Exam 1 Review/Instructions 370-Operations-Mgmt 2018/9/16 Exam 1 Review/Instructions Content: Chapters 1, 2, 5, and 6S Write your full name and ID on both the exam and the answer sheet. Use a No.2 pencil to fill in the answer sheet. Return the exam and the answer sheet when you finish the examination. The duration of this examination is the same as one regular class duration. All multiple-choice questions, 25 questions in total. Closed books, notes, and computers. You may bring a 8.5 in*11in double-sided cheat sheet and a calculator. Homework 1 is due in the discussion ahead of Exam 1 .

Chap 1: Introduction to Operations Management 370-Operations-Mgmt 2018/9/16 Chap 1: Introduction to Operations Management What is Operations? What is Operations Management? Why study Operations Management? Trends in Operations Management

Three Key Functions in an organization 370-Operations-Mgmt 2018/9/16 Three Key Functions in an organization Operations Finance Marketing

Continuum of Types of Firms 370-Operations-Mgmt 2018/9/16 Continuum of Types of Firms More like a manufacturing organization More like a service organization Physical, durable product Output that can be inventoried Low customer contact Long response time Regional, national, or international markets Large facilities Capital intensive Quality easily measured Intangible, perishable product Output that cannot be inventoried High customer contact Short response time Local markets Small facilities Labor intensive Quality not easily measured

Chap 2 Competitiveness, Strategy, and Productivity Mission / Goals / Strategies / Tactics Productivity A measure of the effective use of resources, usually expressed as the ratio of output to input Productivity = Outputs Inputs

Productivity Partial measures Multi-factor measures Total measure output/(single input) Multi-factor measures output/(multiple inputs) Total measure output/(total inputs)

Measures of Productivity Table 2.5 Partial Output Output Output Output measures Labor Machine Capital Energy Multifactor Output Output measures Labor + Machine Labor + Capital + Energy Total Goods or Services Produced measure All inputs used to produce them

Chap 5: Capacity Planning 370-Operations-Mgmt 2018/9/16 Chap 5: Capacity Planning Explain capacity planning and its importance Discuss ways of defining and measuring capacity. Describe the determinants of effective capacity. Steps for Capacity Planning Identify capacity alternatives Bottleneck operation analysis - handout Conduct financial analysis Cost-volume analysis - handout

Capacity Planning Capacity is the upper limit or ceiling on the load that an operating unit can handle. Capacity also includes Equipment Space Employee skills The basic questions in capacity planning are: What kind of capacity is needed? How much is needed? When is it needed?

Efficiency and Utilization Actual output Efficiency = Effective capacity Utilization = Design capacity Both measures expressed as percentages

Efficiency/Utilization Example Design capacity = 50 trucks/day Effective capacity = 40 trucks/day Actual output = 36 units/day Actual output 36 units/day Efficiency = = = 90% Effective capacity 40 units/ day Utilization = Actual output = 36 units/day = 72% Design capacity 50 units/day

Production units have an optimal rate of output for minimal cost. Figure 5.4 Production units have an optimal rate of output for minimal cost. Minimum cost Average cost per unit Rate of output Minimum average cost per unit

Economies of Scale Economies of scale Diseconomies of scale If the output rate is less than the optimal level, increasing output rate results in decreasing average unit costs Diseconomies of scale If the output rate is more than the optimal level, increasing the output rate results in increasing average unit costs

Cost-Volume Relationships Figure 5.6c Amount ($) Q (volume in units) = FC / (R-v) BEP units Profit Total revenue Total cost

Break-Even Problem with Step Fixed Costs Textbook Page 195 Example 4 A manager has the option of purchasing one, two, or three machines. Number of Total Annual Corresponding Machines Fixed Costs Range of Output 1 $ 9,600 0 to 300 2 15,000 301 to 600 3 20,000 601 to 900 Variable cost is $10 per unit, and revenue is $40 per unit. Determine the break-even point for each range. QBEP1=FC1/(R-v)=9600/(40-10)=320 QBEP2=15,000/30=500, QBEP3=20,000/30 ≈667

Break-Even Problem with Step Fixed Costs Figure 5.7b b. If projected annual demand is between 580 and 660 units, how many machines should the manager purchase? Answer: 2 machines TR $ BEP 3 BEP 2 TC TC 667 TC 500 Q 300 600 900

370-Operations-Mgmt Linear Programming 2018/9/16 Chapter 6 Supplement Linear Programming

Linear Programming LP is a set of linear mathematical representations of constrained optimization problems. It consists of 4 components: Objective function Decision variables Constraints Parameters

LP: Formulation Max (or Min) Objective subject to (s.t.): constraint 1 … Feasible solution: If a solution satisfies all the constraints, then it is feasible; Otherwise, i.e. it doesn’t not satisfy one or more of the constraints, then it is infeasible.

Example: LP Formulation 370-Operations-Mgmt 2018/9/16 Example: LP Formulation X1: number of tables produced X2: number of chairs produced Max 7*X1+5*X2 S.t. 4X1+3X2<=240 (Carpentry time) 2X1+ X2<=100 (Painting time) X1, X2>=0 Carpentry time=210 hours Painting time= 90 hours X1=30, X2=30 Profit=$360 (Feasible) Carpentry time= 240 hours Painting time=120 hours X1=60, X2=0 Profit=$420 (Infeasible)

Graphical Linear Programming Set up objective function and constraints in mathematical format Plot the constraints Identify the feasible solution space Plot the objective function Determine the optimum solution

Find Optimal Solution Approach 1: Isoprofit Line Method Let profit equal some arbitrary but small dollar amount. Graph the profit line. Move the profit line farther from the 0 origin to reach the highest possible profit. Approach 2- Corner Point Method Optimal Solution to any LP problem will lie at a corner point, or extreme point, of the feasible region. Calculate the values of the variables at each corner Find the maximum profit or optimal solution at one of the corner points.

Feasible Region Approach 1: Isoprofit Line Method Objective Function 4X1+3X2=240 Objective Function Optimal Solution 2X1+ X2=100 Feasible Region

Approach 2: Corner Point Method 370-Operations-Mgmt 2018/9/16 Approach 2: Corner Point Method X2 Corner Point A: X1=0, X2=80 100 Profit = 7X1+5X2 =400 Corner Point B: X1=0, X2=0 A 80 Profit=7*0+5*0=0 60 Corner Point C: X1=50, X2=0 Profit=7*50+5*0=350 D 40 Corner Point D: X1=30, X2=40 Feasible Region Profit=7*30+5*40=410 20 Feasible Region D is optimal solution B 20 40 C 60 80 100 X1

What is Sensitivity Analysis Sensitivity analysis is a means of assessing the impact of potential changes to the parameters (the numerical values) of an LP model. Change in the objective function coefficient Change in the right-hand side (RHS) values of constraints Change in constraint coefficients (not required)

Determine the range of optimality [Coefficient-allowable decrease, Coefficient+allowable increase] e.g. Max 7X1+5X2 Range of optimality for the objective function coefficient of X1: [7-0.33, 7+3] = [6.67,10], for X2: [5-1.5, 5+0.25] = [3.5, 5.25].

Determine the range of optimality Sensitivity Analysis - objective function coefficient Determine the range of optimality [Coefficient-allowable decrease, Coefficient+allowable increase] e.g. Max 7X1+5X2, opt. solution: (X1,X2)=(30,40) Range of optimality for the objective function coefficient of X1: [7-0.33, 7+3] = [6.67,10], -If “Max 8X1+5X2”, then 6.67 ≤ 8 ≤ 10, so opt. (X1,X2)=(30,40), not changed - If “Max 11X1+5X2”, then 11 > 10, so (X1, X2)=(50,0), changed

Sensitivity Analysis Example - RHS of constraints Shadow price: Amount by which the value of the objective function would change with a one-unit change in the RHS value of a constraint. Range of feasibility: Range of values for the RHS of a constraint over which the shadow price remains the same. -The range is determined by [RHS-allowable decrease, RHS+allowable increase]. For Carpentry time: [240-40,240+60]=[200, 300] Original cons. is “4X1+3X2<=240”, opt. profit =410. Within the range of feasibility : [200, 300], If “4X1+3X2<=200 ”, opt. profit = 410+1.5*(200-240) = 350; If “4X1+3X2<=250 ”, opt. profit = 410+1.5*(250-240) = 425; If “4X1+3X2<=300 ”, opt. profit = 410+1.5*(300-240) = 500; Out of the range of feasibility: If “4X1+3X2<=160 ”, opt. X1=40, X2=0, profit = 280 ; Using the shadow price of 1.5, the opt. profit = 410+1.5*(160-240) = 290 which is wrong!