1. TM 661Risk AnalysisSolutions 1 The officers of a resort country club are thinking of constructing an additional 18-hole golf course. Because of the northerly location of the resort, there is a 60% chance of a 120-day golf season, a 20% chance of 150 days of golfing weather, and a 20% chance of a 165-day season. The course will be utilized by an estimated 350 golfers each day of the 120-day season but only by 100 per day for each extra day in the golfing season. The course will cost $1,000,000 to construct. Annual maintenance costs are $100,000 and green fees would be set at $20 per person. The club uses a 10 year planning horizon and a before tax MARR of 15%. You are to perform supplementary analysis and create a decision tree for this problem to determine if the club should invest. No invest Invest 165 150 120 46,500 golfers @ 0.2 3,255,915 45,000 golfers @ 0.2 3,015,015 42,000 golfers @ 0.6 2,713,889 For 165 day season, A t = 350 golfers * 20 fees * 120 days + 100 * 20 * 45 - 100,000 = 848,000 / yr. NPW 165 = -1,000,000 + 848,000 (P/A,15,10) = 3,255,915 0

2. TM 661Risk AnalysisSolutions 2 You are given the following decision tree. Evaluate the tree to determine which decision(s) the company should select. Be complete. Value 100 300 -100 500 100 0.6 0.4 0.8 0.2 100 50 100 140 212 312 Solution

3. TM 661Risk AnalysisSolutions 3 You are given the following cash flow diagram. where A 1, A 2 are identically independently distributed with the following probability A i p(A i ) 5,000 0.6 7,000 0.4 Compute the probability that the NPW is greater than 0. MARR is 10%. Soln: NPW = -10,000 + 5,000 (1.1) -1 + 5,000 (1.1) -2 = -1,322with probability 0.6 * 0.6 = 0.36 NPW = -10,000 + 5,000 (1.1) -1 + 7,000 (1.1) -2 = 330with probability 0.6 * 0.4 = 0.24 NPW = -10,000 + 7,000 (1.1) -1 + 5,000 (1.1) -2 = 496with probability 0.4 * 0.6 = 0.24 NPW = -10,000 + 7,000 (1.1) -1 + 7,000 (1.1) -2 = 2,149with probability 0.4 * 0.4 = 0.16 10,000 A 1 A 2

4. TM 661Risk AnalysisSolutions 4 You are given the following cash flow diagram. where the A i ’s are identically independent normals with mean  = 7,000 and standard deviation  = 300. Compute the probability that the NPW (MARR = 15%) is greater than 0. Soln:  NPW = -30,000 + 7,000 (P/A,15,5) = -6,535  NPW 2 = 300 2 [(1.15) -2 + (1.15) -4 + (1.15) -6 + (1.15) -8 + (1.15) -10 ] = 300 2 (P/A, 32.25, 5) = 300 2 (2.3343) = 210,088 30,000 A 1 A 2 A 3 A 4 A 5

5. TM 661Risk AnalysisSolutions 5 You are given the following cash flow diagram. where Ai are identically independent exponentials with scale parameter = 3,000. The cumulative is then given by F(x) = 1 - e -x/3000, x > 0. You are given the first three random numbers U(0,1) as follows: U 1 = 0.8 U 2 = 0.3 U 3 = 0.5 Using U t and the inverse method, compute one realization of the NPW for a simulation of the investment project. Soln: A1A1 A2A2 A3A3 10,000 )1ln(000,3 ii UA  )1ln(000,3/ ii UA  1,3/ i A Ue i   1,3/ A i eU i   A 1 = 4,828 A 2 = 1,070 A 3 = 2,079 NPW = -10,000 + 4,828 (1.1) -1 + 1,070 (1.1) -2 + 2,079 (1.1) -3 = - 3,165