1 / 27

Chapters 13-15

Chapters 13-15. Multiple Goal Decision Analysis I. Net value and marginal analysis Decision rules Example Present value analysis Reconciling present and future cash flows Discounting Figures of merit Weighted sum Delivered system capability. d(NV). MNV(X) =. dx. d(TV). dC. =. -.

kay-mccormick
Download Presentation

Chapters 13-15

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. Chapters 13-15 Multiple Goal Decision Analysis I Net value and marginal analysis • Decision rules • Example Present value analysis • Reconciling present and future cash flows • Discounting Figures of merit • Weighted sum • Delivered system capability

  2. d(NV) MNV(X) = dx d(TV) dC = - dx dx Marginal Analysis: Definitions X – Activity level of an alternative C(X) – Cost of alternative TV(X) – Total value of alternative (in same units as cost) NV(X) – Net value of alternative NV(X) = TV(X) – C(X) MNV(X) – Marginal net value

  3. Cost and Total Value TV and C TV NVmax C X1 Xmax X2 Activity level x (a)

  4. MNV = d(TV)/dx – dC/dx (b) Marginal Net Value Activity level Xmax X

  5. NVmax NV = TV - C X Xmax X2 X1 Over- investment (b) Net Value vs. Activity Level Investment Profitable

  6. Marginal Net Value Decision Rule • In the “profitable” segment • If MNV > 0, Increase activity level • If MNV < 0, Decrease activity level • If MNV = 0, Activity level is optimal MNV = d(TV) / dx – dC/dx • For Option B, with VT = value of each TR/sec, C(N) = 1007 + 20N; dC / dN = 20 TV(N) = VT(880N – 40N2); d(TV) / dN = VT(880 –80N) 20 = VT (880 – 80Nmax) Nmax = (880 VT – 20) / 80 VT = 11 – 1/(4VT )

  7. Optimal Number of Processors vs. Transaction Value 10 8 6 Nmax = 11 – 1 /(4 VT) Nmax 4 2 VT, $K 0.1 0.2 0.3 0.4

  8. TPS Decision Problem 3 • Assuming use of composite option, we will acquire 5 processors/system and run option A for 2 years • Which acquisition option should we choose:-A1: Rent processors for 2 years at $2400/Mo.-A2: Purchase processors for $100,000. Resell them for $50,000 after 2 years.

  9. Interest Calculations • Option A2 ties up $50K for 24 months • How much would this be worth at an interest rate of .75%/month?V($50 K, 1) = $50 K (1.0075) V($50 K, 2) = $50 K (1.0075) (1.0075) … V($50 K, 24) = $50 K (1.0075)24 = $59,820 • For any sum X,V(X, 24) = X(1.0075)24

  10. Present Value Calculations • What is the present value X of the $50K we will receive in 24 months? V (X, 24) = X (1.0075)24 = $50K $50K X = = $41,792 (1.0075)24 $50K PV ($50K, .0075, 24) = (1.0075)24 F PV (F, r, n) = (1 +r)n

  11. Present Value • Of a future cash flow F • At an interest rate per time period r • Over a number of time periods n • Using the discount rate D = PV (F, D, n) = FDn F PV (F, r, n) = (1 +r)n 1 1 + F

  12. In option A1, we pay $2400 at the beginning of each month How much is this worth in present value?PVS ($2400, D, 1) = $2400PVS ($2400, D, 2) = $2400 + $2400 DPVS ($2400, D, 3) = $2400 (1 + D + D2) . . . PVS ($2400, D, 24) = $2400 (1 + D + … + D23) = $2400 (1 - D24) / (1 – D ) For D = 1/1.0075 = .9925558, PVS ($2400, 1/1.0075, 24) = $52,928 Present Value, Series of Cash Flows

  13. Present Value of A Series of Cash Flows • m equal cash flows or payments P • At the beginning of each time period • Constant discount rate D PVS (P, D, m) = P [(1-Dm)/(1-D)]

  14. Present Value Analysis Results

  15. 60622 57600 58208 60 55972 55640 54420 52908 52928 50 51494 50000 Present value, $ 40 0 0.25 0.5 0.75 1.0% Interest rate per month Present Value Comparison vs. Interest Rate Option A2 Option A1

  16. TPS Decision Problem 4 • Choose best vendor operating system • System A – Standard OS • System A Plus • Better diagnostics, maintenance features • $40K added cost

  17. Alternative TPS Operating System Characteristics

  18. Meta-Decision Problems Lowercost More insight The system decision problem Lower Cost More Stability The system analysis decision problem

  19. Weighted Sum Figure of Merit • Assign weight Wi to criterion i ΣiWi = 1 • For each option j and criterion i, assign rating riJ (0 < rij < 10) • Compute figure of merit for each option j Fj = Σ Wi rij

  20. TPS Operating System Figure of Merit Calculation

  21. Delivered System Capability (DSC)Figure of Merit DSC = (SC) (DC) (AV) • SC: System Capability = Σ Wi rij • DC: Delivered Capacity • AV: Availability

  22. The DSC Figure of Merit • Dimensionless • Covers effectiveness only • SC component handles many criteria • DC, AV components apply multiplicatively

  23. Gains from System A Plus - I • Reduction of $150K maintenance cost – Maintenance Support: 20% ($30K) – Diagnostics, Error Msgs.: 10% (15K) – Documentation (worse): -10% (-15K) $30K – Net Cost = $650K + 40 – 30 = $660K • Delivered capacity increase via measurement: 5% 2400 tr/sec (1.05) 2520 tr/sec • Availability increase via diagnostics, error messages, trace capability: 50% less downtime 0.95 0.975

  24. Gains from System A Plus - II • System Capability System A System A Plus Criteria Weight Rating Score Rating Score Basic TPS Functions 0.95 1.0 0.950 1.0 0.950 Accounting Systems 0.01 0.6 0.006 1.0 0.010 Usage Summaries 0.01 0.0 0.000 0.8 0.008 OS Documentation 0.03 0.8 0.024 0.4 0.012 Total 0.980 0.980

  25. TPS Comparison: Delivered System Capability

  26. Revised TPS Weighted Sum Analysis

  27. Comparison of Weighted Sum and DSC Figures of Merit

More Related