ENGM 661 Engineering Economics for Managers Risk Analysis

Motivation Suppose we have the following cash flow diagram. NPW = -10,000 + A(P/A, 15, 5) A A A A A 10,000 MARR = 15%

Motivation Now suppose that the annual return A is a random variable governed by the discrete distribution: A p p p          ,/,/,/

Motivation A p p p          ,/,/,/ For A = 2,000, we have NPW = -10, ,000(P/A, 15, 5) = -3,296

Motivation A p p p          ,/,/,/ For A = 3,000, we have NPW = -10, ,000(P/A, 15, 5) = 56

Motivation A p p p          ,/,/,/ For A = 4,000, we have NPW = -10, ,000(P/A, 15, 5) = 3,409

Motivation There is a one-for-one mapping for each value of A, a random variable, to each value of NPW, also a random variable. A 2,000 3,000 4,000 p(A) 1/6 2/3 1/6 NPW -3, ,409 p( NPW ) 1/6 2/3 1/6

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Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% Now Suppose the return in each year is a random variable governed by the some probability distribution. NPW = -10,000 + A 1 (1+i) -1 + A 2 (1+i) A 5 (1+i) -5

Risk Analysis    ,aA t t t   1()whereai t t A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10,000 + A 1 (1+i) -1 + A 2 (1+i) A 5 (1+i) -5

Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals 

Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals NPW Normal   Central Limit

Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals NPW Normal NPW N(  NPW,  NPW )   

Mean  NPWt t t E EaA    [][, Recall: E[Z] = E[X 1 ] + E[X 2 ] E[aX+b] = aE[X] + b

Mean  NPWt t t E EaA    [][, Recall: E[Z] = E[X 1 ] + E[X 2 ] E[aX+b] = aE[X] + b    ,[]aEA t t t  NPW

Mean    ,[]aEA t t t  NPW but, E[A t ] = 3,000    ,[]a3,000 t t  NPW

Mean    ,[]a3,000 t t  NPW = -10, ,000 a t t   1 5 (1+i) -t t   1 5 = -10, ,000(P/A, i, 5)

Variance  NPW tt aA   (,) Recall:  2 (z) =  2 (x) +  2 (y)  2 (ax+b) = a 2  2

Variance  NPW tt aA   (,) Recall:  2 (z) =  2 (x) +  2 (y)  2 (ax+b) = a 2  2  NPW t A a t  

Variance but,  ()A t   NPW t a   (A t )  NPW t a   () = (250) 2 [(1+i) -2 + (1+i) (1+i) -10 ]

Variance  NPW t a   () = (250) 2 [(1+i) -2 + (1+i) (1+i) -10 ] Note that [(1+i) -2 + (1+i) (1+i) -10 ] is just a 5 period annuity factor where the period is 2 years. A=(250) P=  2 NPW

Variance  NPW t a   () = (250) 2 [(1+i) -2 + (1+i) (1+i) -10 ] A=(250) P=  2 NPW = (250) 2 (P/A, i eff, 5), i eff = (1+i) 2 -1

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15%  NPW = -10, ,000(P/A, 15, 5) = -10, ,000(3.3522) = $56 A t iid N(3,000, 250) 

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10, ,000(P/A, 15, 5) = -10, ,000(3.3522) = $56 A t iid N(3,000, 250)  i eff = (1.15) = 32.25%

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10, ,000(P/A, 15, 5) = -10, ,000(3.3522) = $56  2 NPW = (250) 2 (P/A, 32.25, 5) = 62,500(2.3343) = 145,894 A t iid N(3,000, 250)  i eff = (1.15) = 32.25%

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = $56  2 NPW = 145,894  = 382 A t iid N(3,000, 250)  NPW  382)

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% A t iid N(3,000, 250)  NPW  382)   N(56, 382) 

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 NPW  382)   N(56, 382)  PNPWP ()      

Risk Analysis A 1 A 2 A 3 A 4 A 5 10,000 NPW  382)   N(56, 382)  PNPWP ()       = P( Z < ) = 0.44

Review

You are given the following cash flow diagram: Class Problem  5,000,0.6 7,,0.4    If A i iid A 1 A 2 10,000 Compute the distribution for the Net Present Worth if the MARR = 15%.

Class Problem

You are given the following cash flow diagram: Class Problem If the MARR = 15%, what is the probability this investment alternative is no good? A 1 A 2 A 3 A 4 A 5 35,000  If A i iid N(10,000, 300)

Class Problem E[NPW] = -35, ,000(P/A, 15, 5) = -35, ,000(3.3522) = - 1,478  2 (NPW)  [(1.15)  2  (1. )  4 ...  (1. )  10 ] = (2.3343) = 210,087 

Class Problem NPW N(-1,478, 458) P{NPW < 0} =  P NPW    0  (  1,478) 458       = P{Z < 3.27}  1.0

A Critical Thunk A 1 A 2 A 3 A 4 A 5 35,000 If A i iid  N(10,000, 300) Max A i  10,900

A Critical Thunk A 1 A 2 A 3 A 4 A 5 35,000 If Max A i  10,900 NPW = -35, ,900(P/A, 15, 5) = -35, ,900(3.3522) = 1,539

If A i iid U( 5000, 7000 ) A Twist Suppose we have the following cash flow diagram. A 1 A 2 10,000  Now how can we compute the distribution of the NPW? MARR = 15%.

Solution Alternatives u Assume normality

Solution Alternatives u Assume normality u Upper/Lower Bounds u Laplace Transforms u Transformation/Convolution u Simulation

Simulation A 1 A 2 10,000 Let us arbitrarily pick a value for A 1 and A 2 in the uniform range (5000, 7000). Say A 1 = 5,740 and A 2 = 6,500.

Simulation A 1 A 2 10,000 5,740 6,500 10,000 NPW = -10, ,740(1.15) ,500(1.15) -2 = (93.96)

Simulation A 1 A 2 10,000 5,740 6,500 10,000 We now have one realization of NPW for a given realization of A 1 and A 2.

Simulation A 1 A 2 10,000 5,740 6,500 10,000 We now have one realization of NPW for a given realization of A 1 and A 2. Choose 2 new values for A 1, A 2.

A 1 = 6,820 A 2 = 6,218 A 1 A 2 10,000 6,820 6,218 10,000 NPW = -10, ,820(1.15) ,218(1.15) -2 =

Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206, Choose 2 new values.

A 1 = 5,273 A 2 = 6,422 A 1 A 2 10,000 5,273 6,422 10,000 NPW = -10, ,273(1.15) ,422(1.15) -2 = (558.83)

Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206, ,2736,422(558.83) Choose 2 new values.

Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206, ,2736,422(558.83). 6,8555,

Simulation With enough realizations of A1, A2, and NPW, we can begin to get a sense of the distribution of the NPW. -1, ,380 NPW Freq.

Simulation What we now need is a formal method of selecting random values for A 1 and A 2 to avoid selection bias.

Simulation What we now need is a formal method of selecting random values for A 1 and A 2 to avoid selection bias. Recall the uniform 5,000 7,000 1/2,000 f(x)

Simulation The uniform has cumulative distribution given by: 5,000 7,000 1 F(x) F(x)  0,x  5,000 1,x  7, x  5, 2,     

Simulation Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1 (Rand Function in Excel). Let P be a random variable uniformly from 0 to 1. P U(0,1) 

Simulation Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1. Let P be a random variable uniformly from 0 to 1. P U(0,1) Algorithm: 1. Randomly generate P 2. Let P = F(x) 3. Solve for x = F -1 (p) 

Simulation 1. Randomly generate P U(0,1). P =.7 5,000 7,000 1 F(x)  F(x)  x  5,000 2, 5,000 < x < 7,000

Simulation 1. Randomly generate P U(0,1). P =.7 2. Let P = F(x). 5,000 7,000 1 F(x) F(x)  x  5,000 2,  5,000 < x < 7,000.7

Simulation 1. Randomly generate P U(0,1). P =.7 2. Let P = F(x). 3. x = F -1 (p). 5,000 7,000 1 F(x) F(x)  x  5,000 2,  5,000 < x < 7, ,400

Formal Derivation Recall, for 5,000 7,000 1 F(x) F(x)  x  5,000 2, 5,000 < x < 7,000. Then P P x    ,,,   x5 2,,

Formal Derivation Solving for x = F -1 (p), 5,000 7,000 1 F(x) P x xP  50002,,

Formal Derivation Solving for x = F -1 (p), 5,000 7,000 1 F(x) P x xP  50002,, Note: 1. P = 0 x = 5, P = 1 x = 7,000

Risk Analysis (Excel)

Class Problem You are given the following cash flow diagram. The A i are iid shifted exponentials with location parameter a = 1,000 and scale parameter = 3,000. The cumulative is then given by 7,000 A 1 A 2 A 3 Fxe x () (,)/,   , x > 1,000

Class Problem You are given the first 3 random numbers U(0,1) as follows: P 1 = 0.8 P 2 = 0.3 P 3 = 0.5 You are to compute one realization for the NPW. MARR = 15%. 7,000 A 1 A 2 A 3 Fxe x () (,)/,  

Class Problem P  1  e  x        e  x         1  P  x       ln(1  P) x  1,000  3, ln(1  P)

Class Problem x  1,000  3, ln(1  P) A 1 = 1, ln(1 -.8) = 5,828 A 2 = 1, ln(1 -.3) = 2,070 A 3 = 1, ln(1 -.5) = 3,079

Class Problem 7,000 5,828 2,070 3,079 NPW = -7, ,828(1.15) ,070(1.15) ,079(1.15) -3 = 1,657

Class Problem You are given the following cash flow diagram. The A i are iid gammas with shape parameter  = 4 and scale parameter  = 3,000. The density function is given by 7,000 A 1 A 2 A 3 fxxe x () () /         1, x > 0

Class Problem You are given the first 3 random numbers U(0,1) as follows: P 1 = 0.8 P 2 = 0.3 P 3 = 0.5 You are to compute one realization for the NPW. MARR = 15%. 7,000 A 1 A 2 A 3

Class Problem For  = integer, the cumulative distribution function is given by Set P = F(x), solve for x

Class Problem For general  (not integer), F(x) = not analytic

Class Problem For general  (not integer), F(x) = not analytic No Inverse

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