1 / 99

Quantitative Technique for Management

Quantitative Technique for Management. MGM3163 PJJ Sem 1 2014-2015. MANISAH BINTI OTHMAN FAKULTI EKONOMI DAN PENGURUSAN Room No. A231 manisah@upm.edu.my manisahoth@gmail.com. Introduction to Modelling. Chapter 1. Chapter Topics. The Management Science Approach to Problem Solving

rleonard
Download Presentation

Quantitative Technique for Management

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Quantitative Technique for Management MGM3163 PJJ Sem 1 2014-2015

  2. MANISAH BINTI OTHMAN FAKULTI EKONOMI DAN PENGURUSAN Room No. A231 manisah@upm.edu.my manisahoth@gmail.com

  3. Introduction to Modelling Chapter 1

  4. Chapter Topics • The Management Science Approach to Problem Solving • Model Building: Break-Even Analysis

  5. Steps in the Management Science Process • Observation - Identification of a problem that exists (or may occur soon) in a system or organization. • Definition of the Problem - problem must be clearly and consistently defined, showing its boundaries and interactions with the objectives of the organization. • Model Construction - Development of the functional mathematical relationships that describe the decision variables, objective function and constraints of the problem. • Model Solution - Models solved using management science techniques. • Model Implementation - Actual use of the model or its solution.

  6. Model Building: Break-Even Analysis (1 of 9) • Used to determine the number of units of a product to sell or produce that will equate total revenue with total cost. • The volume at which total revenue equals total cost is called the break-even point. • Profit at break-even point is zero.

  7. Model Building:Break-Even Analysis (2 of 9) Model Components • Fixed Cost (cf) - costs that remain constant regardless of number of units produced. • Variable Cost (cv) - unit production cost of product. • Volume (v) – the number of units produced or sold • Total variable cost (vcv) - function of volume (v) and unit variable cost.

  8. Model Building:Break-Even Analysis (3 of 9) Model Components • Total Cost (TC) - total fixed cost plus total variable cost. • Profit (Z) - difference between total revenue vp (p = unit price) and total cost, i.e.

  9. Model Building:Break-Even Analysis (4 of 9) Computing the Break-Even Point The break-even point is that volume at which total revenue equals total cost and profit is zero: The break-even point

  10. Model Building: Break-Even Analysis (5 of 9) Example:Western Clothing Company Fixed Costs: cf = $10000 Variable Costs: cv = $8 per pair Price : p = $23 per pair The Break-Even Point is: v = (10,000)/(23 -8) = 666.7 pairs

  11. Model Building: Break-Even Analysis (6 of 9) Figure 1.2 Break-even model

  12. Model Building: Break-Even Analysis (7 of 9) Price for denim jeans increase from $23 to 30 454.5 Figure 1.3 Break-even model with an increase in price

  13. Model Building: Break-Even Analysis (8 of 9) Variable cost increase to $12 555.5 Figure 1.4 Break-even model with an increase in variable cost

  14. Model Building: Break-Even Analysis (9 of 9) Total Cost $13,000 722.2 Figure 1.5 Break-even model with a change in fixed cost

  15. Break-Even Analysis: Excel Solution (1 of 4) Formula for v, break-even point, =D4/(D8-D6) Exhibit 1.1

  16. Break-Even Analysis: Excel QM Solution (2 of 4) Click on “Add-Ins”, then the menu of Excel QM modules Enter model parameters in cells B10:B13 Exhibit 1.2

  17. Break-Even Analysis: Excel QM Solution (3 of 4) Exhibit 1.3 Western Clothing Company in QM

  18. Break-Even Analysis: QM Solution (4 of 4) Exhibit 1.4 QM break-even graph for Western Clothing Company

  19. Modeling with Linear Programming Chapter 2

  20. Chapter Topics • Model Formulation • A Maximization Model Example • Graphical Solutions of Linear Programming Models • A Minimization Model Example

  21. Linear Programming: An Overview • Objectives of business decisions frequently involve maximizing profit or minimizing costs. • Linear programming uses linear algebraic relationships to represent a firm’s decisions, given a business objective, and resource constraints. • Steps in application: • Identify problem as solvable by linear programming. • Formulate a mathematical model of the unstructured problem. • Solve the model. • Implementation

  22. Model Components • Decision variables - mathematical symbols representing levels of activity of a firm. • Objective function - a linear mathematical relationship describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized. • Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables. • Parameters - numerical coefficients and constants used in the objective function and constraints.

  23. Summary of Model Formulation Steps Step 1 : Clearly define the decision variables Step 2 : Construct the objective function Step 3 : Formulate the constraints

  24. LP Model Formulation A Maximization Example (1 of 3) • Product mix problem - Beaver Creek Pottery Company • How many bowls and mugs should be produced to maximize profits given labor and materials constraints? • Product resource requirements and unit profit:

  25. LP Model Formulation A Maximization Example (2 of 3) Resource 40 hrs of labor per day Availability: 120 lbs of clay Decision x1 = number of bowls to produce per day Variables: x2 = number of mugs to produce per day Objective Maximize Z = $40x1 + $50x2 Function: Where Z = profit per day Resource 1x1 + 2x2 40 hours of labor Constraints: 4x1 + 3x2 120 pounds of clay Non-Negativity x1  0; x2  0 Constraints:

  26. LP Model Formulation A Maximization Example (3 of 3) Complete Linear Programming Model: Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0

  27. Graphical Solution of LP Models • Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). • Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

  28. Coordinate Axes Graphical Solution of Maximization Model (1 of 12) X2 is mugs Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 X1 is bowls Figure 2.2 Coordinates for graphical analysis

  29. Labor Constraint Graphical Solution of Maximization Model (2 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.3 Graph of labor constraint

  30. Labor Constraint Area Graphical Solution of Maximization Model (3 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.4 Labor constraint area

  31. Clay Constraint Area Graphical Solution of Maximization Model (4 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.5 Theconstraint area for clay

  32. Both Constraints Graphical Solution of Maximization Model (5 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.6 Graph of both model constraints

  33. Feasible Solution Area Graphical Solution of Maximization Model (6 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.7 The feasible solution area

  34. Objective Function Solution = $800 Graphical Solution of Maximization Model (7 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.8 Objective function line for Z = $800

  35. Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (8 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.9 Alternative objective function lines for profits Z of $800, $1,200, and $1,600

  36. Optimal Solution Graphical Solution of Maximization Model (9 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.10 Identification of optimal solution point

  37. Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.11 Optimal solution coordinates

  38. Extreme (Corner) Point Solutions Graphical Solution of Maximization Model (11 of 12) Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2  40 4x2 + 3x2  120 x1, x2  0 Figure 2.12 Solutions at all corner points

  39. LP Model Formulation – Minimization (1 of 7) • Two brands of fertilizer available - Super-gro, Crop-quick. • Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate. • Super-gro costs $6 per bag, Crop-quick $3 per bag. • Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?

  40. LP Model Formulation – Minimization (2 of 7) Decision Variables: x1 = bags of Super-gro x2 = bags of Crop-quick The Objective Function: Minimize Z = $6x1 + 3x2 Where: $6x1 = cost of bags of Super-Gro $3x2 = cost of bags of Crop-Quick Model Constraints: 2x1 + 4x2 16 lb (nitrogen constraint) 4x1 + 3x2 24 lb (phosphate constraint) x1, x2 0 (non-negativity constraint)

  41. Constraint Graph – Minimization (3 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x2 + 3x2  24 x1, x2  0 Figure 2.16 Constraint lines for fertilizer model

  42. Feasible Region– Minimization (4 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x2 + 3x2  24 x1, x2  0 Figure 2.17 Feasible solution area

  43. Optimal Solution Point – Minimization (5 of 7) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x2 + 3x2  24 x1, x2  0 The optimal solution of a minimization problem is at the extreme point closest to the origin. Figure 2.18 The optimal solution point

  44. Surplus Variables – Minimization (6 of 7) • A surplus variable is subtracted from a  constraint to convert it to an equation (=). • A surplus variable represents an excess above a constraint requirement level. • A surplus variable contributes nothing to the calculated value of the objective function. • Subtracting surplus variables in the farmer problem constraints: 2x1 + 4x2 - s1 = 16 (nitrogen) 4x1 + 3x2 - s2 = 24 (phosphate)

  45. Graphical Solutions – Minimization (7 of 7) Minimize Z = $6x1 + $3x2 + 0s1 + 0s2 subject to: 2x1 + 4x2 – s1= 16 4x2 + 3x2 – s2 = 24 x1, x2, s1, s2 0 Figure 2.19 Graph of the fertilizer example

  46. Solving Linear Programming Models Chapter 3

  47. Chapter Topics • Computer Solution • Sensitivity Analysis

  48. Computer Solution Early linear programming used lengthy manual mathematical solution procedure called the Simplex Method (See web site Module A). Steps of the Simplex Method have been programmed in software packages designed for linear programming problems. Many such packages available currently. Used extensively in business and government. Text focuses on Excel Spreadsheets and QM for Windows.

  49. Beaver Creek Pottery Example QM for Windows (1 of 4) Set number of constraints and decision variables. Click here when finished. Exhibit 3.6 Data entry screen

  50. Beaver Creek Pottery Example QM for Windows (2 of 4) Exhibit 3.7 Data table

More Related