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Engineering Economic Analysis

Engineering Economic Analysis. Chapter 10  Uncertainty in Future Events. Gambling. If You Bet on a Horse, That’s Gambling. If You Bet You Can Make 3 Spades, That’s Entertainment. If You Bet the Structure Will Last for 100 Years, That’s Engineering.

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Engineering Economic Analysis

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  1. Engineering Economic Analysis Chapter 10  Uncertainty in Future Events rd

  2. Gambling. If You Bet on a Horse, That’s Gambling. If You Bet You Can Make 3 Spades, That’s Entertainment. If You Bet the Structure Will Last for 100 Years, That’s Engineering. See the Difference? rd

  3. Combinations How many ways can one select r objects from n objects? nCr = n! / (r!)(n-r)! = {A B C} AB AC BC Binomial coefficient 1 20 1 1 21 1 2 1 22 1 3 3 1 23 1 4 6 4 1 24 ……………………. 2n = 24 rd

  4. Permutations How many ways can one place 5 different books on a shelf where order matters? 5P5 = 5! /(5-5)! = 120 Find the number and permutations of {a b c} taken 2 at a time. 3P2 = 6 => (Permutation '(a b c) 2) ((A B) (A C) (B A) (B C) (C A) (C B)) nPr = r! nCr (PP (upto 9))  362,880 = 9! permutations of the integers 1-9. rd

  5. Probability An experiment is performed where outcomes are occurrences of events. S ~ Sample Space of Events or Sets X ~ Random variable (RV) and x is its value P(X = x) is the probability that the RV X = x. Experiment: Tossing a fair die Sample Space {x: 1 2 3 4 5 6} P(X = 2) = 1/6 … continued  rd

  6. rd

  7. Statistician Probability Number Favorable Total Numbers Compute the probability of an even number occurring on the roll of die. p = Number favorable = 3 ~ (2 4 6) Total Numbers = 6 ~ (1 2 3 4 5 6) = 1/2 rd

  8. Axioms of Probability For an event A, 0  P(A)  1 P(S) = 1, P() = 0 And for disjoint sets Ai, rd

  9. Probability of Fair Die X 1 2 3 4 5 6 P(X) 1/6 1/6 1/6 1/6 1/6 1/6 a) P(X < 4) = P(X <= 3) = 3/6 b)  = E(X) is the expected value computed as 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = 3.5 • E(X2) = (12 + 22 32 + 42 + 52 + 62)/6 = 91/6 • d) 2 = V(X) = E[(X - )2] = E(X2) – E2(X) = 91/6 – 12.25 = 35/12 = 2(6.25 + 2.25 + 0.25)/6 e)  = 1.707825 = 17.5/6 = 35/12  is the mean,  is the standard deviation, 2 is the varianceof the distribution. rd

  10. Conditional Probability Fair Die Experiment The probability that X = 2 given that the event EVEN occurred may be written as P(X=2|Even) or in set notation as P(A|B). P(A|B) = P(AB)/P(B) where P(AB) is P(A and B) = P(A,B) P(X=2|Even) = P(X=2, Even)/P(Even) = P(X=2)/P(Even) = (1/6) / (3/6) = 1/3. rd

  11. Two Fundamental Principles of Counting (FPC) and the Pigeonhole Principle If there are n1 different items in set 1, n2 different items in set 2, etc. for r disjoint sets, then the number of ways to select an item from one of the r sets is n1 + n2 + … + nr. Addition principle. If there are n1 outcomes for the first decision in stage 1 followed by n2 outcomes for the second decision in stage 2 followed by n3 for the 3rd decision in stage 3, etc. for r stages, where each decision is independent of all prior decisions, then all together there are n1 * n2 *... * nr. Multiplication principle. Pigeonhole Principle: If the number of pigeons exceeds the number of pigeonholes, then some pigeonhole has at least 2 pigeons. rd

  12. Multiplication Rule How many 4-letter words are possible from the 5 letters {E F G H I} without replacing? 5 * 4 * 3 * 2 = 120 = 5P4 = 5! /(5 – 4)! How many end with a vowel? 4 * 3 * 2 * 2 = 48 How many end with G? 4 * 3 * 2 * 1 = 24 (E F H G) (E F I G) (E H F G) (E H I G) (E I F G) (E I H G) (F E H G) (F E I G) (F H E G) (F H I G) (F I E G) (F I H G) (H E F G) (H E I G) (H F E G) (H F I G) (H I E G) (H I F G) (I E F G) (I E H G) (I F E G) (I F H G) (I H E G) (I H F G)) rd

  13. Fundamental Principles of Counting & Pigeon Hole a) At a picnic of 50 people, 30 had hamburgers, 25 had hotdogs, and 15 had both. How many had neither? (30 – 15) + (25 -15) + 15 + x = 50 => X = 10 had neither.b) How many ways can a 20-question true/false test be answered? 220 = 1,048,576 c) In a drawer are 12 black socks and 12 white socks. What is the minimum number of socks one can randomly pull from the drawer to ensure getting a matching pair? d) In an urn are 3 red, 4 white and 5 blue marbles. How many ways can a sample of 4 marbles be drawn with at least one of each color? 60 + 90 + 120 = 270 rd

  14. Probability Class Work 1A 1. If repetitions are not permitted, (Permutations) a) How many 3 digit numbers can be formed from the 6 digits in {2, 3, 5, 6, 7, 9}? 6 * 5 * 4 = 120 b) How many are less than 400? 2 * 5 * 4 = 40 c) How many are even? 5 * 4 * 2 = 40 d) How many are odd? 5 * 4 * 4 = 80 e) How many are multiples of 5? 5 * 4 * 1 = 20 rd

  15. 2. Solve #1 above if repetitions are permitted? a. 6 * 6 * 6 = 215 b. 2 * 6 * 6 = 72 c. 6 * 6 * 2 = 72 d. 6 * 6 * 4 = 144 e. 6 * 6 * 1 = 36 3. Consider the 3 letters in the set {a, b, c}. a) List the combinations for the 3 letters taken 3, 2, 1, and 0 at a time and for each combination its permutations. Combinations Permutations 3C3 = {abc} 3P3 =6 => abc, bca, cab, acb, bac, cba 3C2= {ab, ac, bc} 3P2 = 6 => ab,ba ac,ca, bc,cb 3C1 = {a, b, c} 3P1 = 3 => a, b, c 3C0 = 1 3P0= 1 => Don't take any rd

  16. a) An octagon has how many diagonals? • 8C2 - 8 = 28 - 8 = 20 • b) An n-gon has how many diagonals? • nC2 - n = n(n-1)/2 - n = n(n - 3)/2 • c) What regular polygon has the same number of diagonals as sides? • nC2 - n = n => n(n-1)/2 = 2n => n = 0 or 5 => Pentagon rd

  17. 5. Determine the probability p for each of the following. a) an even number appears in the toss of a die; 3/6 b) a jack appears in drawing a card from a deck; 4/52 c) at least one heads appears in tossing 3 coins; 7/8 d) a red marble is drawn from an urn containing 4 white, 5 red, and 6 blue marbles. 5/15 e) both cards are hearts drawn from a deck. (13/52)(12/51) rd

  18. 6. Of all couples who have two children, what is the probability that both are a) Boys? 1/4 BB, BG, GG, GB b) Boys given that at least one is a boy? 1/3BB, BG, GG, GB c) Boys given that the older child is a boy? 1/2 BB, BG, GG, GB 7. Three fair coins are tossed. Find the probability of exactly 2 heads. Sample Space = {HHH HHTHTH HTT THH THT TTH TTT} 3/8 rd

  19. A coin with P(H) = 2/3 is tossed. If heads appears, a number is selected at random from the set 1 - 9. If tails appears, a number is selected at random from the set 1 - 5. Find the probability that an even number is selected. • P(Even) = P(Even | H)P(H) + P(Even | T)P(T) • = (4/9)*(2/3) + (2/5)*(1/3) = 58/135 = 0.429629 rd

  20. 9. Box A contains 9 cards numbered 1 through 9; Box B contains 5 cards numbered 1 through 5. A box is chosen at random and a card is drawn. If the number is even, find the probability that the card came from Box A? Assume P(Box A) = P(Box B) P(A) = P(B) = 1/2 P(Even) = P(Even|A)P(A) + P(Even|B)P(B) = (4/9) * (1/2) + (2/5) * (1/2) = (1/2) * 38/45 = 19/45 = 0.42222.   P(A | Even) = P(A, Even)/P(Even) = [(1/2)*(4/9)] / (19/45) = 10/19. rd

  21. The probability of 4 aces in 5-card poker is 48/52C5. • 12. The probability that the 4 aces occur consecutively in a deck of 52 cards is 49/52C5. •   1-4, 2-5, … 49-52 • The probability of 4 aces in a bridge hand (13 cards) is [4C4 *48C9]/ 52C13 = 0.002641. • The probability of 4 aces and 4 kings in a bridge hand is • [4C4 * 4C4 * 44C5 ]/ 52C13 = 11/6431950 = 1.7102E-6 rd

  22. 15. The probability of 4 aces or 4 kings in a bridge hand is • P(4A + 4K) = P(4A) + P(4K) – P(4A, 4K) • [48C9 /52C13 + 48C9 /52C13] - 44C5 / 52C13. • = 0.0052804 • 16. The probability of a full house in a toss of 5 fair dice is 0.03858. [xxx yy] • [6C2 * 2C1 *5C3 * 2C2] / 65 = 300/7776 = 0.03858. rd

  23. 17. Compute the probability of winning at the game of craps, where a sum of 7 or 11 on the first roll of a pair of fair dice wins and a sum of 2, 3 or 12 loses; any other sum becomes your point, and you must make your point before a sum of 7 is thrown. P(win) = P(7)+P(11)+2[P(4),P(4|4+7)+P(5),P(5|5+7)+P(6),P(6|6+7)] = 1/6 + 1/18 + 2[(3/36) (3/9) + (4/36)(4/10) + (5/36)(5/11)] = 0.4929292. P(4|4+7) means probability point is 4 and that a sum of 4 occurs before a sum of 7. (sim-wins-in-craps 10000) Digit 2 3 4 5 6 7 8 9 10 11 12 Count 271 546 856 1152 1391 1671 1347 1121 843 518 284 Number of wins = 4919 rd

  24. 18. What is the probability of at least 2 people sharing a birthday (day and month) in a group of n people? P(n people not sharing) = P(sharing) = 1 - (- 1 (/ (perm 365 23) (expt 365 23))) rd

  25. 19. A fair pair of dice are rolled 7 times. Compute the probability of 3S7, 3S6,and S5 (multinomial distribution). P(S7) = 1/6; P(S6) = 5/36; P(S5) = 4/36 p = = 0.0005788. rd

  26. Complement • A fair coin is tossed 6 times. Find the probability that at least one head appears. P(X  1) = 1 – P(X = 0) = 1 – (1/2)6 = 63/64. P(A) = 1 – P(Ac) rd

  27. There are n balls in an urn. One is imperceptibly lighter or heavier than the others. A balance scale is to be used to find the culprit in a minimum number of weighings. Solve for n = 12 balls. Event Balance No Balance X n – 2x x P(X) (n – 2x)/n 2x/n E(X) = (n – 2x)2/n + 2x2/n for 0  X  n/2. E’(X) = [-4(n – 2x)2 +4x]/n = 0 when x = n/3. rd

  28. For n = 12 balls 1 2 3 4 5 6 7 8 3 6 7 4 8 G 1 2 5 rd

  29. Discrete Density Distribution X 0 1P(X) q p E(X) = 0 * q + 1 * p = p E(X2) = 02 * q = 12 * p = p V(X) = E(X2) – E2(X) = p – p2 = p(1 - p) = pq rd

  30. Variance Variance = Example: Variance of the integers from 1 to 6 is [(1- 3.5)2 + (2 - 3.5)2 + (3 - 3.5)2 + (4 – 3.5)2 + (5 – 3.5)2 + (6 – 3.5)2]/6 (Var '(1 2 3 4 5 6)) 35/12 Variance is a measure of dis.persion of values about the mean rd

  31. Project A has the following distribution of expected present worth. Compute the expected value and standard deviation of net present worth.E(NPW), E(NPW2), and V(NPW). • Net Present Worth -12K 4K 12K 20K 30K • Probability 1/10 2/10 4/10 2/10 1/10 • E(NPW) = -12K*0.1 + 4K*0.2 + 12K * 0.4 + 20K * 0.2 + 30K * 0.1 • = (-1.2 + 0.8 + 4.8 + 4 + 3)K / 10 = $11.4K first moment around origin • E(NPW2) = (14.4 + 3.2 + 57.6 + 80 + 90)M = 245.2M second moment around origin • V(NPW) = E(NPW2) – E2(NPW) = 245.2M – (11.4K)2 2nd moment around mean = 115.24M square dollars. Variance is denoted as 2. • Standard Deviation is the square root of the variance =>  = $10,735. rd

  32. Optimistic Most Likely Pessimistic First cost $950 $1000 $1150Net-AB $210 200 170Life (yrs) 12 10 8Salvage value $100 0 0MARR is 10%. (UIRR 950 210 12 100) 19.82%(UIRR 1000 200 10 0) 15.10%(UIRR 1150 170 8 0) 3.89% Mean first cost = [950 + 4(1000) + 1150]/6 = 1016.67Mean Net-AB = (210 + 800 + 170)/6 = 196.7Mean life = (12 + 40 + 8)/6 = 10Mean salvage = 100/6 = 16.7 (UIRR 1016.67 196.7 10 16.7)  14.35% or (19.82 + 4*15.1 + 3.89)/6 = 14.02% rd

  33. Example 10-1 A B First cost $1000 $2000 NAB 150 250Life 10 10 Salvage 100 400 NPWA (3.5%) = -1000 + 150(P/A, 3.5%, 10) + 100(P/F,3.5%, 10) = $318.38 NPWB (3.5%) = -2000 + 250(P/A, 3.5%, 10) + 400(P/F,3.5%, 10) = $362.72 and B is chosen. Suppose B’s salvage value was $300. Then NPWB (3.5%) = -2000 + 250(P/A, 3.5%, 10) + 300(P/F,3.5%, 10) = $291.83 and A is chosen. Breakeven at $339 = S rd

  34. Example 10-5,6 First cost is $25K; MARR 10% Annual Benefit $5K $8K $10KP(AB) 0.3 0.6 0.1 Life 6 9 P(Life) 2/3 1/3 Joint Density DistributionAB 5 8 10 5 8 10Life 6 6 6 9 9 9 P 0.2 0.4 0.067 0.1 0.2 0.033 PW -3224 9842 18,553 3795 21,072 32,590 Expected value = $10,204.42 rd

  35. Risk versus Return IRR Project IRR St-Dev %1 13.1% 6.5% 6 2 12.0 3.9 15 7 8 3 7.5 1.5 2 4 6.5 3.5 10 1 5 9.4 8.0 3 5 6 16.3 10.0 4 7 15.1 7.0 5 8 15.3 9.4 F F 4.0 0.0 Efficient frontier Standard Deviation rd

  36. Simulation Uniform Distribution a b rd

  37. Example 10-15 (sim-scale 10)  Life [12-16] First Cost N(1500, 150) IRR 12 1540 12.15 14 1660 11.91 12 1540 12.14 15 1710 11.97 14 1460 14.57 16 1250 18.74 14 1200 19.06 16 1590 13.71 15 1410 15.69 13 1510 13.25 The simulated Rate of Return is 14.32 with standard error 2.71 (sim-scaler 1000) (13.99 1500.65 14.05) (1.416 148.192 2.221)) Life FirstCost IRR standard deviations rd

  38. Fair Coin Toss 1 2 3 4 5 6 7 8 9 10 #Heads #Flips P(H) 488 516 496 496 520 479 506 486 518 460 4965 10000 0.4965 1028 983 969 1024 991 987 1020 1037 1004 987 10030 20000 0.5015 1527 1471 1489 1499 1525 1513 1446 1516 1468 1505 14959 30000 0.498633 2011 2033 2022 2036 2001 2042 1989 1971 1999 2017 20121 40000 0.503025 2476 2429 2487 2503 2436 2475 2511 2508 2510 2527 24862 50000 0.49724 3012 3049 2991 3016 3021 2994 2996 3007 2988 2941 30015 60000 0.50025 3562 3487 3531 3519 3436 3561 3502 3482 3461 3505 35046 70000 0.500657 3997 3985 4033 3985 3972 4010 3942 3937 4005 4049 39915 80000 0.498938 4538 4435 4566 4565 4506 4524 4469 4541 4498 4465 45107 90000 0.501189 4956 5037 5067 5015 4963 5068 5028 5009 4949 5002 50094 100000 0.50094(sim-coins-1-1 100 1/2 10) rd

  39. Problem 10-5 P(S7 + S11) = 1/6 + 1/18 = 2/9 P(win) = P(S7) + P(S11) + 2{P[S4, S4 |(S4 or S7)] + P[S5,S5 |(S5 or S7)] + P[S6,S6 |(S6 or S7)]}   = 6/36 + 2/36 + 2[3/36 * 3/9 + 4/36 * 4/10 + 5/36 * 5/11]   = 0.491 P(win) = P(S7) + P(S11) + 2 rd

  40. Problem 10-6 X 20K 30K 40KP(X) 0.3 0.5 0.2 E(X) = 0.3 * 20K + 0.5 * 30K + 0.2 * 40K = $29K rd

  41. Problem 10-9 Accident in next 3 years costs $600 and probability of accident is 20%. X at least 1 0 1 2 3 P(X ≥ 1)($600) 0 0.2 0.36 0.488  ≠ 1 P(Y1 + Y2) = P(Y1 + P(Y2) – P(Y1, Y2) = 0.2 + 0.2 – 0.04 = 0.36 P(Y1 + Y2 + Y3) = P(Y1) + P(Y2) + P(Y3) – P(Y1, Y2) – P(Y1, Y3) - P(Y2, Y3) + P(Y1, Y2, Y3) = 3(0.2) – 3(0.04) + 0.008 = 0.488 rd

  42. Problem 10-11 A man is considering a $1K investment with the following risks. Should he invest if he wants his money back in one year? X $2000 1500 1000 500 0 P(X) 0.3 0.1 0.2 0.3 0.1 E(X) = 2000 * 0.3 + 1500 * 0.1 + 1000 * 0.2 + 500 * 0.3 + 0 * 0.1 = $1,100 > $1,000 => Yes rd

  43. Problem 10-13 P(S8) = 5/36 in getting a sum of 8 by rolling two fair dice. The 5 ways are: (3,5) (4,4) (5,6) (6,5) (4,3) => 5/36 P(4,4)|P(S8) = 1/5 X (3,5) (4,4) (5,3) (6,2) (2,6)Y (4,4) othersP(X) 1/5 1/5 1/5 1/5 1/5 P(Y) 0.2 0.8 It costs $1 to play and try to win $4. E(Bet) = 0.2 * $5 - 0.8 * $1 = -$0.20 E(Win) = 0.2 * $4 = $0.80 but by risking $1 to do so, in the long run will lose $0.20/bet rd

  44. Problem 10-18 at 12% for 50 years HgtProb Damage E(D) 1st cost A/P Total D 2m 1/3 300K 100K $100K 12042 $112,0422.5 1/8 300K 37.5K 165K 19868 57,3683.0 1/25 300K 12K 300K 36125 48,125*3.5 1/50 300K 6K 400K 48167 54,1674.0 1/100 300K 3K 550K 66229 69,229 rd

  45. Problem 10-19 10-year life with MARR set at 12% for First cost P Net Revenue P $300K 0.2 $70K 0.3 400K 0.5 90K 0.5 600K 0.3 100K 0.2 E(FC) = -(60K + 200K + 180K) = -$440K E(Annual revenue) = 21K + 45K + 20K = $86K NPW = -440K + 86K(P/A, 12%, 10) = $45.92K => Accept project rd

  46. Problem 10-22 Need $30K with MARR at 10% given the following: Bad OK GreatProbability 0.3 0.5 0.2Net revenue -$15K 15K 20KLife 5 5 10 PWRev -$56,861.8 $56,861.8 $122,891.34 E(PW) = -$30K + [0.3 * 56862 + 0.5 * 56862 + 0.2 *122891] = -$30,000 + 35,950.60 = $5950.60 E(PWRev2) = (EV '(0.3 0.5 0.2) (mapcar #' square '(-56862 56862 122891)))  5.61e9 => V(PWRev) = 5.61e9 – 59512 = 5.571655e9 =>  = 74,643.52 rd

  47. Problem 10-24 Given an $80K first cost with the MARR at 9% and the density functions of savings and life, find the expected worth. Savings $15K 30K 45KLife 3 5P(savings) 0.3 0.5 0.2 P(life) 0.6 0.4 -80K + 15K(P/A, 9%, 3)  -42030.48 with probability 0.3*0.6 = 0.18 -80K + 15K(P/A, 9%, 5)  -21655.23 with probability 0.3*0.4 = 0.12-80K + 30K(P/A, 9%, 3)  -4061.16 with probability 0.5*0.6 = 0.30 -80K + 30K(P/A, 9%, 5)  36689.54 with probability 0.5*0.4 = 0.20-80K + 45K(P/A, 9%, 3)  33908.26 with probability 0.2*0.6 = 0.12-80K + 45K(P/A, 9%, 5)  95034.30 with probability 0.2*0.4 = 0.08 E(return) = 0.18*-42030 + 0.12*-21655 + 0.3*-4062 + 0.2*36689.54 + 0.12*33908 + 0.08*95034 = $7626.88. rd

  48. Problem 10-26 Al ~ N(60, 400, 65) Bill ~ N(12, 16, 14) (normal 60 400 65)  0.59871 Al ~ x + (5/20) sd (normal 12 16 14)  0.69142 => Bill ranked higer. Bill ~ x + (2/4) sd rd

  49. Problem 10-28 First cost is $25K for a 4-year machine with no salvage value. MARR = 12%. Determine return and risk given Annual savings $7K $8.5K $9.5KProbability 0.3 0.4 0.3 E(AS) = 0.3*7K + 0.4* 8.5K + 0.3*9.5K = 8.35K E(PW) = -25K + 8.35K(P/A,12%,4) = $361.87PW1 = -25K + 7.0K(P/A,12%,4) = -$3738.55PW2 = -25K + 8.5K(P/A,12%,4) = $817.47PW3 = -25K + 9.5K(P/A,12%,4) = $3854.82E(PW) = (EV '(-3738.55 817.47 3854.82) '(0.3 0.4 0.3)) = $361.87 E(PW2 ) = 13,976,756*0.3 + 668,257*0.4 + 14,859,637 = 8,918,221.72 => V(PW) = 8,918,222 – 361.872 = 8787178 =>  = $2,964 risk(UIRR 25 8.35 4)  12.69% rd

  50. Problem 10-34 rd

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