D2 June 2016 Q7

EdexcelOld spec15 marksDynamic Programming

7. Remy builds canoes.

He can build up to five canoes each month, but if he wishes to build more than three canoes in any one month he has to hire an additional worker at a cost of £400 for that month.

In any month when canoes are built, the overhead costs are £150

A maximum of three canoes can be held in stock in any one month, at a cost of £25 per canoe per month.

Canoes must be delivered at the end of the month.

The order book for canoes is

Month	January	February	March	April	May
Number ordered	2	2	5	6	4

There is no stock at the beginning of January and Remy plans to have no stock after the May delivery.

(a) Use dynamic programming to determine the production schedule that minimises the costs given above. Show your working in the table provided in the answer book and state the minimum cost. (13)

The cost of materials is £200 per canoe and the cost of Remy’s time is £450 per month. Remy sells the canoes for £700 each.

(b) Determine Remy’s total profit for the five-month period. (2)

Mark scheme (a)

Stage	State	Action	Dest	Value	Marks
May	3	1	0	75 + 150 = 225*
(4)	2	2	0	50 + 150 = 200*
	1	3	0	25 + 150 = 175*
	0	4	0	150 + 400 = 550*
April	3	3	0	75 + 150 + 550 = 775*
(6)		4	1	75 + 150 + 400 + 175 = 800
		5	2	75 + 150 + 400 + 200 = 825	M1 A1 A1 (April)
	2	4	0	50 + 150 + 400 + 550 = 1150
		5	1	50 + 150 + 400 + 175 = 775*
	1	5	0	25 + 150 + 400 + 550 = 1125*
March	3	3	1	75 + 150 + 1125 = 1350*
(5)		4	2	75 + 150 + 400 + 775 = 1400	M1 A1ft A1 (March)
		5	3	75 + 150 + 400 + 775 = 1400
	2	4	1	50 + 150 + 400 + 1125 = 1725
		5	2	50 + 150 + 400 + 775 = 1375*
	1	5	1	25 + 150 + 400 + 1125 = 1700*
Feb	3	0	1	75 + 1700 = 1775
(2)		1	2	75 + 150 + 1375 = 1600
		2	3	75 + 150 + 1350 = 1575*
	2	1	1	50 + 150 + 1700 = 1900
		2	2	50 + 150 + 1375 = 1575
		3	3	50 + 150 + 1350 = 1550*
	1	2	1	25 + 150 + 1700 = 1875	M1 A1ft A1 (February)
		3	2	25 + 150 + 1375 = 1550*
		4	3	25 + 150 + 400 + 1350 = 1925
	0	3	1	150 + 1700 = 1850*
		4	2	150 + 400 + 1375 = 1925
		5	3	150 + 400 + 1350 = 1900
Jan	0	2	0	150 + 1850 = 2000
(2)		3	1	150 + 1550 = 1700*	M1 A1 (January)
		4	2	150 + 400 + 1550 = 2100
		5	3	150 + 400 + 1575 = 2125

Scheme

Marks

Month	January	February	March	April	May
Number made	3	3	5	5	3

Minimum cost: (£) 1700

(13)

Notes

All M marks – must bring optimal result from previous stage into calculations so for the first stage (April) if none of 225, 200, 175 or 550 (the optimal results from May) are used then M0. Ignore extra rows. Condone and credit rows that have been crossed out if they can still be read. Must have right ‘ingredients’ (storage costs, overhead costs, additional worker cost) at least once per stage (as an example for the six rows in April we must see at least one of these rows having a calculation that has either three or four values). Must have values in two of the three colums (State, Action, Dest). If no working seen then the number stated in the Value column must be correct to imply the correct method has been used

a1M1: First stage (April) completed. At least 6 rows, ‘something’ in each cell (but see M mark guidance above)

a1A1: Any two states correct (condone extra rows)

a2A1: CAO for first stage. No extra rows

a2M1: Second stage (March) completed. At least 6 rows, something in each cell (see M mark guidance above)

a3A1ft: Any two states correct – ft their * values/their smallest value from previous stage (condone extra rows)

a4A1: CAO for second stage. No extra rows

a3M1: Third stage (February) completed. At least 12 rows, something in each cell (see M mark guidance above)

a5A1ft: Any two states correct – ft their * values/their smallest value from previous stage (condone extra rows)

a6A1: CAO for third stage. No extra rows

a4M1: Fourth stage completed. At least 4 rows, something in each cell (see M mark guidance above)

a7A1: CAO (no ft) for fourth stage. No extra rows

a1B1: CAO – but must have scored all previous M marks

a2B1: CAO – condone lack of units - but must have scored all previous M marks

Mark scheme (b)

Scheme	Marks
\(700 \times 19 - (6050 + \text{their } 1700) = (\text{£})\ 5550\)	M1 A1
	(2)
	15 marks

Notes

b1M1: \(700 \times 19 - (5 \times 450 + 19 \times 200 + \text{their } 1700)\) or 7250 – their 1700. Must have scored at least two M marks in (a)

b1A1: CAO (condone lack of units) – correct answer with no working can score both marks in this part (but is still dependent on at least two M marks awarded in (a))