Chapter 3 Solutions 1 2 (b) Δw (1) = (4, 2, 10) (4, 0, 7) = (0, 2, 3), Δw Δw (3) = ( 2, 4, 5) (4, 2, 10) = ( 6, 2, 5),

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1 Optimization in Operations Research, nd Edition Ronald L. Rardin Solution Manual Completed download: () Chapter 3 Solutions (b) Δw () = (4,, 0) (4, 0, 7) = (0,, 3), 3-. (a) x () is feasible and only a local max because all constraints are satisfied and no nearby point has better objective value, but there are better such as (, 3). x () is infeasible and thus no sort of optimum because it violates constraint x 0. x (3) is feasible because it satisfies all constraints, but no sort of optimum because it can be improved in the neighborhood. x (4) is feasible and both a local and a global max because it satisfies all constraints and has better objective value than any other point in the feasible region. (b) x () is infeasible and thus no sort of optimum because it violates constraint x + x 4. x () is feasible because it satisfies all constraints, but no kind of optimum because it can be improved in a variety of directions. x (3) is feasible and a local minimum because it cannot be improved in the neighborhood, but not global because other feasible points such as x (4) have better objective values. x (4) is feasible and both a local and a global minimum because it satisfies all constraints and has better objective value than any other point in the feasible region. 3-. (a) y () = (, 0, 5) + (3,, 0) = (8,, 5), y () = (8,, 5) + 5(,, ) = (3, 8, 0), y (3) = (3, 8, 0) + /(0, 6, 0) = (3,, 0) (b) y () = (, 0, 5) + (, 3, ) = (4, 6, ), y () = (4, 6, ) + /(, 0, ) = (4 /, 6, ), y (3) = (4 /, 6, ) + (4, 3, ) = (5 /, 4, 6) 3-3. (a) Δw () = (4,, 7) (0,, ) = (4,, 6), Δw () = (4, 3, 9) (4,, 7) = (0,, 8), Δw (3) = (3, 3, ) (4, 3, 9) = (, 0, 3) Supplement to the nd edition of Optimization in Operations Research, by Ronald L. Rardin, Pearson Higher Education, Hoboken NJ, c 07. As of October, 05 Δw Δw (3) = (, 4, 5) (4,, 0) = ( 6,, 5), = (5, 5, 5) (, 4, 5) = (7,, 0) 3-4. (a) Nonimproving because the objective value worsens in the neighborhood along this direction. (b) Nonimproving because the objective decreases in the neighborhood along this direction. (c) Improving because the objective improves in the neighborhood along this direction. (d) Improving because the objective improves in the neighborhood along this direction. (e) Nonimproving because x (3) is a local minimum. (f ) Improving because the objective value improves in the neighborhood along this direction (a) Feasible because movement along this direction retains feasibility in the neighborhood. (b) Infeasible because any movement along this direction produces infeasibility. (c) Feasible because movement along this direction retains feasibility in the neighborhood. (d) Feasible because movement along this direction retains equality of the only active constraint. (e) Feasible because any movement along this direction violates no constraints immediately. (f ) Infeasible because movement along this direction immediately violates x (a) (4 λ) (0 + 3λ) + 3(6 λ) 5 is satisfied for all positive λ; (4 λ) 0 is satisfied for λ 4; (0 + 3λ) 0 is satisfied for all positive λ; (6 3λ) 0 is satisfied for λ 3. Thus the proper step is λ = min{4, 3} = 3. The model is not unbounded because this λ is finite. (b) (8 λ) (5 + λ) + 3(3 λ) 5 is satisfied for all positive λ; (8 λ) 0 is satisfied for λ 4; (5 + λ) 0 is satisfied for all positive λ; (3 λ) 0 is satisfied for λ 3. Thus the ope tep is pr r s λ in 4, 3 = 3. The model is not = m { } unbounded because this λ is finite. (c) (0 + λ) (0 + 3λ) + 3(4 + λ) 5 is satisfied for all positive λ; (0 + λ) 0 is satisfied for all positive λ; (0 + 3λ) 0 is

2 satisfied for all positive λ; (4 + λ) 0 is satisfied for all positive λ. Thus λ = +, and the model is unbounded. (d) (0 + λ) (4 + 7λ) + 3(3 + 4λ) 5 is satisfied for all positive λ; (0 + λ) 0 is satisfied for all positive λ; (4 + 7λ) 0 is satisfied for all positive λ; (3 + 4λ) 0 is satisfied for all positive λ. Thus the model is unbounded because steps in this direction can be arbitrarily large without losing feasibility (a) f (, 3, 4, 0, 6) = (4, 0,, 0, ) and f Δy = (4, 0,, 0, ) (, 3, 4, 0, 6) = 6, so improving for a maximize. (b) f (, 0, 9, 0, 0) = (, 0, 7, 0, ) and f Δy = (, 0, 7, 0, ) (, 0,, 0, 4) = 7 > 0, so improving for a maximize. (c) f (3, ) = ( + (3), 3 + 4) and f Δy = (5, 7) ( 7, 5) = 0, so more information is needed. (d) f (, ) = (() +, + 4) and f Δy = (5, 6) (, 3) = 3 < 0, so nonimproving for a minimize. (e) f (4, ) = ((4 5), ( + )) and f Δy = (, 4) (, ) = >0, so improving for a maximize. (f ) f (, ) = (( ) +, + ( 3)) and f Δy = (, 3) (3, ) = 0, so more information is needed (a) Δw = f (, 0, 5, ) = (3,, 0, ) (b) Δw = f (,,, 0) = (0, 4, 5, ) (c) Δw = f (3, ) = ((3 + ), 3) = ( 8, 3) (d) Δw = f (, ) = ( 4, 9 + 4()) = ( 4, 7) 3-9. (a) (4 ) + (0 ) = 5<0 so [i] is inactive; (4) (0) = 8 so [ii] is active; (4)>0 so [iii] is inactive; (0) = 0 so [iv] is active. (b) (6 ) + (4 ) = 5 so [i] is active; (6) (4) = 8 so [ii] is active; (6)>0 so [iii] inactive; (4)>0 so [iv] inactive (a) Active constraints are 3y y + 8y 3 = 4 and y 0. For these a () Δy = 3,, 8) (0, 4, ) = 0 and a (4) Δy = 0,, 0) (0, 4, ) = 4 0 as required for feasibility. (b) Active constraints are 3y y + 8y 3 = 4 and y 0. For these a () Δy = (3,, 8) (0, 4, ) = 6, infeasible for an = constraint. a (4) Δy = (0,, 0) (0, 4, ) = 4 0 infeasible for a constraint. (c) Active constraints are 3y y + 8y 3 = 4 and y 0. In the first, a () Δy = (3,, 8) (, 0, ) = 4 = 0 as required so infeasible. (d) Active constraints are 3y y + 8y 3 = 4 and y 0. For these a () Δy = (3,, 8) (,, ) = 0 as required for feasibility of an = constraint. a (3) Δy = (, 0, 0) (,, ) = infeasible fore a 0 constraint. 3-. (a) Active constraints are w + 3w 3 = 8, w + w + w 3 = 4, and w 0. Thus conditions are Δw + 3 Δw 3 = 0, Δw + Δw + Δw 3 = 0, Δw 0 (b) Active constraints are w + 3w 3 = 8,and w + w + w 3 = 4. Thus conditions are Δw + 3 Δw 3 = 0, Δw + Δw + Δw 3 = 0, (c) Active constraints are w + w = 0 and w w 8. Thus conditions are Δw + Δw = 0, Δw Δw 0 (d) Only active constraint is w + w = 0. Thus conditions are Δw + Δw = 0, 3-. (a) Δy + 5 Δy <0 (b) Direct substitution (c) Active constraints are y 0 and y + y 3. (d) Active constraints yield conditions Δy + 0 Δy 0, Δy + Δy 0. (e) Direct supstitution shows direction is feasible. Maximum step is λ = where constraint y 0 is encountered.

3 3 3 (f ) (c) z z 0 z () = (5/,3) z (3) = (3,3) 3 z (0) = (0,0) z 4 z 0 z 4 3 z ( = ) (4,0) z 3-5. (a) f Δz () = (, ) (0, 5) = 5<0 and 3-3. (a) 3 Δx 3 Δx 3 <0 (b) () f Δz = (, ) (, ) = <0 as 3( ) 3() = 6<0 as required. (c) Only first constraint with (3) + 4(9) = 69, and fourth constraint x = 0 are active at (3, 0, 9). (d) Δx + 3 Δx + 4 Δx 3 = 0 and Δx 0. (e) Direct substitution; Checking limits on λ, (3 λ) + (0 + λ) + (9 + λ) 6 required for a minimize. (b) With both directions improving at all z, the only considerations are when directions are feasible. At z (0) = (5, 3), only Δz () is feasible because Δz () has a Δz () = (0, ) (, ) = 0 for active gives λ, 3 λ 0 gives λ 3. z 3. A maximum feasible step follows Combining λ min{, 3} =. Then () () for λ = 3/5 to z x () (3, 0 +, 9 + 4) = (,, 3) Δz () = (5, 0). At () 3-4. (a) f Δz () = (4, 7) (, 0) = 8>0 z = (5, 0), further pursuit of Δz is () and f Δz () = (4, 7) (, 4) = 0>0 as infeasible, but Δz = (, ) is now feasible. required for improving directions in a maximize. (b) With both directions improving at all z, the only considerations are when directions are feasible. At z (0) = (0, 0), only Δz () is feasible because Δz () has A maximum feasible step follows Δz () for λ = 5/ to z () = (0, 5/). At z () = (0, 5/), further pursuit of Δz () is infeasible, but Δz () = (0, 5 is again feasible. A maximum feasible step follows Δz () for λ = /0 to a Δz () = (, 0) (, 4) = 0 for active z 0. A maximum feasible step follows Δz () for λ = to z () = (4, 0). At z () = (4, 0), further pursuit of Δz () is z (3) = (0, ). At z (3), both directions are infeasible, and the search terminates. (c) z z 0 infeasible, but Δz () = (, 4) is feasible. A maximum feasible step follows Δz () for λ = 3/4 to z () = (5/, 3). At z () = (5/, 3), further pursuit of Δz () is infeasible, but Δz () = (, 0) is again feasible. A maximum 3 (3) z = (0,) z () = (0,5/) z 3 z (0) = (5,3) z 0 z 5 z z() = (5,0) feasible step follows Δz () for λ = /4 to z (3) = (3, 3). At z (3), both directions are infeasible, and the search terminates (a) (3,, 0) + λ( 3, 3, 9), λ [0, ]. Setting λ = /3 in this z (3) expression yields ; no λ gives z (4). (b) (6, 4, 4) + λ(4, 4, 3), λ [0, ]. Setting λ = 3/4 in this expression yields z (3) ; no λ gives z (4). 3-7.

4 4 (a) x (b) 0 x 0 8 6 4 x 0 4 6 8 violation x From the graph, the set is not convex because part of the line segment from x () = (0, 3) to x () = (3, 0) lies outside the feasible region.

4 4 4 (a) x (b) 0 x x violation x From the graph, the set is not convex because part of the line segment from x () = (0, 3) to x () = (3, 0) lies outside the feasible region. From the graph, the set is convex because the line segment between every pair of feasible solutions lies entirely within the feasible region. (c) Convex because all constraints are linear. (d) Convex because all constraints are linear. (e) Not convex because fractional solutions between say x () = (0, 0, 0, 4) and x () = (0, 0, 0, 5) are infeasible. (f ) Not convex because fractional solutions between the all x j = and all x j = 0 solutions are infeasible (a) Only the first and third constraints are violated at w = 0. Adding nonnegative artificial variables w 4 in the first and w 5 in the third, and minimizing their sum, produces Phase I model: min w 4 + w 5, s.t. 40w + 30w + 0w 3 + w 4 = 50, w w 0, 4w + w 3 + w 5 0, w, w, w 3, w 4, w 5 0. Setting w 4 = 50 40(0) 30(0) 0(0) = 50 and w 5 = 0 4(0) (0) = 0 (or any higher value) completes a starting (artificially) feasible solution. (b) Only the second and third constraints are violated at w = 0. Adding nonnegative artificial variable w 3 in the second, and w 4 in the third, then minimizing their sum, produces Phase I model: min w 3 + w 4, s.t. w + w 3, w + w 3, w + w 4 w +, w, w, w 3, w 4 0. Setting w 3 = and w 4 = (or any higher values), completes a starting (artificially) feasible solution.

5 5 5 (c) All three constraints are violated at w = 0. Subtracting nonnegative artificial variable w 4 in the first, adding w 5 in the second, adding w 6 in the third, and minimizing their sum, produces Phase I model: min w 3 + w 4 + w 5, s.t. (w 3) + (w 3) w 3 4, w + w + w 4 = 5, w + w 5 3, w 3, w 4, w 5 0. Setting w 3 = (3 0) + (3 0) 4 = 4 (or any larger value), w 4 = 5 (0) = (0) = 5, and w 5 = 3 (0) = 3 (or any larger value) completes a starting (artificially) feasible solution. (d) Only the third constraint is violated at w = 0. Adding nonnegative artificial variable w 3 there and minimizing its value produces Phase I model: min w 3, s.t. w w 9, w = 4w, w + w 3, w, w 3 0. Setting w 3 = (0) = (or any greater value) completes a starting (artificially) feasible solution (a) Constraints w and w make it impossible to also satisfy w + w 5. (b) Only the first constraint is violated at w = 0. Adding nonnegative artificial variable w 3 there and minimizing its value produces Phase I model: min w 3, s.t. w + w + w 3 5, 0 w, 0 w, w 3 0. (c) Setting w 3 = 5 (or any greater value) completes a starting (artificially) feasible solution. (d) The optimal solution is w = w =, w 3 =. Phase optimal value w3 = >0 proves the original model is infeasible (a) Stop and conclude the model is infeasible because all artificial variables cannot be eliminated. (b) Drop artificial variables and proceed with Phase II from initial solution y = (6, 3, ) which is feasible because both artificials = 0 at the end of Phase I. (c) Drop artificial variables and proceed with Phase II from initial solution y = (, 3, ) which is feasible because both artificials = 0 at the end of Phase I. (d) With one artificial positive, the current (y, y, y 3 ) solution is not feasible in the original model. But there may still be such a solution because the optimum is only local. Repeat Phase I from a new starting point. 3-. Needed artificial variables and their starting (artificially) feasible values are exactly as in Exercise??. (a) For a maximize model, subtract a large multiple of the artificial variables in the objective to obtain big-m model: max w w + 5w 3 M (w 4 + w 5 ), s.t. 40w + 30w + 0w 3 + w 4 = 50, w w 0, 4w + w 3 + w 5 0, w, w, w 3, w 4, w 5 0; w 4 = 50, w 5 = 0 (b) For a minimize model, add a large multiple of the artificial variables in the objective to obtain big-m model: min w + 5w + M (w 3 + w 4 ), s.t. w + w 3, w + w 3, w + w 4 w +, w, w, w 3, w 4 0. (c) For a minimize model, add a large multiple of the artificial variables in the objective to obtain big-m model: min w + 3w + M (w 3 + w 4 + w 5 ), s.t. (w 3) + (w 3) w 3 4, w + w + w 4 = 5, w + w 5 3, w 3, w 4, w 5 0; w 3 = 4, w 4 = 5, w 5 = 3 (d) For a maximize model, subtract a large multiple of the artificial variables in the objective to obtain big-m model: max (w ) + w() M w 3, s.t. w w 9, w = 4w, w + w 3, w, w (a) Stop and conclude the model is infeasible if M is big enough because artificial variables remain positive. Otherwise, increase M and repeat the search. (b) Stop and conclude y = (6, 3, ) is a global optimum for the original model because it is feasible with all artificials =0 and optimal because a global optimum was obtained with the big-m model. (c) Conclude y = (, 3, ) is a local optimum for the original model because it is feasible with all artificials =0 but possibly not a global optimum because the big-m search yielded only a local. If desired, repeat the big-m search from a new starting point. (d) Conclude nothing be-

6 6 6 cause only a local optimum has been obtained and some artificials remain positive. Repeat the big-m search from a new starting point and/or using a larger value of M. More download link optimization in operations research solution manual pdf optimization in operations research rardin pdf download optimization in operations research by ronald l. rardin pdf optimization in operations research (nd edition) pdf ronald l rardin optimization in operations research free download optimization in operations research pdf operations research - wayne winston ronald rardin optimization in operations research pdf

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