Linear Programming Problems : Dr. Purnima Pandit
18 likes3,770 views
Linear programming problems involve optimizing an objective function subject to constraints on variables. They can be modeled and solved using techniques like the simplex method. The simplex method works by moving from one basic feasible solution to an adjacent extreme point through an exchange of variables in and out of the basis. It begins with an initial basic feasible solution and proceeds iteratively until an optimal solution is reached.
Linear Programming Problems : Dr. Purnima Pandit
- 1. Linear Programming Problems Dr. Purnima Pandit Assistant Professor Department of Applied Mathematics Faculty of Technology and Engg. The Maharaja Sayajirao University of Baroda
- 3. Elements Variables Objective function Constraints Obtain values of the variables that optimizes the objective function ( satisfying the constraints )
- 4. Linear Programming (LP) LP is a mathematical modeling technique used to achieve an objective, subject to restrictions called constraints
- 5. Types of LP
- 6. LP Model Formulation Decision variables mathematical symbols representing levels of activity of an operation Objective function a linear relationship reflecting the objective of an operation most frequent objective of business firms is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost Constraint a linear relationship representing a restriction on decision making
- 7. LP Model Formulation (cont.) Max/min z = c1x1 + c2x2 + ... + cnxn subject to: a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1 a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2 : am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm xj = decision variables bi = constraint levels cj = objective function coefficients aij = constraint coefficients
- 8. LP Model: Example RESOURCE REQUIREMENTS Labor Clay Revenue PRODUCT (hr/unit) (lb/unit) ($/unit) Bowl 1 4 40 Mug 2 3 50 There are 40 hours of labor and 120 pounds of clay available each day Decision variables x1 = number of bowls to produce x2 = number of mugs to produce
- 9. LP Formulation: Example Maximize Z = $40 x1 + 50 x2 Subject to x1 + 2x2 40 hr (labor constraint) 4x1 + 3x2 120 lb (clay constraint) x1 , x2 0 Solution is x1 = 24 bowls x2 = 8 mugs Revenue = $1,360
- 10. Graphical Solution Method 1. Plot model constraint on a set of coordinates in a plane 2. Identify the feasible solution space on the graph where all constraints are satisfied simultaneously 3. Plot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function
- 11. Convex Hull A set C is convex if every point on the line segment connecting x and y is in C. The convex hull for a set of points X is the minimal convex set containing X.
- 12. Graphical Solution: Example x2 50 – 40 – 4 x1 + 3 x2 120 lb 30 – 20 – Area common to both constraints 10 – x1 + 2 x2 40 hr 0– | | | | | | 10 20 30 40 50 60 x1
- 13. Computing Optimal Values x2 x1 + 2x2 = 40 40 – 4x1 + 3x2 = 120 4 x1 + 3 x2 120 lb 4x1 + 8x2 = 160 30 – -4x1 - 3x2 = -120 5x2 = 40 20 – x1 + 2 x2 40 hr x2 = 8 10 – 8 x1 + 2(8) = 40 0– | | 24 | | x1 x1 = 24 10 20 30 40 Z = $50(24) + $50(8) = $1,360
- 14. Extreme Corner Points x1 = 0 bowls x2 x2 =20 mugs x1 = 224 bowls Z = $1,000 40 – x2 =8 mugs Z = $1,360 x1 = 30 bowls 30 – x2 =0 mugs 20 – A Z = $1,200 10 – B 0– | | | C| 10 20 30 40 x1
- 15. Objective Function x2 40 – 4x1 + 3x2 120 lb 30 – Z = 70x1 + 20x2 Optimal point: x1 = 30 bowls 20 –A x2 =0 mugs Z = $2,100 10 – B x1 + 2x2 40 hr 0– | | | C | 10 20 30 40 x1
- 16. Minimization Problem CHEMICAL CONTRIBUTION Brand Nitrogen (lb/bag) Phosphate (lb/bag) Gro-plus 2 4 Crop-fast 4 3 Minimize Z = $6x1 + $3x2 subject to 2x1 + 4x2 16 lb of nitrogen 4x1 + 3x2 24 lb of phosphate x 1, x 2 0
- 17. Graphical Solution x2 14 – x1 = 0 bags of Gro-plus 12 – x = 8 bags of Crop-fast 2 10 – Z = $24 8–A Z = 6x1 + 3x2 6– 4– B 2– C 0– | | | | | | | 2 4 6 8 10 12 14 x1
- 18. Two type of Constraints Binding and Nonbinding constraints: A constraint is binding if the left-hand and right- hand side of the constraint are equal when the optimal values of the decision variables are substituted into the constraint. A constraint is nonbinding if the left-hand side and the right-hand side of the constraint are unequal when the optimal values of the decision variables are substituted into the constraint. 18
- 19. Special Cases of Graphical Solution • Some LPs have an infinite number of solutions (alternative or multiple optimal solutions). • Some LPs have no feasible solution (infeasible LPs). • Some LPs are unbounded: There are points in the feasible region with arbitrarily large (in a maximization problem) z-values. 19
- 20. Alternative or Multiple Solutions X2 60 B D 50 Any point (solution) Feasible Region falling on line segment 40 AE will yield an optimal solution with the same E 30 objective value z = 100 z = 120 20 z = 60 10 F A C 10 20 30 40 50 20 X1
- 21. No feasible solution X2 60 No Feasible Region 50 x1 >= 0 40 Some LPs have no x2 >=0 solution. Consider 30 the following formulation: 20 10 No feasible region exists 10 20 30 40 50 21 X1
- 22. Unbounded LP X2 Feasible Region 6 D The constraints are 5 B satisfied by all points z=4 bounded by the x2 axis 4 and on or above AB and 3 CD. 2 There are points in the feasible region which 1 z=6 will produce arbitrarily A C large z-values 1 2 3 4 5 6 X1 (unbounded LP). 22
- 23. Higher Dimensional Problems Graphical Method is useful for two dimensional Problems. Simplex method for 2 and higher dimensional problems
- 24. Cannonical form Standard form is a basic way of describing a LP problem. It consists of 3 parts: A linear function to be maximized maximize c1x1 + c2x2 + … + cnxn Problem constraints subject toa11x1 + a12x2 + … + a1nxn < b1 a21x1 + a22x2 + … + a2nxn < b2 … am1x1 + am2x2 + … + amnxn < bm Non-negative variables x1, x2 ,…, xn > 0
- 25. Obtain Standard Form IMPORTANT: Simplex only deals with equalities General Simplex LP model: min (or max) z = ci xi s.t. Ax=b x0 In order to get and maintain this form, use slack, if x b, then x + slack = b surplus, if x b, then x - surplus = b artificial variables (sometimes need to be added to ensure all variables 0 )
- 26. Different "components" of a LP model LP model can always be split into a basic and a non-basic part. “Transformed” or “reduced” model is another good way to show this. This can be represented in mathematical terms as well as in a LP or simplex tableau.
- 27. Simplex A simplex or n-simplex is the convex hull of a set of (n +1) . A simplex is an n- dimensional analogue of a triangle. Example: a 1-simplex is a line segment a 2-simplex is a triangle a 3-simplex is a tetrahedron a 4-simplex is a pentatope
- 28. Movement to Adjacent Extreme Point Given any basis we move to an adjacent extreme point (another basic feasible solution) of the solution space by exchanging one of the columns that is in the basis for a column that is not in the basis. Two things to determine: 1) which (nonbasic) column of A should be brought into the basis so that the solution improves? 2) which column can be removed from the basis such that the solution stays feasible?
- 29. Entering and Departing Vector (Variable) Rules General rules: The one non-basic variable to come in is the one which provides the highest reduction in the objective function. The one basic variable to leave is the one which is expected to go infeasible first. NOTE: THESE ARE HEURISTICS!! Variations on these rules exist, but are rare.
- 30. Example Problem Maximize Z = 5x1 + 2x2 + x3 subject to x1 + 3x2 - x3 ≤ 6, x2 + x3 ≤ 4, 3x1 + x2 ≤ 7, x1, x2, x3 ≥ 0.
- 31. Simplex and Example Problem Step 1. Convert to Standard Form a11 x1 + a12 x2 + ••• + a1n xn ≤ b1, a11 x1 + a12 x2 + ••• + a1n xn + xn+1 = b1, a21 x1 + a22 x2 + ••• + a2n xn ≥ b2, a21 x1 + a22 x2 + ••• + a2n xn - xn+2 = b2, am1 x1 + am2 x2 + ••• + amn xn ≤ bm, am1 x1 + am2 x2 + ••• + amn xn + xn+k = bm, In our example problem: x1 + 3x2 - x3 ≤ 6, x1 + 3x2 - x3 + x4 = 6, x2 + x3 ≤ 4, x2 + x3 + x5 = 4, 3x1 + x2 ≤ 7, 3x1 + x2 + x6 = 7, x1, x2, x3 ≥ 0. x1, x2, x3, x4, x5, x6 ≥ 0.
- 32. Simplex: Step 2 Step 2. Start with an initial basic feasible solution (b.f.s.) and set up the initial tableau. In our example Maximize Z = 5x1 + 2x2 + x3 x1 + 3x2 - x3 + x4 = 6, x2 + x3 + x5 = 4, 3x1 + x2 + x6 = 7, cB Basis cj Constants 5 2 1 0 0 0 x1, x2, x3, x4, x5, x6 ≥ 0. x1 x2 x3 x4 x5 x6 0 x4 1 3 -1 1 0 0 6 0 x5 0 1 1 0 1 0 4 0 x6 3 1 0 0 0 1 7 c row 5 2 1 0 0 0 Z=0
- 33. Step 2: Explanation Adjacent Basic Feasible Solution If we bring a nonbasic variable xs into the basis, our system changes from the basis, xb, to the following : x1 + ā1sxs= b1 x = b a for i =1, …, m i i is xr + ārsxr= b xs = 1 r xj = 0 for j=m+1, ..., n and js xm + āmsxs= b s The new value of the objective function becomes: m Z c (b a i 1 i i is ) c s Thus the change in the value of Z per unit increase in xs is cs = new value of Z - old m m value of Z = c (b a i 1 i m i is ) c s c b i 1 i i This is the Inner Product rule = c c a s i is i 1
- 34. Simplex: Step 3 Use the inner product rule to find the relative profit coefficients cB Basis cj Constants 5 2 1 0 0 0 x1 x2 x3 x4 x5 x6 0 x4 1 3 -1 1 0 0 6 0 x5 0 1 1 0 1 0 4 0 x6 3 1 0 0 0 1 7 c row 5 2 1 0 0 0 Z=0 c j c j cB Pj c1 = 5 - 0(1) - 0(0) - 0(3) = 5 -> largest positive c2 = …. c3 = …. Step 4: Is this an optimal basic feasible solution?
- 35. Simplex: Step 5 Apply the minimum ratio rule to determine the basic variable to leave the basis. The new values of the basis variables: xi = bi a is x s for i = 1, ..., m bi max xs min a is 0 a is In our example: cB Basis cj Constants 5 2 1 0 0 0 Row Basic Variable Ratio x1 x2 x3 x4 x5 x6 1 x4 6 0 x4 1 3 -1 1 0 0 6 2 x5 - 0 x5 0 1 1 0 1 0 4 3 x6 7/3 0 x6 3 1 0 0 0 1 7 c row 5 2 1 0 0 0 Z=0
- 36. Simplex: Step 6 Perform the pivot operation to get the new tableau and the b.f.s. cB Basis cj Constants 5 2 1 0 0 0 x1 x2 x3 x4 x5 x6 0 x4 1 3 -1 1 0 0 6 0 x5 0 1 1 0 1 0 4 0 x6 3 1 0 0 0 1 7 c row 5 2 1 0 0 0 Z=0 New iteration: find entering cB Basis cj Constants variable: 5 2 1 0 0 0 c j c j c B Pj x1 x2 x3 x4 x5 x6 0 x4 0 8/3 -1 1 0 0 11/3 cB = (0 0 5) 0 x5 0 1 1 0 1 0 4 5 x1 1 1/3 0 0 0 1/3 7/3 c2 = 2 - (0) 8/3 - (0) 1 - (5) 1/3 = 1/3 c row 0 1/3 1 0 0 -5/3 Z=35/3 c3 = 1 - (0) (-1) - (0) 1 - (5) 0 = 1 c6 = 0 - (0) 0 - (0) 0 - (5) 1/3 = -5/3
- 37. Final Tableau cB Basis cj Constants x3 enters basis, 5 2 1 0 0 0 x5 leaves basis x1 x2 x3 x4 x5 x6 0 x4 0 8/3 -1 1 0 0 11/3 0 x5 0 1 1 0 1 0 4 5 x1 1 1/3 0 0 0 1/3 7/3 c row 0 1/3 1 0 0 -5/3 Z=35/3 cB Basis cj Constants 5 2 1 0 0 0 x1 x2 x3 x4 x5 x6 0 x4 0 11/3 0 1 1 0 23/3 1 x3 0 1 1 0 1 0 4 5 x1 1 1/3 0 0 0 1/3 7/3 c row 0 -2/3 0 0 -1 -5/3 Z=47/3
- 38. Computational Considerations Unrestricted variables (unboundedness) Redundancy (linear dependency, modeling errors) Degeneracy (some basic variables = 0) Round-off errors
- 39. Simplex Variations Various variations on the simplex method exist: "regular" simplex two-phase method: Phase I for feasibility and Phase II for optimality Condensed / reduced / revised method: only use the non-basic columns to work with (revised) dual simplex, etc.
- 40. Limitations of Simplex 1. Inability to deal with multiple objectives 2. Inability to handle problems with integer variables Problem 1 is solved using Multiplex Problem 2 has resulted in: Cutting plane algorithms (Gomory, 1958) Branch and Bound (Land and Doig, 1960) However, solution methods to LP problems with integer or Boolean variables are still far less efficient than those which include continuous variables only
- 41. Primal and Dual LP Problems •Economic theory indicates that scarce (limited) resources have value. In LP models, limited resources are allocated, so they should be, valued. •Whenever we solve an LP problem, we implicitly solve two problems: the primal resource allocation problem, and the dual resource valuation problem. •Here we cover the resource valuation, or as it is commonly called, the Dual LP.
- 42. Max c X j j Primal s.t. a X j ij j bi for all i j Xj 0 for all j Min b Y i i Dual s.t. i a Y i ji i c j for all j Yi 0 for all i