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ISEN 315 Spring 2011 Dr. Gary Gaukler. Operations Finance Marketing Functional Areas of the Firm.

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Presentation on theme: "ISEN 315 Spring 2011 Dr. Gary Gaukler. Operations Finance Marketing Functional Areas of the Firm."— Presentation transcript:

1 ISEN 315 Spring 2011 Dr. Gary Gaukler

2 Operations Finance Marketing Functional Areas of the Firm

3 Time Horizons for Decisions 1. Long Term Decisions 2. Intermediate Term Decisions 3. Short Term Decisions

4 The Elements of Production and Operations Strategy

5 History of POM Early years of the Industrial Revolution 1780-1890 –Shift from low-volume production to large scale operations Mass Production 1890-1920 –Concept of the assembly line –Frederick Taylor champions the idea of scientific management –The first IE Golden Age of IE 1920-1960 –Operations Research and IE influence industry –Inventory control, quality control, scheduling

6 History of POM 1960-1990 –Sophisticated concepts: Quality Circles, JIT –US companies lose competitive edge 1990-today –Supply chain management –New technologies: internet, wireless, RFID

7 How Do Firms Differentiate Themselves from Competitors? Low Cost Leaders: Some examples include High Quality (and price) Leaders. Ex:

8 Along What Other Dimensions Do Firms Compete?

9 The Product Life-Cycle Curve


11 A First Operations Model: Capacity Strategy Fundamental issues: –Amount. When adding capacity, what is the optimal amount to add? Too little Too much –Timing. What is the optimal time between adding new capacity? –Type. Level of flexibility, automation, layout, process, level of customization, outsourcing, etc.

12 Three Approaches to Capacity Strategy Policy A : Try not to run short. Here capacity must lead demand, so on average there will be excess capacity. Policy B : Build to forecast. Capacity additions should be timed so that the firm has excess capacity half the time and is short half the time. Policy C : Maximize capacity utilization. Capacity additions lag demand, so that average demand is never met.

13 Capacity Leading and Lagging Demand

14 Determinants of Capacity Strategy Highly competitive industries (commodities, large number of suppliers, limited functional difference in products, time sensitive customers) – here shortages are very costly. Monopolistic environment where manufacturer has power over the industry: (Intel, Lockheed/Martin). Products that obsolete quickly, such as computer products. Want type C policy, but in competitive industry, such as computers, you will be gone if you cannot meet customer demand. Need best of both worlds: Dell Computer.

15 Mathematical Model for Timing of Capacity Additions (Capacity leads demand) Let D = Annual Increase in Demand x = Time interval between adding capacity r = annual discount rate (compounded continuously) f(y) = Cost of expanding by capacity y

16 Mathematical Model (continued) A typical form for the cost function f(y) is: Where k is a constant of proportionality, and a measures the ratio of incremental to average cost of a unit of plant capacity.

17 Capacity Expansion Cost

18 Dynamic Capacity Expansion Suppose demand exhibits a linear trend: y: current demand (= current capacity) D: rate of increase per unit time

19 Dynamic Capacity Expansion Capacity leads demand

20 Dynamic Capacity Expansion Question: What value of x minimizes the net present value of capacity expansion costs? x = time interval between capacity expansions

21 Model Assumptions Infinite planning horizon Demand grows linearly Capacity expansion allowed at any time point Any size capacity expansion allowed No shortages allowed Continuous discounting at rate r Capacity expansion is instantaneous Expansion cost for expanding by size x is f(x)=kx a (0<a<1)

22 Continuous Discounting The present value of $1 incurred in t years, assuming continuous compounding at a rate r, is e -rt Lets derive this!



25 Optimal Expansion Size Need to satisfy all demands x is the time interval between expansions Hence, at the time of expansion, the expansion size should be: Cash flows:

26 Sum of Discounted Costs Cost = C(x) = f(xD) + f(xD)e -rx + f(xD)e -2rx +... After some algebra: – Cost = C(x) = f(xD)/(1-e -rx ) Want to find: min C(x) s.t. x>=0 Result: rx / (e rx -1) – a = 0 Numerical solution only!

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