Review Exercise a Chemical A 5x+ y 0 Chemical B x+ y [ x+ y 6] b Chemical C 6 [ ] x+ y x+ y x, y 0 c T = x+ y d ( x, y) = (, ) T = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a Maximise P= 00x+ 500y b Finishing.5x+ y56 x+ 8y (o.e.) Packing x+ y0 x+ y 0 (o.e.) c d For example, point testing Test all corner points in feasible region. Find profit at each and select point yielding maximum. profit line Draw profit lines. Select point on profit line furthest from the origin. e Using a correct, complete method. Making 6 Oxford and York gives a profit = 500 integer (6, ) 500 (.,.) (, ) 00 (6, 0) 800 (0, 0) 5000 f The line.5x+ y = 9 passes through (6, ) so reduce finishing by hours. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a Objective: maximise P= 0x+ 0 y (or P= 0.x+ 0. y ) subject to: x+ y00 x+ y500 0 x ( x+ y ) xy 00 x 0 ( x+ y ) x y 00 x, y 0 b c Visible use of objective line method objective line drawn or vertex testing all vertices tested Vertex testing (0, 60) 600 (80, 0) 00 (00, 00) 9 000 (00, 00) 8000 Intersection of y= x and x+ y = 500 (00, 00) profit 90 (or 9000 p) Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a b Visible use of objective line method objective line drawn or vertex testing. 5, 5 8, (, 8) 8(, 6) 6 6 6 5 Optimal point, 6 with value 5 6 c Visible use of objective line method objective line drawn, or vertex testing all vertices tested. 5, 6 8, not an integer try (, ) 0 (, 8) 8 not an integer try (8, ) (, 6) Optimal point (8, ) with value, so Becky should use kg of bird feeder and.5 kg of bird table food. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
5 a Objective: maximise P= 0.x+ 0. y ( P= 0x+ 0 y ) subject to: x6.5 y8 x+ y yx x, y0 b Visible use of objective line method objective line drawn (e.g. from (, 0) to (0, )) or all 5 points tested. vertex testing (0, 0) 0;(, 8).;(, 8).;(6.5, 5.5).;(6.5, 0).6 [ ] Optimal point is (6.5, 5.5) 6 500 type X and 5500 type Y c P = 0.(6500) + 0.(5500) = 00 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 a Maximise P = 50x + 80y + 60z Subject to x+ y+ z 0 x+ y+ z0 x+ y+ z50 where x, y, z 0 b Initialising tableau b.v. x y z r s t value r 0 0 0 s 0 0 0 t 0 0 50 P 50 80 60 0 0 0 0 Choose correct pivot, divide R by State correct row operation R R, R R, R + 80R, R c The solution found after one iteration has a slack of 0 units of black per day d i b.v. x y z r s t value r 0 0 0 y 0 0 0 t 0 0 0 0 P 0 0 0 0 0 0 600 (given) b.v. x y z r s t value z 0 0 6 R y 0 0 6 R R t 0 0 0 0 R no change P 0 0 0 R + 0R ii Not optimal, as there is a negative value in profit row iii x= 0 y= 6 z = 6 P=. r= 0, s= 0, t = 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6
a Objective: Maximise P= x+ 5y+ z Subject to x+ y+ z 5 x+ y+ z0 b x+ y+ z b.v. x y z r s r 0 5 0 s 0 0 y 0 0 P 0 0 0 t value R R R R 6 R 5 0 R + 5R b.v. x y z r s 5 x 0 0 s 0 0 8 y 0 0 8 P 0 0 0 8 P= x=, y=, z = 0 t value 8 8 89 8 R R + R R R R + R c There is some slack on s, so do not increase blending: therefore increase Processing and Packing which are both at their limit at present. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
8 a x+ y+ z b i x+ y+ z+ s = ii s ( 0) is the slack time on the machine in hours c euro d b.v. x y z r s value r 0 R 6R z 0 6 R P 0 0 0 R + R b.v. x y z t s value y 0 R 5 5 z 0 R R 8 8 P 0 0 R + R Profit = euros y = z =.5 x = r = s = 0 e Cannot make a lamp f e.g. (0, 0, 0) or (0, 6, ) or (,, ) checks in both inequalities Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8
9 a Small (x) Board (m) Time (R) 0 Medium (y) 0 0 Large (z) 5 50 Available 00 000 Board Time 0 5 00 x+ y+ z x+ y+ 6z0 0x+ 0y+ 50z000 x+ y+ 5z00 b P= 0x+ 0y+ 8z c b.v. x y z r s values r 6 0 0 s 5 0 00 P 0 0 8 0 0 0 θ = 0, θ = 50; pivot d Row b.v. x y z r s value operation y 0 0 R s 0 0 R R P 5 0 5 0 600 R + 0R θ = 0, θ = 80;pivot Row b.v. x y z r s value operation y 0 0 R R x 0 80 R P 0 0 0 0 000 R + 5R e This tableau is optimal as there are no negative numbers in the profit line. f Small 80, medium 0; large 0 Profit 000 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9
0 a The constraints include a mixture of and variables. b x + y + z + s = 0 x + y + z + s = 5 x + y + z s + a = c The purpose is to maximise I = a which =x+ y+ zs So Ixy z+ s = This gives the final row of the tableau. d The smallest value in the bottom row is and the smallest θ value is a (6) so this is the pivot and we obtain: b.v. x y z s s s a value s 0 0.5.5 0 0.5 0.5 s 0.5 0.5 0 0.5 0.5 9 x 0.5 0.5 0 0 0.5 0.5 6 P 0.5.5 0 0 0.5 0.5 6 I 0 0 0 0 0 0 0 e There are no negative values in the bottom row, so the optimal value of I is 0 when a = 0 f There is a negative value in the bottom row. g b.v. x y z s s s Value z 0 0 y 0 0 x 0 0 P 0 0 0 5 5 The maximum value of P is 0 = 6 which occurs when x = 5, y=, z=, s = s = s = 0 5 0 Row operation R R R R R R+ R Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0
a x + y + z + s = 8 x + y + z s + a = 6 x + 5y + z s + a = 0 b New objective is maximise I = (a + a ) a = x + y + z s 6 a = x + 5y + z s 0 In terms of non-basic variables, the new objective is maximise I = 5x + 6y + z s s 6 c b.v. x y z s s s a a value s 0 0 0 0 8 a 0 0 0 6 a 5 0 0 0 0 P 0 0 0 0 0 0 I 5 6 0 0 0 6 d st iteration Pivot is y-column a row Row b.v. x y z s s s a a Value operation s 0 0 0 5 5 5 5 0 R R 5 a 0 0 5 5 5 5 R R 5 y 0 0 0 5 5 5 5 5 R 6 P 0 0 0 0 5 5 5 5 R R 5 6 6 I 0 0 0 R5 + R 5 5 5 5 5 nd iteration Pivot is x column a row Row b.v. x y z s s s a a Value operation s 0 0 5 5 8 R R x 0 5 5 0 5 0 R y 0 8 0 R R P 0 0 6 5 5 8 0 R + R I 0 0 0 0 0 0 0 R5+ R Basic feasible solution is x= 0, y=, z= 0, s 8 =, s = s = a = a = 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a x + y + z + s = 5x + y + z + s = 60 x s + a = b P = x + y + z Ma = x + y + z M( x + s ) P = x( + M) + y + z M Ms P ( + M)x y z + Ms = M c b.v. x y z s s s a value s 0 0 0 s 5 0 0 0 60 a 0 0 0 0 P ( + M) 0 0 M 0 M d The most negative value in the P row is in the x-column so, in the first iteration, x enters the basic variables. e st iteration x column a row is the pivot b.v. x y z s s s a Value Row operation s 0 0 8 R R s 0 0 5 5 50 R 5R x 0 0 0 0 P 0 0 0 ( + M) R+ ( + M) R nd iteration z column s row is the pivot b.v. x y z s s s a Value Row operation z 0 0 8 s 0 0 R R x 0 0 0 0 P 0 5 0 0 M R+ R All entries in the P row are non-negative so the tableau represents the optimal solution. x =, y = 0, z = 8, s = 0, s =, s = 0, a = 0 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a x + y + z + s = 6 x + z + s = 5 x + y s + a = 0 b Maximise P = x + y z Ma = x + y z M(0 x y + s ) = x(m ) + y(m + ) z 0M Ms Rearranging gives P (M )x (M + )y + z + Ms = 0M c b.v. x y z s s s a value s 0 0 0 6 s 0 0 0 0 5 a 0 0 0 0 P (M ) (M + ) 0 0 M 0 0M d st iteration y column a row is the pivot b.v. x y z s s s a Value Row operation s 0 0 6 R R s 0 0 0 0 5 y 0 0 0 0 P 5 0 0 0 M + 0 R + ( M + ) R nd iteration s column s row is the pivot Row b.v. x y z s s s a Value operation s 0 0 R s 0 0 0 0 5 y 0 0 0 R + R P 6 0 0 0 M 6 R+ R Maximum value of P = 6 Minimum value of C = 6 This occurs when x = 0, y =, z = 0, s = 0, s = 5, s = Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
5 a 6 a b Here we have that I and J depend only on E, whereas H depends on C, D, E and F. Hence we need separate nodes with a dummy. b D will only be critical if it lies on the longest path Path A to G Length A B E G A C F G 5 A C D E G + x So we need + x to be the longest, or equal longest +x5 x Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a b Total float on A = 0 0 5 = 5 Total float on H = 9 = Total float on B = 0 0 0 = 0 Total float on I = 9 = 0 Total float on C = 0 = Total float on J = 5 8 = Total float on D = 0 0 8 = Total float on K = 5 = 0 Total float on E = 0 0 = Total float on L = 5 8 = 0 Total float on F = 9 0 9 = 0 Total float on M = 5 = 9 Total float on G = 8 = c Critical activities: B, F, I, K and L length of critical path is 5 days d New critical path is B F H L length of new critical path is 6 days 8 a x = 0 y = [latest out of ( + ) and (5 + )] z = 9 [Earliest out of ( ) and (9 ) and (6 )] b Length is Critical activities: B, D, E and L c i Total float on N = = 5 ii Total float on H = 6 5 = 8 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5
9 a For example, it shows dependence but it is not an activity. G depends on A and C only but H and I depend on A, C and D. b c So B, C, E, F, I, J and L d Total float on A = 0 9 = Total float on H = 5 = Total float on D = = Total float on K = 5 6 = Total float on G =8 5 = e Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6
0 a Critical activities are B, F, J, K and N length of critical path is 5 hours I is not critical. b Total float on A = 5 0 = Total float on H = 6 = Total float on C = 9 0 6 = Total float on I = 6 9 5 = Total float on D = = 5 Total float on L = = Total float on E = 9 = Total float on M = 6 = Total float on G = 9 = Total float on P = 5 8 = c a d Look at 6.5 in the chart in c. F, E and G b 6 days, workers c Delay the start of H until time d Activity H would have to take place on its own so the project will be delayed by at least days. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
a b 8 days c ADFIKN d e workers f e.g. delay the start times of: E to time G to time H to time J to time 0 L to time M to time 6 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 8
a b A, C, G, H, J, K and L All critical activities have a zero total float. c Total float = 5 = d Either 6 8 =.6 ( d.p.) so at least workers needed (here 6 is the total number of hours required for all the activities) or 69 hours into the project activities J, K, I and M must be happening so at least workers will be needed. e New shortest time is 89 hours. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 9
a b Critical activities: A, C, F and H; length of critical path = c Total float on B = 0 5 = Total float on E = = Total float on D = 9 = Total float on G = 9 8 = d e For example; Minimum time for workers is days. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 0
Challenge a Optimal values x = y = 5 b x =, y = P = x = 5, y = P = 5 c x =, y = P = d If the gradient of the objective line is similar to the gradient of a constraint that runs through the optimal vertex, then the optimal integer solution may not lie close to the optimal vertex. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
Challenge We need to maximise D= x+ y z subject to the constraints x+ y+ z6 x+ z5 x+ y0 We rewrite these using slack, surplus and artificial variables to obtain: x+ y+ z+ s = 6 x+ z+ s = 5 x+ y s+ a = 0 The new objective function is: I = a = x+ ys 0 So Ix y+ s =0 So the initial tableau is: b.v. x y z s s s a Value s 0 0 0 6 s 0 0 0 0 5 a 0 0 0 0 D 0 0 0 0 0 I 0 0 0 0 0 Both the x and y columns have the same value in the I row. But sin the x-column has the smallest θ value (9) so we use that as the pivot. We then obtain: b.v. x y z s s Row s a Value operation x 0 0 0 9 R s 0 R R a 0 5 R R D 0 9 R R I 0 0 0 R5 + R Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
Challenge continued The y-column still has a negative value in the I row and only the x-row has a positive θ value so we need to use that as the pivot giving. Row b.v. x y z s s s a Value operation y 0 0 0 R s 0 0 0 0 0 5 R+ R a 5 0 0 6 R+ 5R D 6 0 0 0 0 6 R+ 6R I 0 0 0 R5 + R There are now no negative entries in the bottom two rows so we have reached the optimal solution: y =, x = z = 0 and C = 6 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.