Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

NCEA Answers – Linear Programming 2003 Sample Q2b 2bVertex (2,6) (3,6) (4,8) (3,12) (2,13) R = 54 000m + 36 000c 324 000 378 000 504 000 594 000 576 000 maximum return, R = $594 000 when m = 3 and c = 12. Q7a x = number of Xvi bought y = number of Yco bought The constraints are: x + y ≤ 60 1 NCEA Answers – Linear Programming 60x + 120y ≤ 4800 x and y non-negative Objective function: R = 10x + 12y Vertex R = 10x + 12y (0,0) (60,0) (40,20) 640 (0,40) 0 600 480 maximum return, R = $640 when Xvi = 40 and Yco = 20. Q7b New objective function: R = 10x + 20y Vertex (0, 0) (60, 0) (40, 20) (0, 40) R = 10x + 20y 0 600 800 800 maximum return, R = $800 when you choose any integer (x, y) with 0 ≤ x ≤ 40 and x + 2y = 80. Multiple solutions because the objective function and a constraint have the same slope. 2004 Q3 Must calculate return at the vertices (0,320), (600,120), (720,0) or use the “parallel line test” to identify the optimal vertex. (i) s = 600 and b = 120. (ii) Maximum return R = $2400 2 NCEA Answers – Linear Programming Q4 (a) The constraints are: 100a + 100b ≤ 60 000 50a ≤ 25 000 50a + 100b ≤ 45 000 a and b non-negative. Objective function: P = 2.45a + 1.95b Calculate R at the vertices (0,450), (300,300) and (500,100), or use the “parallel line test” to identify the optimal vertex. Maximum return R = $1420 when a = 500 and b = 100. (b) Gradient of the objective function is now the same as the first constraint and is valid between 300 ≤ a ≤ 500 and a + b = 600. 2005 Q3 (a) Vertex P = 9x + 12y (20,60) 900 (40,50) 960 (56,40) 984 (b) Maximum profit, P = $984. ([Accept P = $975 for (55,40) or P = $993 for (57,40).] Q7 x = number of medium frames made per week. y = number of large frames made per week. The constraints are: 4 x + 6 y ≤ 480 6 000 x + 12 000 y ≤ 900 000 20 ≤ x ≤ 60 20 ≤ y ≤ 60 Objective function: P = 6 x + 10 y 3 NCEA Answers – Linear Programming (a) Vertex P = 6 x + 10 y (60,40) 760 (30,60) 780 maximum profit when x = 30 and y = 60. (b) Moving the line 4x + 6y = 480 upwards, until it reaches the edge of the region bounded by the other constraints, leads to it crossing at the point where x = 60. ⇒ x = 60 and y = 45. ⇒ P = 810 >780, so this is the new maximum profit. ⇒ minimum hours = 8.5 h (8 hours & 30 minutes, 510min) OR Removing the Time constraint introduces the point (60, 45) into the feasible region. ⇒ P = 810 >780, so this is the new maximum profit. ⇒ minimum hours = 8.5 h (8 hours and 30 minutes, 510min) 2006 Q3 (b) (b) Identify vertices and value of I (15,12) $35.55 (15,40) $74.75 (30,30) $79.50 (45,12) $73.05 maximum income when 30 of each soap is produced. OR Use parallel line method to identify the point (30,30) as the optimal solution along with statement of number of each soap needed. [Evidence needed – maybe on the graph.] 4 NCEA Answers – Linear Programming Q6 (a) (i) The constraints are: 30x + 40y ≤ 1800 2x + 5y ≤ 190 x ≥ 10 y ≥ 15 (ii) Objective function is P = 8x + 12y EITHER calculate P at the vertices Vertex P = 8x + 12y (10,15) 260 (10,34) 488 (20,30) 520 (40,15) 500 OR Use the parallel line test to identify the optimal vertex. Therefore maximum profit is when Vili produces 20 sun shelters and 30 tents. (b) There will now be multiple solutions as the objective function is now parallel to the constraint 30x + 40y = 1800. Therefore all solutions will be integer points on the line 30x + 40y = 1800 between x = 20 and x = 40. Need • more than 2 solutions • where the solutions are • boundaries. Desirable • Whole number solutions. 2007 Q3 (a) (b) Vertex P = 9x + 16y (0,0) 0 (0,45) 720 5 NCEA Answers – Linear Programming (20,30) 660 (40,0) 360 OR Parallel line method to identify point (0,45) as optimal solution. [evidence needed – maybe on graph] AND So need to make 0 gnomes and 45 frogs . [Accept x = 0 and y = 45 as x and y defined in question]. [A clear logical argument as to why “as many frogs as possible” may gain A]. 2007 Q6(a) (i) x = no. of standard butterflies y = no. of monarch butterflies Constraints: 40x + 90y ≤ 2160 8x + 12y ≤ 336 x≥9 y≥9 (ii) Objective Function P = 18x + 60y EITHER Vertex P (9,9) 702 (9,20) 1362 (18,16) 1284 (28,9) 1044 [or (28.5,9) giving P = $1053 or (27,10) giving P = $1086] Max. profit when 9 standard and 20 monarch produced each day. OR Shown by parallel line method. (b) Vertex (9,20) no longer in feasible region. Maximum profit of $1218 now at point (21,14) Ie: 21 standard and 14 monarch . May use parallel line method. 2008 Q7 6 NCEA Answers – Linear Programming (a) (b) Line passing through (0,90) and (90,0) evident and feasible region shown Points for optimisation: (30,45) I = $1 425 000 (60,30) I = $1 350 000 (80,10) I = $1 050 000 OR Parallel line method The grower should plant 30 hectares of tomatoes and 45 hectares of artichokes Q8 x = sheep, y = cows 0.3 x + 0.4 y ≤ 120 180 x + 150 y ≤ 54 000 y ≤ 250 Intersection points for feasible region are Points of particular interest are at (0, 250), æ 2 ö æ ö 1 ç 66 , 250 ÷ , ç133 , 200 ÷ , (300, 0) è 3 ø è ø 3 (66,250) = $258 700 income (67,249) = $258 550 income (133,200) = $263 100 income (132,201) = $263 250 income The farmer should run 132 sheep and 201 cows (accept also 133 sheep and 200 cows). (b) Optimisation equation has changed to 560x + 850y . New income is $249 730 (from 68 sheep and 249 cows) OR New income is $249 460 (from 66 sheep and 250 cows). 2009 Q3 Only need to focus on two key points (others have to be smaller by coordinates).(20, 255/11)⇒ I = $101 363 7 NCEA Answers – Linear Programming (accept 25.5, 25.45 ⇒ I =$101 500, $101 350) (13, 30) ⇒ I = $107 187.50 (accept $107 187.50, $107 188) He should grow 13 . tonnes of corn and 30 tonnes of tomatoes. (Must have 13. Or 13.75.) (b) New constraint: 9c + 8t ≤ 360 (or equivalent) New vertices are (13⅓, 30) ⇒ I = $131 666 and (20, 22 ) ⇒ I = $118 750 Maximises income using13⅓ tonnes of corn and 30 tonnes of tomatoes . (Do not accept rounded answers.) (c) For multiple solutions to occur, the function to be optimised is parallel to one of the constraints. (Do not accept the constraints are all in the same place.) E.G. 2010 Q2 (a) C = 4.5a + 8b (b) Vertex Cost in $ (60,60) 750 (20,60) 570 (40,20) 340 (60,10) 350 OR Parallel line method to identify point (40,20) as optimal solution. [Evidence needed – maybe on graph.] So need to make 40 trips in Van A and 20 Trips in Van B. (b) New objective function C = 5a + 10b has the same gradient as the constraint 6a + 12b ≥ 480. Therefore any integer (whole) coordinates along the line 6a + 12b = 480, for 40 ≤ a ≤ 60, will minimise the fuel cost. If answer is correct except no mention of integer (whole) number solutions. 2011 8 NCEA Answers – Linear Programming (a) Vertices of feasible region are: Point Vertex (m,a) A (5,10) 40 minutes B (8,7) 37 minutes C (8,10) 46 minutes So therefore the assessment should have 7 algebra Q and 8 measurement questions . (b) Constraints are: a≥m a + m ≤ 20 a + m ≥ 11 a≥ 8 Vertices of feasible region are: Point Vertex (a,m) Object fn = 4a + 6m A (10,10) 100 minutes B (20,0) 80 minutes C (11,0) 44 minutes D (8,3) 50 minutes E (8,8) 80 minutes therefore minimum time is 44 minutes. (c) Test would now have between 6 and 20 algebra questions and between 0 and 10 measurement questions OR similar 9