Linear Programming Review.ppt
- 2. 2 •Linear Programming (LP) Problems Both objective function and constraints are linear. Solutions are highly structured and can be rapidly obtained. Linear Programming (LP) •Has gained widespread industrial acceptance since the 1950s for on-line optimization, blending etc. •Linear constraints can arise due to: 1. Production limitation e.g. equipment limitations, storage limits, market constraints. 2. Raw material limitation 3. Safety restrictions, e.g. allowable operating ranges for temperature and pressures. 4. Physical property specifications e.g. product quality constraints when a blend property can be calculated as an average of pure component properties: n 1 i i iP y P Chapter 7
- 3. 3 5. Material and Energy Balances - Tend to yield equality constraints. - Constraints can change frequently, e.g. daily or hourly. •Effect of Inequality Constraints - Consider the linear and quadratic objective functions on the next page. - Note that for the LP problem, the optimum must lie on one or more constraints. •Generic Statement of the LP Problem: subject to: •Solution of LP Problems - Simplex Method (Dantzig, 1947) - Examine only constraint boundaries - Very efficient, even for large problems n 1 i i ix c f max 1 0 1,2,..., 1,2,..., i n ij j i j x i n a x b i n Chapter 7
- 4. 4 Figure The effect of an inequality constraint on the maximum of quadratic function, f(x) = a0 +a1 x + a2 x2. The arrows indicate the allowable values of x. Chapter 7
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- 6. 6 x1 x3 x4 x2 x5 x6 Refinery input and output schematic. Chapter 7
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- 8. 8 Chapter 7 Solution Let x1 = crude #1 (bbl/day) x2 = crude #2 (bbl/day) Maximize profit (minimize cost): y = income – raw mat’l cost – proc.cost Calculate amounts of each product Produced (yield matrix): gasoline x3 = 0.80 x1 + 0.44 x2 kerosene x4 = 0.05 x1 + 0.10 x2 fuel oil x5 = 0.10 x1 + 0.36 x2 residual x6 = 0.05 x1 + 0.10 x2 Income gasoline (36)(0.80 x1 + 0.44 x2) kerosene (24)(0.05 x1 + 0.10 x2) fuel oil (21)(0.10 x1 + 0.36 x2) residual (10)(0.05 x1 + 0.10 x2)
- 9. 9 So, Income = 32.6 x1 + 26.8 x2 Raw mat’l cost = 24 x1 + 15 x2 Processing cost = 0.5 x1 + x2 Then, the objective function is Profit = f = 8.1 x1 + 10.8 x2 Constraints Maximum allowable production: 0.80 x1 + 0.44 x2 < 24,000 (gasoline) 0.05 x1 + 0.10 x2 < 2,000 (kerosene) 0.10 x1 + 0.36 x2 < 6,000 (fuel oil) and, of course, x1 > 0, x2 > 0 Chapter 7
- 10. 10 Chapter 7 Graphical Solution 1. Plot constraint lines on x1 – x2 plane. 2. Determine feasible region (those values of x1 and x2 that satisfy maximum allowable production constraints. 3. Find point or points in feasible region that maximize f = 8.1 x1 + 10.8 x2; this can be found by plotting the line 8.1 x1 + 10.8 x2 = P, where P can vary, showing different profit levels.
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- 21. 21 Chapter 7 Convert inequalities to equalities using slack variables
- 22. 22 Chapter 7 Minimize: f = cTx (7.6) Subject to: Ax = b (7.7) and I < x < u (7.8)
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- 24. 24 Chapter 7 DEFINITION 1: A feasible solution to the linear programming problem is a vector x = (x1, x2, …., xn) that satisfies all constraints and bounds (7.8). DEFINITION 2. A basis matrix is an m x m nonsingular matrix formed from some m columns of the constraint matrix A. DEFINITION 3. A basic solution to a linear program is the unique vector determined by choosing a basis matrix, and solving the resulting system of equations for the remaining m variables. DEFINITION 4. A basic feasible solution is a basic solution in which all variables satisfy their bounds (7.8). DEFINITION 6. An optimal solution is a feasible solution that also minimizes f in Equation (7.6).
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- 26. 26 Slack variables 1 r ij i i j a x b 1 0 r i j i i i i j a x s b s refinery example: 2 variables r = 2 3 constraints p = 3 (3 slacks) n = r + p = 5 total variables m = q + p = 3 total constraints (q = 0 = no. equality constraints) 3 eqns / 5 unknowns set 2 variables = 0 basic feasible sol’n set (n – m) variables = 0 non-basic m variables ≠ 0 basic (could have infinite # soln’s If variables can assume any value) possible solutions ! = with 2 variables = 0 !( - )! n n m m n m 5 10 possible constraint interactions 3 (constraint intersections) Chapter 7
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- 29. 29 In initiating the simplex algorithm, we treat the objective function As just another equation, that is, The basic variables are the first m, that is x1 … xm and –f. Find values of x1 > 0, x2 > 0, . . . . Xn > 0 and min f satisfying 1 1 2 2 n n f c x c x c x 1 1 2 2 0 n n f c x c x c x (7.11) Chapter 7
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- 31. 31 Chapter 7 Assume that we know that x5, x1, -f can be used as basic variables. We can pivot successively on the terms x5 (first equation) and x1 (second equation)
- 32. 32 Chapter 7 Reduced cost coefficient = -24 (< 0): not optimal Increasing x3 causes f to decrease f = 28 -24 x3 (7.21) Maximum value of x2 ? Check constraints (x2 = x4 = 0) x3 = 5 -3x3 x1 = 3 -2 x3 (7.22) 3 c
- 33. 33 Chapter 7 Is f optimal ? x3 replaces x1 as a basic variable using pivot transformation.
- 34. 34 Chapter 7 5 1 2 4 3 1 2 4 1 2 4 1.5 0.875 0.375 0.5 0.5 0.375 0.125 1.5 12 2 8 x x x x x x x x f x x x (7.25) 5 2 3 2 2 0.5 0.875 = 1.5 0.375 8 x x x x f x is not optimal because 1 Check how much can be increased. 2 2 f c x (7.26)
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- 37. 37 1 2 Ex min f x x 1 2 1 2 3 1 2 1 2 4 1 2 1 2 5 (A) 2 2 2 2 (B) 3 2 3 2 (C) 4 4 x x x x x x x x x x x x x x x start at 1 2 1 2 0, 0 ( 0, 0) x x x x which variable when increased will improve obj. fcn more? 1 2 ( or ) x x 1 x 1 2 f x x How far can be increased? 1 x 2 hold 0 x constraint (1) no limit (2) (3) 1 2.0 limiting constraint x 1 4.0 x (see Figure of feasible region) calculate new basic feasible sol’n and repeat above analysis – iterate until obj. fcn cannot be improved further (row operations) add slacks Chapter 7
- 38. 38 Sensitivity Analysis • How does the value of the optimum solution change when coefficients in the obj. fcn. or constraints change? • Why is sensitivity analysis important? - Coefficients and/or limits in constraints may be poorly known - Effect of expanding capacity, changes in costs of raw materials or selling prices of products. • Market demand of products vary • Crude oil prices fluctuate Sensitivity information is readily available in the final Simplex solution. Optimum does not have to be recomputed. Chapter 7
- 39. 39 Sensitivity Analysis (Constraints) Shadow price: The change in optimum value of obj. fcn. per unit change in the constraint limit. Final Set of Equations of Refinery Blending Problem x3 = 0 x4 = 0 x5 + 0.14 x3 – 4.21 x4 = 896.5 x1 + 1.72 x3 – 7.59 x4 = 26,207 x2 – 0.86 x3 + 13.79 x4 = 6,897 f – 4.66 x3 – 87.52 x4 = -286,765 ↑ gasoline constraint ↑ kerosene constraint Chapter 7
- 40. 40 Sensitivity Analysis 3 4 5 x = 0 gasoline constraint active x = 0 kerosene constraint active x = 896.5 fuel oil constraint active Which constraint improves obj. fcn. more (when relaxed)? • D = 1 bbl (x3 = -1) $4.66 Df = 4.66 Dx3 (x4 = -1) $87.52 Df = 87.52 Dx4 • No effect of fuel oil (x5);x5 ≠ 0 Inactive constraint Shadow prices Chapter 7
- 41. 41 Sensitivity Analysis gasoline capacity is worth $4.66/bbl kerosene capacity is worth $87.52/bbl fuel oil capacity is worth $0/bbl←No effect Capacity limit in original constraints * shadow prices 4.66 (24,000) + 87.52 (2,000) = 286,880 Same as $286,740 Duality (roundoff) Chapter 7
- 42. 42 Sensitivity Analysis (Obj. Fcn.) small changes use solution (matrix) large changes ("ranging" of the coefficients) recompute optimum. From final tableau opt 1 opt 2 x = 26,207 x = 6,897 Crude oil prices change (Coeff. in obj. fcn.) Max. profit = 8.1 x1 + 10.8 x2 $1.00 9.1 x1 or 11.8 x2 ↓ x1 profit coefficient. Chapter 7
- 43. 43 Duality • One dual variable exists for each primal constraint • One dual constraint exists for each primal variable • The optimal solution of the decision variables (i.e., the Dual Problem) will correspond to the Shadow Prices obtained from solution of the Primal Problem. • Commercial Software will solve the Primal and Dual Problems. i.e., it provides sensitivity information. Chapter 7
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