Sensitivity Analysis with Data Tables. 10% annual interest now =$110 one year later. 10% annual interest now =$121 one year later

Transcription

1 Sensitivity Analysis with Data Tables Time Value of Money: A Special kind of Trade-Off: 10% annual interest now =$110 one year later 10% annual interest now =$121 one year later 10% annual interest now =$121 two years later General Formulation: PV(X) = Present Value of dollar amount X R = interest rate per period (e.g. R = 10% = 0.10) FV n (X) = Future Value of X after n periods n FV ( X ) = (1 + R) PV ( X ) n PV ( X ) = FV n ( X ) (1 + R) n Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 1

2 Note difference between: 10% Annual Interest Rate and 10% Annual Interest Rate compounded monthly 1. Annual Interest = $110 dollars after one year 2. 10% Annual Interest compounded monthly over 12 periods = *$100 = $ *100 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 2

3 Stream of Cash Flow: Interest Rate 10.00% At the End of Year Future Value Factor Present Value 0 -$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $38.55 NPV $ Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 3

4 Constant Interest Rate Formulation: X k = cost/revenue at the end of period k R% = interest rate per period NPV ( X, X, L X ) 0 1 n = n k = 0 X k ( 1+ R) k Varying Interest Rate Formulation: X k = cost/revenue at the end of period k R k % = interest rate during period k NPV ( X, X, L X ) 0 1 n = n k k k = 0 ( 1+ R j ) j= 1 X Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 4

5 Note: cost X k negative ; revenue X k positive Comparison of Two Streams of Cash Flows Cash Flow for Two Projects $ $ $50.00 $0.00 -$ $ $ Project 1 Project 2 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 5

6 Cash Flow Characteristics Project 1: Small Start-Up Cost and Increasing Profits. Cash Flow Characteristics Project 2: Large Start-Up Costs and Decreasing Profits. WHICH ONE WOULD YOU PREFER? ANSWER: DEPENDS ON THE INTEREST RATE Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 6

7 Interest 10.00% Rate Time Project 1 Project 2 0 -$ $ $10.00 $ $20.00 $ $30.00 $ $40.00 $ $50.00 $ $60.00 $ $70.00 $ $80.00 $ $90.00 $ $ $5.00 NPV $ $ Is there an Interest Rate at which Project 2 would be preferred? Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 7

8 R NPV 1 NPV 2 NPV1 - NPV 2 PREFERRED $ $ $ % $ $ $ NPV % $ $ $ NPV % $ $ $90.04 NPV % $ $ $75.38 NPV % $ $ $62.23 NPV % $ $ $50.45 NPV % $ $ $39.88 NPV % $ $ $30.41 NPV % $ $ $21.93 NPV % $ $ $14.33 NPV % $ $ $7.53 NPV % $ $ $1.45 NPV % $ $ $3.98 NPV % $ $ $8.83 NPV % $ $ $13.15 NPV % $ $ $16.99 NPV % $ $ $20.41 NPV % $ $ $23.43 NPV % $ $ $26.11 NPV % $ $ $28.46 NPV 2 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 8

9 NPV $ $ $50.00 Difference between NPV 1 and NPV 2 as a function of Interest Rate Project 1 preferred Project 2 preferred $0.00 -$ % 3.00% 5.00% 7.00% 9.00% 11.00% 13.00% Interest Rate 15.00% 17.00% 19.00% NPV1 - NPV 2 It appears that NPV 1 and NPV 2 break even between 12% and 13 %. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 9

10 The Break Even Point for the interest rate can be calculated exactly using the GOALSEEK function in Excel. GOALSEEK allows to search for root of the equation where F(x) is a continuous function. F(x)=0, GOALSEEK Method is similar to Bisection Method or Newton-Raphson Method Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 10

11 BISECTION METHOD: Starting at interval [a 1,b 1 ] established such that F(a 1 )*F(b 1 )<0 EXAMPLE BISECTION METHOD a b 2 b 4 b 2 =b 3 b 1 a 1 = a 2 a 3 a b 1 Stop when a + k b F k 2 < δ or b k a k < δ Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 11

12 NEWTON-RAPHSON METHOD: Requires Best-Guess and being able to calculate first order derivative EXAMPLE NEWTON RAPHSON METHOD F(a 1 ) a k+ 1 = a k F( ak ) d F( a dx k ) a 1 a 2 a 3 a 4 d F( a1) ( x a1) + F( a1) dx Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 12

13 Stop when a + k b F k 2 < δ QUESTION: DOES NEWTON-RAPHSON ALWAYS WORK? ANSWER: NO! EXAMPLE NEWTON RAPHSON METHOD F(a 1 ) a k+ 1 = a k F( ak ) d F( ak ) dx a 4 a 3 a 1 a 2 d dx F( a1) ( x a1) + F( a1) Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 13

14 Newton-Raphson may not converge at all or may converge to another solution of the equation F(x)=0. GOALSEEK FUNCTION of EXCEL is similar to the Newton-Raphson method. If a solution is not found, does not necessarily mean that none exists. Try different starting values! Be careful when using this method!!! Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 14

15 Interest 12.26% Rate Time Project 1 Project 2 NPV1 - NPV 2 0 -$ $ $ $10.00 $ $20.00 $ $30.00 $ $40.00 $ $50.00 $ $60.00 $ $70.00 $ $80.00 $ $90.00 $ $ $5.00 NPV $ $ Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 15

16 HOME WORK 1: 1. Consider the cash flows of Project 1 and Project 2. Assume that at the beginning of the project the interest rate equals 10%, and that over the duration of the project (10 years), the interest rate increases each year by 0.5%. Which Project is preferred based on Net Present Value? 2. Determine the rate of increase in interest at which you would be indifferent between Project 1 and Project 2 based on Net Present Value. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 16

17 Two Way Data Tables Description Case Study: A product costs $3.00 per unit to make. Demand for the product is determined by two factors: The price of the product, i.e. the lower the price, the higher the demand. The advertising budget, i.e. the more you advertise the higher the demand. It is estimated (e.g. using market research) that Demand D (in thousands of units) is: 3 D = A 7P + A* 1 P where A = advertising budget (in $000 s) and P = Price per Unit. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 17

18 Profit can be calculated as a function of A and P as well: 1. Profit=Total Revenue Total Cost Advertising Budget : A $ Price per Unit : P $ Total Revenue = 1000*D*P Cost per Unit : C $ Total Cost = Production Cost + Advertising Cost 4. Production Cost = 1000*D*$3 5. Advertising Cost = 1000*A Demand (in 1000's) Revenue $356, Production Cost $106, Advertising Cost $120, Total Cost $226, Profit $129, Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 18

19 Using Data Tables in Excel we can produce a graph of the DEMAND as a function of A and P. Demand Advertising Budget Price Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 19

20 Demand Price Advertisin g Budget Note that: Demand decreases when Price Increases Demand increases when you spend more money on advertising Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 20

21 Using Data Tables we can produce a graph of the Profit as a function of A and P. Profit $129, $20.00 $40.00 $60.00 $80.00 $ $ $ $ $ $ $156, $180, $203, $225, $247, $269, $291, $313, $335, $ $82, $105, $127, $149, $171, $192, $213, $234, $256, $ $20, $40, $60, $80, $100, $120, $140, $160, $180, $ 4.00 $31, $15, $1, $19, $37, $55, $73, $91, $110, $ 5.00 $71, $59, $46, $31, $16, $ $15, $31, $47, $ 6.00 $98, $92, $83, $72, $60, $47, $34, $20, $5, $ 7.00 $113, $113, $109, $102, $93, $83, $73, $62, $50, $ 8.00 $115, $122, $123, $120, $116, $109, $102, $94, $85, $ 9.00 $105, $119, $126, $128, $127, $125, $121, $116, $110, $ $82, $104, $116, $124, $127, $129, $129, $127, $124, $ $47, $77, $95, $108, $116, $122, $126, $128, $129, $ $1, $37, $62, $80, $94, $104, $112, $118, $122, $ $63, $15, $17, $41, $59, $74, $86, $96, $105, Max Profit $115, $122, $126, $128, $127, $129, $129, $128, $129, Note that: For a fixed advertising budget there is a price at which profit is optimal For a fixed price there is an advertising budget at which profit is optimal (less clear from the figure) Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 21

22 Is there a combination of Advertising Budget and Price at which profit is maximized? $1.00 $5.00 Price $9.00 $13.00 $20.00 $60.00 $ $ $ $150,000 $100,000 $50,000 $0 -$50,000 -$100,000 -$150,000 -$200,000 -$250,000 -$300,000 -$350,000 Profit Advertising Budget Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 22

23 Using Solver to Maximize Profit General Formulation of Optimization Problems: Max (or Min): F(x 1, x n ) Subject to : G 1 (x 1, x n ) 0 G 2 (x 1, x n ) 0 M G m (x 1, x n ) 0 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 23

24 Function F(x 1, x n ) is called the objective function with variables x 1, x n. Functions G j (x 1, x n ) j=1,,m are called the constrained functions (or constraints). The set S = n {( x1, L, xn ) G1( x1, L, xn ) 0, L, Gm ( x1, L, x ) 0} is called the feasible set (informally, the set of allowable solutions). IMPORTANT: The feasible set is bounded Optimization problem has a solution Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 24

25 (i.e. there is a feasible solution that globally maximizes the objective function) Objective Function Local Maxima Global Maximum x a 0 b x 0 Constraint Function Constraint Function Feasible Set is Bounded Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 25

26 Linear Optimization Problems: Both the objective function and the constrained functions are of the form a 1 x 1 + a 2 x 2 + L+ a n x n Non-Linear Optimization Problems: Objective function or constrained functions are non-linear. An optimization method solving a Non-Linear Optimization problem may get trapped in a local maximum. QUESTION? Under what conditions on the objective function and the constraint functions is a typical maximization method guaranteed to find a global maximum? Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 26

27 ANSWER: When the objective function is CONCAVE and constraint functions are CONVEX A function is concave on [a,b] when xy, [ ab, ] : f ( λ x + (1 λ) y) λ f ( x) + (1 λ) f ( y), λ [0,1] f ( λ x + (1 λ) y) f ( y) f (x) λ f ( x) + (1 λ) f ( y) x y Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 27

28 A function is convex on [x,y] when xy, [ ab, ] : f ( λ x + (1 λ) y) λ f ( x) + (1 λ) f ( y), λ [0,1] λ f ( x) + (1 λ) f ( y) f ( y) f (x) f ( λ x + (1 λ) y) x y Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 28

29 QUESTION? Under what conditions on the objective function and the constraint functions is a typical minimization method guaranteed to find a global minimum? ANSWER: When the objective function is CONVEX and constraint functions are CONVEX An optimization problem is convex if: 1. The constraint functions G j (x 1, x n ) are convex functions. 2. If the optimization problem is a minimization problem the objective function F(x 1, x n ) is convex. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 29

30 3. If the optimization problem is a maximization problem the objective function F(x 1, x n ) is concave. Practical Implications: If you can show that the optimization problem is convex and your optimization algorithm finds an optimal solution, your solution is a global optimum. If you cannot show that the optimization problem is convex and your optimization algorithm finds an optimal solution, your solution is a local optimum and you can never guarantee global optimality. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 30

31 Graph 3D-Convex Maximization Problem. F(x,y) y x Constraint 1 Constraint 3 Feasible Set Constraint 2 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 31

32 Basic approach to solving non-linear maximization problem: 1. Start at a feasible solution, choose a feasible search direction of ascent and follow that direction until you go downhill. 2. Choose a new search direction of ascent and follow that until you go downhill. 3. Stop when you cannot find a feasible search direction of ascent. (E.g. Steepest Ascent Method, Conjugate Gradient Method). Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 32

33 Graph 3D- Linear Maximization Problem. F(x,y) Constraint 5 y x Constraint 1 Global Optimum Constraint 4 Constraint 2 Feasible Set Constraint 3 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 33

34 Note that: Global Optimum of a Linear Optimization Problem is attained at a corner point of the feasible set Corner points of a feasible set in a Linear Optimization Problem are called Extreme Points (or Vertices). Each Linear Optimization problem has a finite number of extreme points. Basic approach to solving linear optimization problem: Enumerate extreme points, evaluate the objective function at extreme points and stop when you cannot improve. (E.g. Simplex Method). Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 34

35 QUESTION? WHY CAN YOU NOT USE A LINEAR OPTIMIZATION METHOD TO SOLVE A NON-LINEAR OPTIMIZATION PROBLEM? ANSWER: LINEAR-OPTIMIZATION METHODS ARE TYPICALLY LIMITED TO ENUMERATING EXTREME POINTS AND A GLOBAL OPTIMUM OF A NON- LINEAR OPTIMIZATION PROBLEM DOES NOT HAVE TO BE ATTAINED IN AN EXTREME POINT. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 35

36 QUESTION? CAN YOU USE A NON-LINEAR OPTIMIZATION METHOD TO SOLVE A LINEAR OPTIMIZATION PROBLEM? ANSWER: YES, BUT THIS MAY OR MAY NOT BE COMPUTATIONALLY EFFICIENT. EXAMPLE: INTERIOR POINT METHOD OF KARMAKAR HAS BETTER THEORETICAL COMPUTATIONAL COMLEXITY THAN SIMPLEX METHOD Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 36

37 Going back to the optimization problem of our case study: Maximize : Profit = (1000*P 3000)*D 1000*A Subject to : P 13 Where: 3 D = A 7P + A* 1 P IS THIS OPTIMIZATION PROBLEM LINEAR? IS THIS OPTIMIZATION PROBLEM CONVEX? Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 37

38 Using SOLVER in EXCEL and the GRG Optimization method 1, we find the following optimal solution: Advertising Budget : A $ Price per Unit : P $10.09 Cost per Unit : C $3.00 Demand (in 1000's) Revenue $371, Production Cost $110, Advertising Cost $131, Total Cost $241, Profit $129, GRG Optimization method is the Generalized Reduced Gradient optimization method, a non-linear optimization approach. Details of this approach are beyond the scope of this course. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 38

39 HOMEWORK 2: PROVE OR PROVIDE COUNTER EXAMPLE IS THE SUM OF TWO CONCAVE FUNCTIONS CONCAVE? IS THE PRODUCT OF TWO CONCAVE FUNCTIONS CONCAVE? Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 39

40 Two-way Data Tables and Decision Making Uncertainty Description Case Study: Eli Daisy must decide on monthly storage capacity for a new drug. The drug will sell for a period of 10 years at a price of $7 per unit. The production cost per drug unit is $4. One additional unit of storage capacity will cost $75 to build per drug unit. Storage capacity cost $1 annually per drug unit for maintenance. Production can only occur at the beginning of the month (Set-up Cost for production are exorbitant). You always produce the full storage capacity as FDA has a monthly expiration date on the drug. Clearly, the most ideal (profitable scenario) would be to produce the same amount as the monthly demand for the drug. The monthly demand for the drug will be constant over this 10 year period but is uncertain. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 40

41 Probability (0.1) (0.2) (0.3) Monthly Demand 100, , ,000 Note that: (0.2) (0.1) (0.1) 400, , ,000 You would prefer not to produce more on a monthly basis than your monthly demand as demand is constant. You do not want excess storage capacity as even unused storage space cost $1 annually in maintenance cost. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 41

42 Profit Calculation given fixed Storage Capacity and a known Monthly Demand for 10 years: Profit = Total Revenue Total Cost Total Revenue = 10*(Units Sold per Year)*Price Unit Sold Per Year = 12* Min(Storage Capacity, Monthly Demand) Total Cost = Building Cost + 10 *Annual Maintenance Cost + 10 *Annual Production Cost Building Cost = Storage Capacity*$15 Annual Maintenance Cost = Storage Capacity*$1 Annual Production Cost = 12*Storage Capacity*$4. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 42

43 Capacity Planning (10 Year Period) Variable Value Unit Production Cost 4 Sales price 7 Building cost per Unit Storage Capacity 75 Annual maintenance cost per unit of Storage Capacity 1 Storage Capacity Level 300,000 Monthly demand 100,000 Planning Horizon 10 Revenue over Planning Horizon Unit Monthly sales 100,000 Total Revenue $ 84,000,000 Total Costs over Planning Horizon Building Cost (Fixed) $ 22,500,000 Maintenance Cost (Variable) $ 3,000,000 Production Costs (Variable) $ 144,000,000 Total costs $ 169,500,000 Profit -$85,500,000 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 43

44 Possible Capacity Levels are 100, ,000. What amount of storage capacity should Eli build for this 10 year period? Solve Decision Problem using Expected Monetary Value (EMV) Expected Value of discrete random variable Y: E Y [ Y ] = n i= 1 y i Pr( Y = y i ) = n i= 1 y i p i Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 44

45 EMV= $4.5 Trade Ticket -$1 Keep Ticket $0 EMV= $4 EMV= $4.5 Max Profit Win (0.20) $24 $25 Lose (0.80) -$1 $0 Win (0.45) $10 $10 Lose (0.55) $0 $0 y Pr(Y=y) y*pr(y=y) $ $4.80 -$ $0.80 $4.00 =EMV y Pr(Y=y) y*pr(y=y) $ $4.50 $ $0.00 $4.50 =EMV Interpretation: Playing the lottery a lot of times will result in an average payoff equal to the EMV. Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 45

46 Returning to our Case Study: probabilities profit Monthtly Demand $ (85,500,000.00) Mean profit $27,500, $27,500, $27,500, $27,500, $27,500, $27,500, $ 27,500, $29,000, $55,000, $55,000, $55,000, $55,000, $55,000, $ 46,600,000 Annual $85,500, $1,500, $82,500, $82,500, $82,500, $82,500, $ 48,900,000 Capacity $142,000, $58,000, $26,000, $110,000, $110,000, $110,000, $ 26,000, $198,500, $114,500, $30,500, $53,500, $137,500, $137,500, $ (13,700,000) $255,000, $171,000, $87,000, $3,000, $81,000, $165,000, $ (61,800,000) Probability Monthly Demand Probability Monthly Demand EMV = $27,500K (0.1) (0.2) (0.3) 100, , ,000 EMV = -$61,800K (0.1) (0.2) (0.3) 100, , ,000 (0.2) (0.1) (0.1) 400, , ,000 (0.2) (0.1) (0.1) 400, , ,000 FIRST ROW LAST ROW Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 46

47 10 Year Capacity 100, , , , , ,000 EMV 27,500,000 EMV 46,600,000 EMV 48,900,000 EMV 26,000,000 EMV -13,700,000 EMV -61,800,000 CONCLUSION: SET STORAGE CAPACITY AT 300,000 Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 47

48 Variance : VARIANCE AND STANDARD DEVIATION OF Y: Var Y E Y E Y 2 ( ) = σ Y = [ [ ] ] ( ) 2 n = i i i= 1 p *( y E[ Y]) 2 Standard Deviation : σ 2 Y = σ Y Informal Interpretation: Standard deviation is the best guess distance from the mean for an arbritrary outcome Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 48

49 Squared deviation Variance Std dev E E E E E E E+00 $ E E E E E E E+14 $25,200, E E E E E E E+15 $55,719, E E E E E E E+15 $84,000, E E E E E E E+16 $104,915, E E E E E E E+16 $119,090, CONCLUSION: A Storage Capacity of 300,000 yields the highest expected profit of $48,900,000 over 10 years with a standard deviation of $55,719,296. IS THIS A GOOD INVESTMENT OPPORTUNITY? ANSWER DEPENDS ON DECISION MAKER S RISK AVERNESS! Pr( EMV < 0 Storage Capacity=100,000)=0% Pr( EMV < 0 Storage Capacity=200,000)=10% Pr(EMV < 0 Storage Capacity=300,000)=30% Lecture Notes by: Dr. J. Rene van Dorp Chapter 1 - Page 49