Notes

(Corrected from the printed mark scheme: the value in the P row of the first tableau is printed as 1800; it is \(60 \times 300 = 18000\), consistent with 20400 in the second tableau.)

Mark scheme (c)

Scheme	Marks
\(\text{P} = 20400\quad x = 0\quad y = 240\quad z = 120\)	M1
\(r = 0\quad s = 0\quad t = 1200\)	A2ft, A1ft, 0
	(3)
	(16 marks)

Notes

(The printed mark scheme shows a part total of 2 here; the codes M1 A2 make 3, as on the paper.)