D1 June 2006 Q6

EdexcelOld spec16 marksLinear Programming

6. 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)

Mark scheme (a)

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

Notes

B2, B1 First 3 equations c.a.o. – 1 each error, but penalise only 1 error per equation. Inequalities get B0

B2 c.a.o. (B1 for a ‘little slip’)

Mark scheme (b)

Scheme

Marks

b.v.	\(x\)	\(y\)	\(z\)	\(r\)	\(s\)	\(t\)	value	Row ops.
\(r\)	2	\(\tfrac{5}{2}\) (pivot)	0	1	\(-\tfrac{5}{6}\)	0	600	\(R_1 - 10R_2\)
\(z\)	\(\tfrac{1}{2}\)	\(\tfrac{3}{4}\)	1	0	\(\tfrac{1}{12}\)	0	300	\(R_2 \div 12\)
\(t\)	0	0	0	0	\(-\tfrac{1}{3}\)	1	1200	\(R_3 - 4R_2\)
\(P\)	\(-5\)	\(-10\)	0	0	5	0	18000	\(R_4 + 60R_2\)

M1 A1
M1 A1ft
B1 (5)

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	\(R_1 \div \tfrac{5}{2}\)
\(z\)	\(-\tfrac{1}{10}\)	0	1	\(-\tfrac{3}{10}\)	\(\tfrac{1}{3}\)	0	120	\(R_2 - \tfrac{3}{4}R_1\)
\(t\)	0	0	0	0	\(-\tfrac{1}{3}\)	1	1200	\(R_3\) stet
\(P\)	3	0	0	4	\(\tfrac{5}{3}\)	0	20400	\(R_4 + 10R_1\)

M1 A1ft
M1 A1 (4)

(9)

Notes

M1 Correct pivot chosen and some attempt to deal with whole row

A1 pivot row correct c.a.o. including b.v.

M1 correct row operations used (all 3) – at least 1 non-zero or 1 term correct in each row

A1ft non-pivoted rows correct; ft on error in pivot choice only

B1 Row operations correctly stated (condone lack of \(R_2 \div 12\)); must be in terms of new pivot row

M1ft Correct pivot chosen + some attempt to deal with whole row, ft from previous tableau. No negatives in value of previous tableau or M0

A1ft c.a.o. including b.v. but ft from previous tableau

M1 Correct row operations used (all 3) – at least 1 non-zero or 1 term correct in each row

A1 c.a.o.

Mark scheme (c)

Scheme	Marks
\(P = 20\,400\quad x = 0\quad y = 240\quad z = 120\) \(r = 0\quad s = 0\quad t = 1200\)	M1 A2ft,1ft,0
	(3)
	(16 marks)

Notes

M1 3 variables stated – must have completed b.v. and value columns (or 1s and zeros) on tableau. If reading top → bottom M0. Must be a final tableau. Any negative M0

A2ft all 7 correct

A1ft at least 4 correct