Chapter 16 The Total LP Formulation is Given by

CHAPTER 16 16.4 (a) The total LP formulation is given by Maximize C = 0.15 X + 0.025Y + 0.05Z {Maximize profit} subject to X +Y + Z ≥ 6 X +Y < 3 X −Y ≥ 0 Z − 0.5Y ≥ 0 {Material constraint} {Time constraint} {Storage constraint} {Positivity constraints} (b) The simplex tableau for the problem can be set up and solved as (c) An Excel spreadsheet can be set up to solve the problem as The Solver can be called and set up as The resulting solution is In addition, a sensitivity report can be generated as (d) The high shadow price for storage from the sensitivity analysis from (c) suggests that increasing storage will result in the best increase in profit. 16.5 An LP formulation for this problem can be set up as Maximize C = 0.15 X + 0.025Y + 0.05Z {Maximize profit} subject to X +Y + Z ≥ 6 X +Y < 3 X −Y ≥ 0 {X material constraint} {Y material constraint} {Waste constraint} Z − 0.5Y ≥ 0 {Positivity constraints} (b) An Excel spreadsheet can be set up to solve the problem as The Solver can be called and set up as The resulting solution is This is an interesting result which might seem counterintuitive at first. Notice that we create some of the unprofitable z2 while producing none of the profitable z3. This occurred because we used up all of Y in producing the highly profitable z1. Thus, there was none left to produce z3. 16.6 Substitute xB = 1 – xT into the pressure equation, (1 − xT ) Psat B + xT PsatT = P and solve for xT, xT = P − Psat B PsatT − Psat B where the partial pressures are computed as PsatB = 10 1211    6. 905−  T + 221   (1) PsatB = 10 1344    6. 953−  T + 219   The solution then consists of maximizing Eq. 1 by varying T subject to the constraint that 0 ≤ xT ≤ 1. The Excel solver can be used to obtain the solution of T = 111.04. 16.7 This is a straightforward problem of varying xA in order to minimize  1 f ( x A ) =  2  (1 − x A )     0 .6  1 + 5  xA    0. 6 (a) The function can be plotted versus xA 25 20 15 10 5 0 0 0.2 0.4 0.6 0.8 1 (b) The result indicates a minimum between 0.5 and 0.6. Using Golden Section search or a package like Excel or MATLAB yields a minimum of 0.564807. 16.8 This is a case of constrained nonlinear optimization. The conversion factors range between 0 and 1. In addition, the cost function can not be evaluated for certain combinations of XA1 and XA2. The problem is the second term, x A1    1−  xA2    (1 − x ) 2  A2     0. 6 If xA1 > xA2, the numerator will be negative and the term cannot be evaluated. Excel Solver can be used to solve the problem: The result is 16.9 Errata: Change B0 to 100. This problem can be set up on Excel and the answer generated with Solver: The solution is 16.10 The problem can be set up in Excel Solver. The solution is 16.11 w A s d θ e The following formulas can be developed: e= w 2 θ = tan −1 (1) d e (2) s = d 2 + e2 (3) P = 2s (4) A= (5) wd 2 Then the following Excel worksheet and Solver application can be set up: Note that we have named the cells with the labels in the adjacent left columns. Our goal is to minimize the wetted perimeter by varying the depth and width. We apply positivity constraints along with the constraint that the computed area must equal the desired area. The result is Thus, this specific application indicates that a 45o angle yields the minimum wetted perimeter. The verification of whether this result is universal can be attained inductively or deductively. The inductive approach involves trying several different desired areas in conjunction with our solver solution. As long as the desired area is greater than 0, the result for the optimal design will be 45o. The deductive verification involves calculus. The minimum wetted perimeter should occur when the derivative of the perimeter with respect to one of the primary dimensions (i.e., w or d) flattens out. That is, the slope is zero. In the case of the width, this would be expressed by: dP =0 dw If the second derivative at this point is positive, the value of w is at a minimum. To formulate P in terms of w, substitute Eqs. 1 and 5 into 3 to yield s = (2 A / w) 2 + ( w / 2) 2 (6) Substitute this into Eq. 4 to give P = 2 (2 A / w) 2 + ( w / 2) 2 (7) Differentiating Eq. 7 yields dP = dw − 8 A 2 / w3 + w / 2 ( 2 A / w) 2 + ( w / 2) 2 =0 (8) Therefore, at the minimum − 8 A 2 / w3 + w / 2 = 0 (9) which can be solved for w=2 A (10) This can be substituted back into Eq. 5 to give d= A (11) Thus, we arrive at the general conclusion that the optimal channel occurs when w = 2d. Inspection of Eq. 2 indicates that this corresponds to θ = 45o. The development of the second derivative is tedious, but results in d 2P A2 = 32 ( 2 A / w) 2 + ( w / 2) 2 dw 2 w4 (12) Since A and w are by definition positive, the second derivative will always be positive. 16.12 w A s d θ b e The following formulas can be developed: e= (1) d tan θ b = w − 2e (2) s = d 2 + e2 (3) P = 2s + b (4) A= w+ b d 2 (5) Then the following Excel worksheet and Solver application can be set up: Note that we have named the cells with the labels in the adjacent left columns. Our goal is to minimize the wetted perimeter by varying the depth, width and theta (the angle). We apply positivity constraints along with the constraint that the computed area must equal the desired area. We also constrain e that it cannot be greater than w/2. The result is Thus, this specific application indicates that a 60o angle yields the minimum wetted perimeter. 16.13 Aends = 2πr 2 Aside = 2πrh Atotal = Aends + Aside Vcomputed = 2πr 2 h Cost = Fends Aends + Fside Aside + Foperate Aoperate Then the following Excel worksheet and Solver application can be set up: which results in the following solution: 16.14 Excel Solver gives: x = 0.5, y = 0.8 and fmin = -0.85. 16.19 100 = π 3 ( 29) r 4 4 L2 35 = πr 2 L L= 4 L2 = π 3 (.29) r 4 L = 1.499r2 35 πr 2 r = 1 .65 m L = 4.08 m 16.20 I1 = 4 I2 = 2 I3 = 2 I4 = 0 I5 = 2 P = 80 16.22 Total cost is C = 2 p1 + 10 p2 + 2 Total power delivered is P = 0.6 p1 + 0.4 p2 Using the Excel Solver: which yields the solution 16.23 This is a trick question. Because of the presence of (1 – s) in the denominator, the function will experience a division by zero at the maximum. This can be rectified by merely canceling the (1 – s) terms in the numerator and denominator to give T= 15s 4 s − 3s + 4 2 Any of the optimizers described in this section can then be used to determine that the maximum of T = 3 occurs at s = 1. 16.27 An LP formulation for this problem can be set up as Maximize C = 0.15 X + 0.025Y + 0.05Z {Minimize cost} subject to X +Y + Z ≥ 6 X +Y < 3 X −Y ≥ 0 Z − 0.5Y ≥ 0 {Performance constraint} {Safety constraint} {X-Y Relationship constraint} {Y-Z Relationship constraint} (b) An Excel spreadsheet can be set up to solve the problem as The Solver can be called and set up as The resulting solution is 16.28 τ= Tc 500 ro ⇒ 20000000 = π ( ro4 − ri4 ) J 2 ri = 4 ro4 − 1.5915 x 10− 5 ro φ= TL  π  ⇒ 2.5 = JG  180  500( 5) π  77 x 109   ro4 − ri4  2 ( ) ri = 4 ro4 − 2.8422 x 10− 7 ro = 29.76 mm ri = 23.61 mm but ro − ri ≥ 8 mm ∴ ro = 29.76 mm, ri = 21.76 mm 16.29 L= h= Re µ = 0.567 ρV 2F C D ρV 2 b = .0779 h = L = 0.567 cm A 1 2 3 4 5 6 7 B C X 1.5 1 1 1 0 0.15 Amount Performance Safety X-Y Z-Y Cost Set target cell: E7 Equal to ❍ max ● min ❍ value of By changing cells B2:D2 Subject to constraints: E3≥F3 E4≤F4 E5≥F5 E6≥F6 A B D Y 1.5 1 1 -1 -0.5 0.025 E Z 3 1 0 0 1 0.05 Total F Constraint 6 3 0 2.25 0.4125 6 3 0 0 0 C D E F 1 2 3 4 5 6 7 X 0 1 1 1 0 0.15 Amount Performance Safety X-Y Z-Y Cost A 1 2 3 4 5 6 B Y 0 1 1 -1 -0.5 0.025 C Z1 4000 1 2.5 1 2500 amount amount X amount Y amount W profit D Z2 3500 1 0 -1 -50 Set target cell: F6 Equal to ● max ❍ min ❍ value of By changing cells B2:E2 Subject to constraints: B2≥0 C2≥0 F3≤G3 F4≤G4 F5=G5 A 1 2 3 4 5 6 B Total Constraint 0 0 0 0 0 6 3 0 0 E Z3 0 0 1 -1 200 F W 500 0 0 -1 -300 total 7500 10000 0 9675000 G constraint 7500 10000 0 0 C Z1 0 1 2.5 1 2500 amount amount X amount Y amount W profit Z 0 1 0 0 1 0.05 D Z2 0 1 0 -1 -50 E Z3 0 0 1 -1 200 F W 0 0 0 -1 -300 total 0 0 0 0 G constraint 7500 10000 0 Microsoft Excel 5.0c Sensitivity Report Worksheet: [PROB1605.XLS]Sheet3 Report Created: 12/12/97 9:47 Changing Cells Cell Name $B$2 amount Product 1 $C$2 amount Product 2 $D$2 amount Product 3 Final Reduced Objective Value Cost Coefficient 150 0 30 125 0 30 175 0 35 Allowable Allowable Increase Decrease 0.833333333 2.5 1.666666667 1 35 5 Final Shadow Constraint Value Price R.H. Side 3000 0.625 3000 55 12.5 55 450 26.25 450 Allowable Allowable Increase Decrease 1E+30 1E+30 1E+30 1E+30 1E+30 1E+30 Constraints Cell Name $E$3 material total $E$4 time total $E$5 storage total A 1 2 3 amount material B Product 1 150 5 C Product 2 125 4 D Product 3 175 10 E total F constraint 3000 3000 4 5 6 time storage profit 0.05 1 30 Set target cell: E6 Equal to ● max ❍ min ❍ value of By changing cells B2:D2 Subject to constraints: E3≤F3 E4≤F4 E5≤F5 A B Product 1 0 5 0.05 1 30 0.1 1 30 0.2 1 35 C Product 2 0 4 0.1 1 30 D Product 3 0 10 0.2 1 35 55 450 14375 55 450 0 E total F constraint 1 2 3 4 5 6 amount material time storage profit Basis P S1 S2 S3 P 1 0 0 0 x1 -30 5 0.05 1 x2 -30 4 0.1 1 x3 -35 10 0.2 1 S1 0 1 0 0 S2 0 0 1 0 S3 0 0 0 1 Solution 0 3000 55 450 Intercept Basis P S1 x3 S3 P 1 0 0 0 x1 -21.25 2.5 0.25 0.75 x2 -12.5 -1 0.5 0.5 x3 0 0 1 0 S1 0 1 0 0 S2 175 -50 5 -5 S3 0 0 0 1 Solution 9625 250 275 175 Intercept Basis P x1 x3 S3 P 1 0 0 0 x1 0 1 0 0 x2 -21 -0.4 0.6 0.8 x3 0 0 1 0 S1 8.5 0.4 -0.1 -0.3 S2 -250 -20 10 10 S3 0 0 0 1 Solution 11750 100 250 100 Basis P x1 x3 x2 P 1 0 0 0 x1 0 1 0 0 x2 0 0 0 1 x3 0 0 1 0 S1 0.625 0.25 0.125 -0.375 S2 12.5 -15 2.5 12.5 S3 26.25 0.5 -0.75 1.25 Solution 14375 150 175 125 0 0 0 0 3000 55 450 300 275 450 100 1100 233.3333 Intercept -250 416.6667 125