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Chapter 2

Chapter 2. Linear Programming. Introduction. In management it is always an important to proper allocation of limited and scarce resources among the competing departments of an organisation.

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Chapter 2

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  1. Chapter 2 Linear Programming

  2. Introduction • In management it is always an important to proper allocation of limited and scarce resources among the competing departments of an organisation. • The output may be measured in the form of profits, costs, effectiveness, production which can be presented as a liner relationship among a number of variables. • Available resources can also be expressed in linear relationship with some variables. • The goal is to optimize (minimize or maximize) the objective function with subject to the set of constraints.

  3. Basic Concept • Decision variables to determine • Objective to optimize • Constraints to satisfy • Technological co-efficient to determine

  4. Application • Product mix problems • Diet planning problems • Cargo loading problems • Capital budgeting problems • Manpower planning problems

  5. General Rules • LPP is a mathematical programming where functions representing the objectives and constraints are linear. • Usually it is defined as Maximize or minimize Z = C1X1+C2X2+---+CnXn Which is subject to constraints, a11x1+a12x2+---+a1nxn –b1 a21x1+a22x2+---+a2nxn – b2 -------------------------------------- am1+x1+am2x2+---+amnxn – bm Where x1>/ , X2>/0 --- Xn >/0

  6. Elementary transformation used for LPP • Minimization of function is mathematically equivalent to maximization of negative expression of this function. I.e minimize Z = 8x1 + 6x2 is equal to maximize –Z = -8x1 - 6x2

  7. Inequality in one direction may be changed to inequality in opposite direction by multiplying both sides of the inequality by –1. I.e. 5 > 3 or –5 < -3 • An equation can be replaced by tow inequalities in opposite direction. I.e. 4x1+9x2 = 12 can be represented as 4x1+9x2 /< 12 and 4x1+9x2 >/ 12 or 4x1+9x2 /< 12 and –4x1-9x2 /< -12

  8. Inequality constraint with left hand side in the absolute form can be changed into two regular inequalities. I.e. 5x1+ 8x2

  9. Example 1 Ex. A small company consisting of two carpenters and a finisher produce and sell two types of tables: type A and type B. The type-A table will result in a profit of $50, and each type-B table will result in a profit of $54. A type-A table requires 3 hours of carpentry and 1 hour of finishing. A type-B table requires 2 hours of carpentry and 2 hours of finishing. Each day there are 16 hours available for carpentry and 8 hours available for finishing. How many tables of each type should be made each day to maximize profit?

  10. Organize the Information: Let x = # type A and y= # type B. The Profit to Maximize (in dollars) is given by: P = 50x + 54y

  11. The constraints are given by: Carpentry Finishing Also so that the number of units is not less than 0: So we have:

  12. Example 2 Ex. A particular company manufactures specialty chairs in two plants. Plant I has an output of at most 150 chairs/month. Plant II has an output has an output of at most 120 chairs/month. The chairs are shipped to 3 possible warehouses - A, B, and C. The minimum monthly requirements for warehouses A, B, and C are 70, 70, and 80 respectively. Shipping charges from plant I (to A, B, and C) are $30, $32, and $38/chair and from plant II (to A, B, and C) are $32, $28, $26. How many chairs should be shipped to each warehouse to minimize the monthly shipping cost?

  13. Organize the Information: Number of Chairs Cost to Ship

  14. We want to minimize the cost function: C = 30x1 + 32x2 + 38x3 + 32x4 + 28x5 + 26x6 Production constraints: Plant I Plant II Warehouse constraints: A B C

  15. So the problem is: Minimize: Subject to: C = 30x1 + 32x2 + 38x3 + 32x4 + 28x5 + 26x6

  16. Examples 3 • A Firm engaged in producing two products. A and B. each unit of A requires 2 kg raw material and 4 labor hours for processing, while each unit of product B requires 3 kg raw material and 3 labor hours. Every week firm has availability of 60 kg raw material and 96 labor hours. Selling cost of each unit of A and B are Rs. 40and 35 respectively. Formulate problem in order to maximize the profit.

  17. Example 4 • The agricultural research institute suggested to a farmer to spread out at least 4800 kg of a special phosphate fertilizer and not less than 7200 kg of a special nitrogen fertilizer to raise productivity of crops in his fields. There are two sources for obtaining these – mixture A and B. the cost of both is Rs. 40 and 24 of 100 kg respectively. A contains 20 and 80kg of phosphate and nitrogen respectively while B contains thee ingredients equivalent of 50 kg each. • Formulate this problem and determine how many bags of each type the farmer should buy in order to obtain the required fertilizer at minimum cost.

  18. Graphical Method of LPP • The main limitation is that it can be used for two variables. • Following steps required - identify the problem, decision variables, obj. function and constraint restrictions. - Draw a graph that includes all the constraints/ restrictions and identify the feasible region. - Obtain the point on the feasible region that optimizes the objective function the optimal solution - Interpret the results

  19. Graphic Analysis • The feasible region is the area on the graph that contains the solutions that satisfy all the constraints simultaneously. • To find the feasible region, first locate the feasible points for each constraint and then the area that satisfies all constraints. • Generally, the following three rules identify the feasible points for a given constraint: 1. For the = constraint, only the points on the line are feasible solutions. 2. For the ≤ constraint, the points on the line and the points below or to the left of the line are feasible. 3. For the ≥ constraint, the points on the line and the points above or to the right of the line are feasible.

  20. Graphing Linear Inequalities Ex. Graph Notice that the line (=) is part of the solution Also, any point in the lower half-plane satisfies the inequality so this region is shaded.

  21. Graphing Linear Inequalities Ex. Graph The line (=) is not part of the solution (so it is dashed. Any point in the lower half-plane satisfies the inequality.

  22. Ex. Graph Dashed since < Test (0,0): 3(0)+2(0) < 6 True so shade region containing (0,0)

  23. Ex. Graph the solution set for the system: Points must satisfy both inequalities (overlap of individual shaded regions)

  24. Ex. Graph Ex. Graph

  25. The Galaxy Industries Production Problem –A Prototype Example • Galaxy manufactures two toy doll models: • Space Ray. • Zapper. • Resources are limited to • 1000 pounds of special plastic. • 40 hours of production time per week.

  26. The Galaxy Industries Production Problem –A Prototype Example • Marketing requirement • Total production cannot exceed 700 dozens. • Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 350. • Technological input • Space Rays requires 2 pounds of plastic and • 3 minutes of labor per dozen. • Zappers requires 1 pound of plastic and • 4 minutes of labor per dozen.

  27. 8(450) + 5(100) The Galaxy Industries Production Problem – A Prototype Example • The current production plan calls for: • Producing as much as possible of the more profitable product, Space Ray ($8 profit per dozen). • Use resources left over to produce Zappers ($5 profit per dozen), while remaining within the marketing guidelines. • The current production plan consists of: • Space Rays = 450 dozen • Zapper = 100 dozen • Profit = $4100 per week

  28. Management is seeking a production schedule that will increase the company’s profit.

  29. A linear programming model can provide an insight and an intelligent solution to this problem.

  30. The Galaxy Linear Programming Model • Decisions variables: • X1 = Weekly production level of Space Rays (in dozens) • X2 = Weekly production level of Zappers (in dozens). • Objective Function: • Weekly profit, to be maximized

  31. The Galaxy Linear Programming Model Max 8X1 + 5X2 (Weekly profit) subject to 2X1 + 1X2£ 1000 (Plastic) 3X1 + 4X2£ 2400 (Production Time) X1 + X2£ 700 (Total production) X1 - X2£ 350 (Mix) Xj> = 0, j = 1,2 (Nonnegativity)

  32. 2.3 The Graphical Analysis of Linear Programming The set of all points that satisfy all the constraints of the model is called a FEASIBLE REGION

  33. Using a graphical presentation we can represent all the constraints, the objective function, and the three types of feasible points.

  34. Graphical Analysis – the Feasible Region The non-negativity constraints X2 X1

  35. The Plastic constraint 2X1+X2 £ 1000 Graphical Analysis – the Feasible Region X2 1000 700 Total production constraint: X1+X2 £700 (redundant) 500 Infeasible Feasible Production Time 3X1+4X2 £2400 X1 500 700

  36. Graphical Analysis – the Feasible Region X2 The Plastic constraint 2X1+X2 £ 1000 1000 700 Total production constraint: X1+X2 £ 700 (redundant) 500 Infeasible Production mix constraint: X1-X2 £350 Feasible Production Time 3X1+4X2£ 2400 X1 500 700 • Boundary points. Extreme points. • Interior points. • There are three types of feasible points

  37. The search for an optimal solution Start at some arbitrary profit, say profit = $2,000... X2 Then increase the profit, if possible... 1000 ...and continue until it becomes infeasible 700 Profit =$4360 500 X1 500

  38. Summary of the optimal solution Space Rays = 320 dozen Zappers = 360 dozen Profit = $4360 • This solution utilizes all the plastic and all the production hours. • Total production is only 680 (not 700). • Space Rays production exceeds Zappers production by only 40 dozens.

  39. Extreme points and optimal solutions • If a linear programming problem has an optimal solution, an extreme point is optimal.

  40. Example 8 • The Crandon Manufacturing Company produces two principal product lines. One is a portable circular saw, and the other is a precision table saw. Two basic operations are crucial to the output of these saws: fabrication and assembly. The maximum fabrication capacity is 4000 hours per month; each circular saw requires 2 hours, and each table saw requires 1 hour. The maximum assembly capacity is 5000 hours per month; each circular saw requires 1 hour, and each table saw requires 2 hours. The marketing department estimates that the maximum market demand next year is 3500 saws per month for both products. The average contribution to profits and overhead is $900 for each circular saw and $600 for each table saw.

  41. Management wants to determine the best product mix for the next year so as to maximize contribution to profits and overhead. Also, it is interested in the payoff of expanding capacity or increasing market share. • Maximize: 900x1 + 600x2 = Z • Subject to: 2x1 + 1x2 ≤ 4,000 (Fabrication) • 1x1 + 2x2 ≤ 5,000 (Assembly) • 1x1 + 1x2 ≤ 3,500 (Demand) • x1, x2 ≥ 0 (Nonnegativity)

  42. Example 5 Graphical Method (2 variables) Maximize P = 4x + 5y Optimal solution (if it exists) will occur at a corner point Subject to Check Vertices: Feasible set – candidates for a solution The maximum is 25 at (5, 1)

  43. Ex. Example 6 Minimize C = 10x + 11y Subject to Check Vertices: Feasible set The minimum is 205 at (15, 5).

  44. Ex. Example 7Unbounded – No Solution Maximize P = 3x + 5y Subject to Unbounded Feasible set No Maximum

  45. Ex. Example 8Unfeasible Problem Maximize P = 10x + 11y Subject to No Feasible set – No overlap

  46. feasible region. Any non-negative value of (x1, x2) (i.e.: x1 >/ 0, x2 >/ 0) is a feasible solution of the LPP if it satisfies all the constraints. The collection of all feasible solutions is known as the feasible region. • convex Set A set X is convex if for any points x1, x2 in X, the line segment joining these points is also in X. By convention, a set containing only a single point is also a convex set. • Extreme point A point x of a convex set X is said to be an extreme point if there do not exist x1, x2 € X (x1 is not equal to x2)

  47. Half- plane : A linear inequality in two variables is known as a half plane. The corresponding equality or the line is known as the boundary of the half- plane. • Convex polygon - A convex polygon is a convex set formed by the inter-section of finite number of closed half-planes. • Note: The objective function is maximized or minimized at one of the extreme points which is the Optimum solution. Extreme points are referred to as vertices or corner points of the convex regions. • Redundant constraint: A redundant constraint is a constraint which does not affect the feasible region. • Abasic feasible solution: A basic feasible solution of a system of m equations and n variables (m < n) is a solution where n variables are non-negative (>/ 0) and n-m variables are zero. • Optimal feasible solution.: Any feasible solution that optimizes the objective function is called an optimal feasible solution.

  48. Summary • LPP with the clear identification of decision variables can be solved easily. • The decision variables interact with each other through some constraints. These constraints occur due to limited resources, stipulation on quality, technical legal or variety of other reasons. • The objective function and the constraints are linear function of decision variables. • LPP with tow decision variables can be solved graphically. • Any non-negative solution which satisfies all the constraints is known as feasible solution of problem. • The collection of all feasible solutions is known as feasible region which is also known as convex set.

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