1. 2/27/2016rd1 Present Worth Analysis 1. Mutually Exclusive at MARR of 7% A B First cost$1500$1600 Salvage value 200 325 Life (years) 5 5 PW A = -1500 + 200(P/F, 7%, 5) = -$1357.40 ** PW B = -1600 + 325(P/F, 7%, 5) = -$1368.28

2. 2/27/2016rd2 2. Mutually Exclusive: Choose the best at the MARR of 10%. A B C First cost-2500-3500-6000 AOC -900 -700 -50 Salvage value 200 350 100 Life (years) 5 5 5 PW A (10%) = -2500 -900(P/A, 10%, 5) + 200(P/F, 10%, 5) = -$5787.53 PW A (10%) = (+ -2500 (PGA -900 10 5) (PGF 200 10 5))  -$5787.53 PW B (10%) = -3500 -700(P/A, 10%, 5) + 350(P/F, 10%, 5) = -$5936.23 PWB(10%) = (+ -3500 (PGA -700 10 5) (PGF 350 10 5))  -$5936.23 PWC(10%) = -6000 -50(P/A, 10%, 5) + 100(P/F, 10%, 5) = -$6127.45 PWC(10%) = (+ -6000 (PGA -50 10 5) (PGF 100 10 5))  -$6127.45 Choose A, least cost

3. 2/27/2016rd3 Three ways to buy a car Plan A: $5,000 cash immediately Plan B: $1,500 down and 36 monthly payments of $116.25 Plan C: $1,000 down and 48 monthly payments of $120.50 Keep car for 5 years with i = 12% compounded monthly; repeat at 6% at i = 12%/ 12 = 1% per month PW A = -$5K PW B (1%) = -1500 – 116.25(P/A, 1%, 36) = -$5000 PW C (1%) = -1000 – 120.50(P/A, 1%, 48) = -$5575.86 Choose A or B at i = 6%/12 = ½% PW A (1/2%) = -$5K PW B (1/2%) = -1500 -116.25,1/2%,36) = -$5321.26 PW C (1/2%) = -1000 – 120.50(P/A,1/2%, 48) = -$6130.93 Choose A

4. 2/27/2016rd4 3. Different life on leasing example at 15% MARR A B First Cost, $-15K-18K Annual cost-3,500-3100 Deposit return, $1,000$2,000 Lease term 6 9 Compare for 18 years, least common multiple and same first costs. PW A (15%) = -15K -14K[(P/F,15%,6) + (P/F,15%,12)] - 3500(P/A 15% 18) + 1000(P/F,15%,18) or in code (+ -15e3 (PGF -14e3 15 6) (PGF -14e3 15 12) (PGF 1e3 15 18) (PGA -3500 15 18))  -$45,036.36 PW B = (+ -18e3 (PGA -3100 15 18) (PGF -16e3 15 9)(PGF 2e3 15 18)) = $ -41,383.29 ** …continued 

5. 2/27/2016rd5 3. Different life on leasing example at 15% MARR A B First Cost, $-15K-18K Annual cost-3,500-3100 Deposit return, $1,000$2,000 Lease term 6 9 Compare for a 5-year study with same deposit return. PW A (15%) = -15K -3.5K(P/F,15%,5) + 1K(P/F,15%,1) = -$25,862.98 PW B (15%) = -18K -3.1K(P/F,15%,5) + 2K(P/F,15%,4) = -$27,248.17 Choose A

6. 2/27/2016rd6 Present Worth 4. Find P at 6% equivalent to the benefits. n1 2 3 4567 cfP100200300400500600 P = 100(P/A, 6%, 6) + 100(P/G, 6%, 6) = G(P/G, 6%, 7)(F/P, 6%, 1) (PGG A G i n)  (PGG 100 100 6 6)  $1637.66

7. 2/27/2016rd7 Capitalized Costs P = A[(1 + i) n -1] / [i(1 + i) n ] As n  infinity, P  A/i PW = CC = A/i or A = CC * i = Pi $10,000 at 10% can generate $1000 per year in perpetuity … N   P

8. 2/27/2016rd8 5. A system costs $150K to install and $50K 10 years later. Annual maintenance is $5K for the first 4 years and $8K thereafter. A $15K recurring upgrade cost every 13 years is required. At i = 5%, find the equivalent annual cost forever. Non-recurring costs  CC1 = (+ -150e3 (PGF -50e3 5 10))  -$180,695.68 Recurring cost $15K every 13 years  A1 = (AGF -15e3 5 13)  -$846.84 CC2 = 5000/0.05 = 100K CC3 = 3000/0.05 = 60K but brought back 4 years with (PGF 60e3 5 4)  = - $49,362.15 CC = (+ 180695.68 100e3 49362.15)  $330,057.84 Annualize  $330,057.84 * 0.05 = $16,502.89 Total annualized cost is then (+ 16502.89 846.84)  $17,349.73

9. 2/27/2016rd9 Problem 5-84 n 012345 6 7 cf0 -120 -60204080 100 60 (list-PGF '(0 -120 -60 20 40 80 100 60) 15)  -$8.13287K Choose the Do Nothing earning at the MARR of 15%.

10. 2/27/2016rd10 Alternatives A B MARR = 10% First cost$205K$235K Ann Maint 29K 27K Salvage 2K 20K Life (years) 2 4 PW A = 205K + 29K(P/A,10%, 4) + 203K(P/F,10%,2) -2K(P/F,10%, 4) = $463,329 PW B = 235K + 27K(P/A,10%, 4) – 20K(P/F, 10%, 4) = $306,926 => Choose B, Less Cost

11. 2/27/2016rd11 a) Select better machine based on PW at 15% per year and state assumptions. Machine AMachine B First Cost11,00018,000 Annual operating cost 3,500 3,100 Salvage 1,000 2,000 Life (years) 6 9 PW A (15%) = 3500(P/A,15%,18) + 11 + 10K(0.43233 + 0.18691)-1000(P/F, 15%, 18) = $38,559.42 (cost) PW B (15%) = 3100(P/A,15%,18) + 18K + 16K(P/A,15%,9) - 2K(P/F,15%,18) b) Select based on 5 year study period. PW A (15%) = 11,000 + 3500(P/A,15%,5) - 1000(P/F,15%,1) = -$21,743.91. PW B (15%) = 18,000 + 3100(P/A, 15%, 5) - 2000(P/F,15%,5) = -$27,248.17 c) Find B's salvage for an indifferent decision in a 6-year study. PW A (15%) = 11,000 + 13,245.69 - 432.33 = 18,000 + 11,731.90 -S(.432328) 23,813.36 = 29,731.90 - 0.432328 S => S = $13,689.93.

12. 2/27/2016rd12 Annual Deposit The annual deposit needed in years 1 to 5 to provide for an annual withdrawal of $1000 for 20 years beginning 6 years from now at i = 10% per year is closest to $1395b) $1457 c) $1685d) > $1750 A(F/A, 10%, 5) = 1000(P/A, 10%, 20) A(6.1051) = 1000(8.513564) = $1394.5

13. 2/27/2016rd13 Combining Factors If $5000 is deposited now, $7000 two years from now and $2000 per year in years 6 to 10, the amount in year 10 at an interest of 10% per year will be closest to < $40Kb) $40,185c) $42,200d) $43,565 F 10 = 5K(1.1) 10 + 7K(1.1) 8 + 2K(F/A, 10%, 5) = $12,968.71 + 15,005.12 + $12,210.2 = $40,184.03

14. Perpetuity With interest at 8% compounded annually, how much money is needed to provide a perpetual income of $14,316 per year? P = (/ 14316 0.08)  $178,950 2/27/2016rd14

15. Education You want to deposit $8,000 per year for 4 years of your child's education on the 18 th birthday. Today is your child's first birthday. At 12% interest, how much to deposit today? a) $24,298b) $3538c) $32,000d) $3963 2/27/2016rd15