Linear Programming Model Answers.

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Presentation transcript:

Linear Programming Model Answers

Type A 2x + 4y + 3z ≤ 360 Type B 3x + 2y + 4z ≤ 270 Type C x + 3y + 5z ≤ 450 Altogether 6x + 9y + 12z ≥ 720 (the minimum you can make it) Simplified 2x + 3y + 4z ≥ 240 (the minimum you can make it) 2x + 4y + 3z (Type A) ≥ 0.4(40%) x (6x + 9y + 12z) 10x + 20y + 15z ≥ 2(6x + 9y + 12z) (multiplying by 5) 10x + 20y + 15z ≥ 12x + 18y + 24z (Expand brackets) You must leave ONE value on the left – as y would be the only value to leave a positive value, bring the y’s over from the right 2y ≥ 2x + 9z

Type A 5x + 4y + 3z ≤ 180 (3 hours is 180 minutes) Type B 12x + 8y + 10z ≤ 240 (4 hours is 240 minutes) Type C 24x + 12y + 18z ≤ 540 (9 hours is 540 minutes) Simplified becomes Type A 5x + 4y + 3z ≤ 180 (No Common Factors) Type B 6x + 4y + 5z ≤ 120 (Divide by 2) Type C 4x + 2y + 3z ≤ 90 (Divide by 6)

x > y y > z x ≥ 0.4 (x + y + z) 5x ≥ 2 (x + y + z) 5x ≥ 2x + 2y + 2z) 3x ≥ 2y + 2z

Type A 2x + 3y + 4z ≤ 360 Type B 3x + y + 5z ≤ 300 Type C 4x + 3y + 2z ≤ 400 Type A > Type B 2x + 3y + 4z > 3x + y + 5z You must leave ONE value on the left – as y would be the only value to leave a positive value, bring the y’s over from the right 2y > x + z

5x + 4y + 9z ≥ 4x + 3y + 2z So x + y + 7z ≥ 0 4x + 3y + 2z ≥ 0.4 (9x + 7y + 11z) 20x + 15y + 10z ≥ 2 (9x + 7y + 11z) 20x + 15y + 10z ≥ 18x + 14y + 22z You must leave at least ONE value on the left – as x and y would leave positive values bring the x and y’s over from the right 2x + y ≥ 12z

Type A 6x + 4y + 2z ≤ 240 Type B 6x + 3y + 9z ≤ 300 Type C 12x + 18y + 6z ≤ 900 Simplified becomes Type A 3x + 2y + z ≤ 120 (Divide by 2) Type B 2x + y + 3z ≤ 100 (Divide by 3) Type C 2x + 3y + z ≤ 150 (Divide by 6) Type C ≥ 2 x Type B 12x + 18y + 6z ≥ 2(6x + 3y + 9z) 12x + 18y + 6z ≥ 12x + 6y + 18z 12y ≥ 12z y ≥ z

So luxury (z) = basic (x) z = x Type A 3x + 2y + z ≤ 120 becomes Type A 4x + 2y ≤ 120 (as z = x) Type A 2x + y ≤ 60 (Simplified) Type B 2x + y + 3z ≤ 100 becomes Type B 5x + y ≤ 100 (as z = x) Type C 2x + 3y + z ≤ 150 becomes Type C 3x + 3y ≤ 150 (as z = x) Type C x + y ≤ 50 (Simplified) Lastly if y ≥ z, then y ≥ x (as z = x)

Slow x ≥ 190 Medium y ≥ 50 Fast z ≥ 50 Total is x + y + z ≥ 300 Cost 2.5x + 2y + 2z ≤ 1000 Cost 5x + 4y + 4z ≤ 2000 As well x ≥ 0.6(x + y + z) 5x ≥ 3(x + y + z) 5x ≥ 3x + 3y + 3z 2x ≥ 3y + 3z

Medium balls (y) = Fast balls (z) so z = y Slow x ≥ 190 (unaffected) Fast y ≥ 50 (as z = y) Total is x + y + z ≥ 300 becomes x + 2y ≥ 300 (as z = y) Cost 5x + 4y + 4z ≤ 2000 becomes 5x + 8y ≤ 2000 (as z = y) As well x ≥ 0.6(x + y + z) = 2x ≥ 3y + 3z becomes 2x ≥ 6y (as z = y) becomes x ≥ 3y Re-arranged to y ≤ ⅓x