Linear Programming Model Answers

Jan 2008

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) 10x + 20y + 15z ≥ 2(6x + 9y + 12z) (multiplying by 5) 10x + 20y + 15z ≥ 12x + 18y + 24z (multiplying by 5) 2x + 4y + 3z (Type A) ≥ 0.4(40%) x (6x + 9y + 12z) 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

Jan 2006

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

Jan 2010

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

Jan 2009

Giles buys x goats, y pigs and z sheep x + y + z ≥ 110 x ≥ y y + z ≤ x + 8y + 24z ≤ 3120 simplified becomes 2x + y + 3z ≤ 390 P = 70x + 30y + 50z

z = 30 x + y + z ≥ 110 becomes x + y ≥ 80 x ≥ y stays the same x ≥ y y + z ≤ 150 becomes y ≤ 120 2x + y + 3z ≤ 390 becomes 2x + y ≤ 300 P = 70x + 30y Write each as an equation and plot on the graph x + y = 80 x = y y = 120 2x + y = 300

y = 120 2x + y = 300 x + y = 80 y = x Feasible Region P = 70x + 30y Objective Line x = 30 and y = 70 As 70x + 30y = 2100 Objective line

Consider all the integer points of intersection in the feasible region. Which co-ordinates give the maximum profit. (90, 120) gives a profit of £ x = 90 x 70, y = 120 x 30, + z = 30 x 50 gives a profit of £ (100, 100) gives a profit of £ x = 100 x 70, y = 100 x 30, + z = 30 x 50 gives a profit of £ goats, 100 pigs and 30 sheep.

Tins 6x + 9y + 9z ≤ 600Simplified 2x + 3y + 3z ≤ 200 Packets 9x + 6y + 9z ≤ 600 Simplified 3x + 2y + 3z ≤ 200 Bottles 6x + 12y + 18z ≥ 480Simplified x + 2y + 3z ≥ 80 Jan 2011

So z = y 2x + 3y + 3z ≤ 200 → 3x + 2y + 3z ≤ 200 → x + 2y + 3z ≥ 80 → As z = y, then put the z and y values together as y’s 2x + 6y ≤ 200 → 3x + 5y ≤ 200 x + 3y ≤ 100 x + 5y ≤ 80

Write each inequality as an equation 3x + 5y ≤ 200 becomes x + 3y ≤ 100 becomes x + 5y ≤ 80 becomes x + 3y = 100 3x + 5y = 200 x + 5y = 80 Also remember that x, y and z are also ≥ 0 Now plot the graph See next slide

x + 5y = 80 x + 3y = 100 3x + 5y = 200 x ≤ 0 y ≤ 0 Feasible Region iii) Maximum is when x = 25 and y = 25 Remember x + 2y as there are 3 variables, x, y and z and z = y So x + 2y = = 75 iv) 25 basic 25 standard 25 luxury ii)

June 2009

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)

Feasible Region

June 2010

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

There would be 320 slow, 50 medium and 50 fast balls

Jan 2007

F. R. Feasible Region shown

Write the inequalities as equations x ≥ 0, y ≥ 0 x + 4y ≤ 36 4x + y ≤ 68 y ≤ 2x y ≥ ¼x Plot these on the graph and shade the UNWANTED region

x + 4y = 36 y = 2x y = ¼x 4x + y = 68 FEASIBLE REGION

Maximum value for P = x + 5y isMax (4, 8) = 44 Maximum value for P = 5x + y isMax (16, 4) = 84

Change all the inequalities to equations y = 20 x + y = 25 5x + 2y = 100 y = 4x y = 2x Plot the lines on the graph Shade the region that doesn’t represents the inequality

y = 20 x + y = 25 5x + 2y = 100 y = 4x y = 2x Feasible Region

You can use co-ordinates or an objective line If you use co-ordinates they have to be integers Minimum for P = x + 2y is x = 5 and y = 2 is P = 45 Minimum for P = -x + y is x = 10 and y = 20 is P = 10

200x y = 5000 → x + 5y ≥ x y = 6000 → 2x + 15y ≥ x y = 4000 → x + 25y ≥ 40 (C =) 2.5x +15y

x + 5y ≥ 25 2x + 15y ≥ 60 x + 25y ≥ 40 Feasible Region Objective Line (C =) 2.5x +15y 2.5x + 15y = 37.5 x = 15, y = 2.5 Objective Line

Feasible Region Objective Line 15 DIY, 2 Trade Cost (C =) 2.5x +15y 2.5 x x 2 = = £67.50

Try all the vertices of the feasible region i) 30 (9, 3) ii) 29.6 (6.4, 5.6) iii) -15 (9, 3)

Feasible Region Objective Line

5x +10y ≤1500 (balloons) ⇒ x + 2y ≤ x + 8y ≤ 4000 (sweets) ⇒ 4x + y ≤ 500 x ≥ 50, y ≥ 50, at least 50 of each x + y ≥ 140, at least 140 in total

4x + y = 500 x + 2y = 300

i) Maximum ( 100,100 ) = £200 ii) Maximum ( 90,50 ) = £132

June 2013

June 2014

Try all the vertices of the feasible region i) 30 (9, 3) ii) 29.6 (6.4, 5.6) iii) -15 (9, 3)

June 2014

≤ 240 (note 240 is the 4 hours changed to minutes) Machine A Machine B ≤ 210 Machine C ≤ 420 Simplified becomes ≤ 120 Machine A Machine B ≤ 30 Machine C ≤ 60