Question

Your cat's summer kitty-cottage needs a new roof. You feel a 15-year analysis period is in line with your cat's remaining lives. (there is no salvage value for old roofs.)

Thatch Slate

First cost $200 $350

Annual upkeep 50 20

Service life, in years 3 5

A. Graph the EUAC or EUAW for the alternatives. Construct a choice table for interest rates from 0% to 100%.

B. Which roof should you choose if your MARR is 12%? What is the actual value of the IRR on the incremental cost?

Answer 1

A.Calculation of EUAC
Interest rate	1%	2%	3%	4%	5%	6%	7%	8%	9%	10%	11%	12%	13%	14%	15%	16%	17%	18%	19%	20%	21%	22%	23%	24%	25%	26%	27%	28%	29%	30%	31%
Annuity factor = (1- (1/(1+interest rate)^3))/interest rate	2.94	2.88	2.83	2.78	2.72	2.67	2.62	2.58	2.53	2.49	2.44	2.40	2.36	2.32	2.28	2.25	2.21	2.17	2.14	2.11	2.07	2.04	2.01	1.98	1.95	1.92	1.90	1.87	1.84	1.82	1.79
EUAC of Thatch = 200/annulty +50	118.00	119.35	120.71	122.07	123.44	124.82	126.21	127.61	129.01	130.42	131.84	133.27	134.70	136.15	137.60	139.05	140.51	141.98	143.46	144.95	146.44	147.93	149.43	150.94	152.46	153.98	155.51	157.04	158.58	160.13	161.68
Annuity factor = (1- (1/(1+interest rate)^5))/interest rate	4.85	4.71	4.58	4.45	4.33	4.21	4.10	3.99	3.89	3.79	3.70	3.60	3.52	3.43	3.35	3.27	3.20	3.13	3.06	2.99	2.93	2.86	2.80	2.75	2.69	2.64	2.58	2.53	2.48	2.44	2.39
EUAC of Slate = 350/annulty factor +20	92.11	94.26	96.42	98.62	100.84	103.09	105.36	107.66	109.98	112.33	114.70	117.09	119.51	121.95	124.41	126.89	129.40	131.92	134.47	137.03	139.62	142.22	144.85	147.49	150.15	152.82	155.52	158.23	160.96	163.70	166.46
Used excel sheet, and one can do in the same manner upto 100% interest And option Slate is better option uoto interest rate of 28% and beyond Thatch is best option.
b. If MARR is 12 percent
IRR of Thatch = -200 - 50(1-(1.12)^-3)/0.12 =-200 - 120.09 = -320.09
IRR of Slate = -350 - 20(1-(1.12)^-5)/0.12 =-350 - 72.09= -422.09
So option Thatch is the best option will less negative but life is three years only.
Formula = present value of annual out flow = annual flow(( 1-(1+MARR)^-N))/MARR, where N= Life