Tugas 1 434m5c

Factors in Engineering Economy

Excel Functions

P = Present Value

= PV (i%,n,A,F)

F = Future Value

= FV (i%,n,A,P)

A = Equal, Periodic Value

= PMT (i%,n,P,F)

n = Number of Periods

= NPER (i%,A,P,F)

i = Compound Interest Rate

= RATE (i%,A,P,F)

EXERCISES 1. How much money should you be willing to pay now for a guaranteed $600 per year for 9 years starting next year, at a rate of return of 16% per year? Answer : A = 600 per year n = 9 years i = 16 % = 0,16 Menggunakan rumus : P=A*

(

) (

P = 600*

) (

+ )

(

)

+

P = 2763,93 Menggunakan fungsi di excel : F(x) = PV(i%;n;A;F) F(x) = PV(0,16;9;600;0) P = $2763,93

2. Determine the value of the A/P factor for an interest rate of 7.3% and n of 10 years, that is, (A/P, 7.3%, 10). Answer : i1 = 7 % A/P factor = 0,14328 i2 = 8 %  A/P factor =0,14903 i = 7,3 %  A/P factor dengan n = 10 years ?

A/P

A/P factor untuk n= 10 tahun 0.15 0.149 0.148 0.147 0.146 0.145 0.144 0.143 0.142

y = 0.575x + 0.103 R² = 1

7%

7%

8%

8%

9%

i

Diperoleh persamaan : y = 0,575x + 0,103 y = (0,575×7,3%)+0,103 = 0,144975 Jadi A/P factor untuk i = 7,3% dengan n = 10 adalah 0,144975.

3. HP has completed a study indicating that $50,000 in reduced maintenance this year on one processing line resulted from improved integrated circuit (IC) fabrication technology based on rapidly changing designs. a. If HP considers these types of savings worth 20% per year, find the equivalent value of this result after 5 years. b. If the $50,000 maintenance savings occur now, find its equivalent value 3 years earlier with interest at 20% per year. Answer : a. P = $50000, i =20%, n = 5 years, F ? Menggunakan rumus : F = P(1+i)n F = 50000(1+0,2)5 F = 12441 Menggunakan fungsi excel : F(x) = FV(i%;n;A;P) F(x) = FV(20%;5;0;50000) F = $124416

b. i = 20% n = 3 tahun F = 50000 P? Menggunakan rumus : P = F *(

+

)

P = 50000 *(

)

+

P = 28935,2 Menggunakan fungsi excel : F(x) = PV(i%;n;A;F) F(x) = PV(0,2;3;0;50000) P = $28935,2

4. Three contiguous counties in Florida have agreed to pool tax resources already designated for county-maintained bridge refurbishment. At a recent meeting, the county engineers estimated that a total $500,000 will be deposited at the end of the next year into an for the repair of old and safety-questionable bridges throughout the three-county area. Further, they estimate that the deposits will increase by $100,000 per year for only 9 years thereafter, then cease. Determine the equivalent: a. Present worth if county funds earn interest at a rate of 5% per year. b. Annual series amounts if county funds earn interest at a rate of 5% per year. Answer : a. A1 = 500000, G = 100000, n = 9 years, i = 5 % Mencari A, A = A1 + A2 A2 = A2 =

*

(

)

*

+ (

)

+

A2 = 409908,5 A = A1 + A2 = 409908,5 + 500000 = 909908,5

Kemudian, mencari P1 dengan n = 10 F(x) = PV(i%;n;A;F) F(x) = PV(5%;10;909908,5) P1 = 7026072

Mencari P, dengan n =1 dan F = 909908,5 F(x) = PV(i%;n;A;F) F(x)= PV(5%;1;0;9099085) P = $6691497

b. A per tahun ? Menggunakan fungsi excel F(x) = PMT(i%;n;P;F) F(x) = PMT(5%;9;6691497;0) A = $805581,95

5. How long will it take for $1000 to double if the interest rate is 5% per year? Answer : P = 1000, F = 2000, i = 5%, n ? Menggunakan rumus : F = P(1+i)n 2000 = 1000(1+5%)n 2 = (1,05)n n = 14,203 Menggunakan fungsi excel : F(x) = NPER (i%;A;P;F) F(x) = NPER (5%;0;1000;2000) n = 14,203 tahun

6. Engineers at SeaWorld, a division of Busch Garden, Inc., have completed an innovation on an existing water sports ride to make it more exciting. The modification costs only $8000 and is expected to last 6 years with a $1300 salvage value for the solenoid mechanisms. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost. The interest rate is 8% per year. Answer : P1 = 8000 (pengeluaran) F1 = 1700, g = 11% F2 = 1300, n = 6 years

i = 8% g’ =

=

= -0,020703

Mencari P2 (pengeluaran), dengan F1 = 1700 : P2 = F1 (

)

*

(

) (

P2 = 1700(

+

) )

*

(

) (

)

+

P2 = 10125,11 Mencari P3 (pemasukan), dengan F2 = 1300, menggunakan fungsi excel : F(x) = PV(i%;n;A;F) F(x) = PV(8%;6;0;1300) P3 = 819,22

Maka, P = P1 + P2 + P3 = -8000 + (-10125,11) + 819,22 = $17305,89

7. If Laurel can make an investment in a friend’s business of $3000 now in order to receive $5000 five years from now, determine the rate of return. If Laurel can receive 7% per year interest on a certificate of deposit, which investment should be made? Answer : a. P = 3000, F = 5000, n = 5 tahun, i ? Menggunakan rumus : F = P(1+i)n 5000 = 3000(1+i)5 1,67 = (1+i)5 i = 0,10733 Menggunakan excel : F(x) = RATE (i%;A;P;F) F(x) = RATE (7%;0;3000;5000) i = 0,10733 =10,733%

b. i = 7,3% P? Menggunakan excel : F(x) = PV(i%;n;A;F) F(x) = PV(7,3%;5;0;5000)

P = $4207,66

8. An engineering technology group just purchased new CAD software for $5,000 now and annual payment of $500 per year for 6 years starting 3 years from now for annual upgrades. What is the present worth of the payments if the interest rate is 8% per year? Answer : P1 = 5000, i = 8% Mencari P2, dengan A = 500 dan n = 6 (starting 3 years from now) Fungsi excel : F(x) = PV(i%;n;A;F) F(x) = PV(8%;6;500;0) P2 = $2311,44 Kemudian, mencari P3 dengan n = 2 dan F = 2311,44 F(x)= PV(i%;n;A;F) F(x) = PV(8%;2;0;2311,44) P3 = $1981,68

Maka, P = P1 + P3 = 5000 + 1981,68 = $6981,68

9. Recalibration of sensitive measuring devices costs $8,000 per year. If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent uniform series at 16% per year. Answer : Mencari P1, dengan A = 8000 dan n = 6 Menggunakan rumus excel : F(x) = PV(i%;n;A;F) F(x) = PV(16%;6;8000;0) P1 = $29477,89

Mencari P, dengan n = 2 dan F = 29477,89 F(x) = PV(i%;n;A;F) F(x) = PV(16%;2;0;29477,89) P = $21906,87

Mencari A pada tahun ke-8 F(x) = PMT(i%;n;P;F) F(x) = PMT(16%;8;21906,87;0) A8 = $5043,49

10. An engineering company in Wyoming that owns 50 hectares of valuable land has decided to lease the mineral rights to a mining company. The primary objective is to obtain long term income to finance ongoing projects 6 and 16 years from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year for 20 years beginning 3 years from now, plus $10,000 six years from now and $15,000 sixteen years from now. Determine the equivalent values listed below at 16% per year: a. Total present worth in year 0. b. Future worth in year 22. c. Annual series over all 22 years. d. Annual series over the first 10 years. e. Annual series over the last 12 years. Answer : a. P di tahun ke 0? Mencari P1, i = 16%, n = 6, F1 = 10000 F(x) = PV(i%;n;A;F) F(x) = PV(16%;6;0;10000) P1 = $4104 Mencari P2, n = 16, F2 = 15000 F(x) = PV(i%;n;A;F) F(x) = PV(16%;6;0;15000) P2 = $4104 Mencari P3, n = 20, A = 20000 F(x) = PV(i%;n;A;F) F(x) = PV(16%;6;0;15000) P3 = $118577 Mencari P3, n = 2, F = 118577 F(x) = PV(i%;n;A;F) F(x) = PV(16%;2;0;118577) P3 = $88122

Maka, Ptot = P1 + P2 + P4 = 4104 + 1396 + 118577 = $93621,92

b. F di tahun ke 22? P = 93621,92, n = 22 F(x) = FV(i%;n;A;P) F(x) = FV(16%;22;0;93621,92) F = $2451620,92

c. A untuk 22 tahun ? F(x) = PMT(i%;n;P;F) F(x) = PMT(16%;22;2451620,92;0) A = $15574,25

d. A untuk 10 tahun pertama ? F(x) = PMT(i%;n;P;F) F(x) = PMT(16%;10;2451620,92;0) A = $19370,48

e. A untuk 12 tahun terakhir ? n = 12, F = 2451620 F(x) = PMT(i%;n;P;F) F(x) = PMT(16%;12;0;2451620)

11. Gerri, an engineer at Fujitsu, Inc., has tracked the average inspection cost on a robotics manufacturing line for 8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have increased consistently by $50 per unit for each of the last 4 years. Gerri plans to analyze the gradient increase using the P/G factor. Where is the present worth located for gradient? What is the general relation used to calculate total present worth in year 0? Answer : a. Untuk menghitung gradient, maka cari P di tahun ke-3 b. Untuk menghitung P di 0, maka P di tahun ke-3 = nilai F P = PG(

(

)

(

) (

)

) ; PG = 50 (P/G,i,G)

12. Chemical engineers at a Coleman Industries plants in the Midwest have determined that a small amount of a newly available chemical additive will increase the water repellency of Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive through a 5-year contract at $7,000 per year, starting 1 year from now. He expects the annual price to increase by 12% per year thereafter for the next 8 years. Additionally, an initial investment of $35,000 was made now to prepare a site suitable for the contractor to deliver the additive. Use i = 15% to determine the equivalent total present worth for all cash flows. Answer : A = 7000, n = 5 years, starting 1 year from now g = 12%, n = 8 years, P0 = 35000, i = 15%, P? g’ =

=

= 0,026786

Mencari P1, dengan A = 7000 dan n = 5 F(x) = PV(i%;n;A;F) F(x) = PV(15%;5;7000;0) P1 = $23465,09 Mencari P2, dengan g’ = 0,026786, F1 = 7000, n = 8 P2 = F1 (

)

*

(

P2 = 7000(

) (

) )

*

+

(

) (

)

+

P2 = $44474,09

Maka, P = P0 + P1 + P2 = 35000 + 23465,09 + 44474,09 = $102939,09

13. Assume that you are planning to invest money at 7% per year as shown by the increasing gradient of the following figure. Further, you expect to withdraw according to the decreasing gradient shown. Find the net present worth and equivalent annual series for the entire cash flow sequence and interpret the results.

i = 7% Periode 1 : A1 = 2000 A = A1 + A2 Mencari A2, dengan G = 500 dan n = 5 A2 =

*

(

*

A2 =

+

) (

+

)

A2 = 5860,39 Sehingga, A = 2000 + 5860,39 = $7860,39 Kemudian menghitung P-nya F(x) = PV(i%;n;A;F) F(x) = PV(7%;5;7860,39;0) P = $32229,16

Periode 2 : Karena diagram cashflow nya menurun maka, A = A1-A2 A1 = 5000 Mencari A2, dengan G = 1000 dan n = 5 n(dari tahun ke-6 s/d 10) A2 = A2 =

*

(

*

+

) (

)

+

A2 = 11720,78 Sehingga, A = A1 - A2 = 5000 + 11720,78 = $-6720,78 Kemudian menghitung P-nya (P1) F(x) = PV(i%;n;A;F) F(x) = PV(7%;5;-6720,78;0) P1 = $27556,54 P dari P1 dengan n = 5 dan F = P1  P2 F(x) = PV(i%;n;A;F) F(x) = PV(7%;5;0;27556,54) P2 = $19647,43

Mencari P3, dengan A = 1000 dan n = 2 (tahun 10 s/d 12) F(x) = PV(i%;n;A;F)

F(x) = PV(7%;2;1000;0) P3 = $1808,02 Mencari P di tahun ke 0 untuk P3 dengan n = 10 dan F =1808,02  P4 F(x) = PV(i%;n;A;F) F(x) = PV(7%;10;0;1808,02) P4 = $919,1

Pperiode2 = P2 + P4 = 19647,43 + 919,1 = 20566,53

Keseluruhan : P = Pperiode1 + Pperiode2 = 32229,16 + 20566,53 = $52795,69 Mencari A dengan n = 12 F(x) = PMT (i%;n;P;F) F(x) = PMT (7%;12;52795,69;0) A = $6647,08