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程序代写案例-ENGR 265

时间：2021-03-25

1

SIE/ENGR 265 – HW 4 Solution

1. The annual equivalent amount of the bi-annual payment is

$11,000 (A/F, 3%, 2) = $11,000 (0.4926) = $5,418.60

The capitalized worth is

CW = A = $5,418 60 = $180,620

i 0 03

So, $180,620 must be deposited now into an account earning 3% per year such that $11,000 can be paid

out every two years indefinitely (starting two years from now).

2. The PW of the incremental investment is:

(8%) = −$400 + (

15,000

19

−

15,000

24ℎ

) (

$3.50

) (/, 8%, 10)

= −$400 + (164.5 gal/yr) ($3.50/gal) (6.7101)

= $3,463

This is a very attractive investment in the Ford truck ($3,463 > 0).

3. The bond pays (0.05)($5,000) = $250 once per year. The yield, or effective annual interest rate, can be

found as follows:

0 = −$5,500 + $250 (P/A, i’%, 10) + $5,000 (P/F, i’%, 10)

i’% = RATE(NPER,PMT,PV,FV) = RATE(10,250,-5500,5000) = 3.78%

$5,000

A = $250 / year

Buyer’s

Viewpoint: 0 1 2 3 4 5 6 7 8 9 10

End of year

2

4. P = $150 (P/A, 2%/qtr., 60 qtr.) + $10,000 (P/F, 2%/qtr., 60 qtr.)

P = $150(34.7609) + $10,000(0.3048)

P = $5,214 + $3,048

P = $8,262

5. FW(15%) = −$10,000 (F/P,15%,5) + ($8,000 − $4000)(F/A,15%,5) − $1,000

= −$10,000 (2.0114) + $4000(6.7424) − $1,000

= $5,855.60

Since FW(15%) ≥ 0, the project is acceptable.

6. Let X = annual savings required.

AW(20%) = 0 = X – [$3,000,000(A/P, 20%, 7) − $300,000(A/F, 20%, 7)]

X = $808,980 / year

Net savings per pallet = ($808,980/year) / (1,000,000 pallets/year) = $0.81 per pallet

$255

7.

$3,000

$3,000 = $255 (P/A, i′, 15)

i′ = RATE(NPER,PMT,PV) = RATE(15,255,-3000) = 3.2% per month

r = 12 × 3.2% = 38.4% compounded monthly (APR)

ieff = (1 + ⁄ )

− 1 = (1 +

0.384

12

)

12

– 1 = 0.459 or 45.9% per year

0

1 2 3 14 15

3

$375

8.

$350

The CFD is from the lender’s viewpoint. The net cash flow will have a $350 outflow at end of month 0,

uniform series of inflows of $25 from at end of months 1-11, and finally a $375 outflow at end of month

12. Using IRR function, we can then find the interest rate as shown in the table below:

EOM

Net Cash

Flow

0 ($350)

1 $25

2 $25

3 $25

4 $25

5 $25

6 $25

7 $25

8 $25

9 $25

10 $25

11 $25

12 $375

IRR 7.14%

The monthly interest rate (rmonthly) = 7.14%;

The nominal annual rate (r) = 12 × 7.14% = 85.68%;

The effective annual interest rate is

ieff = (1 + ⁄ )

− 1 = (1 +

0.8568

12

)

12

– 1 = (1.0714)12 – 1 = 1.288 (128.8%).

Jess’s wife is correct in her worry.

0 1 2 3 11 12

学霸联盟

SIE/ENGR 265 – HW 4 Solution

1. The annual equivalent amount of the bi-annual payment is

$11,000 (A/F, 3%, 2) = $11,000 (0.4926) = $5,418.60

The capitalized worth is

CW = A = $5,418 60 = $180,620

i 0 03

So, $180,620 must be deposited now into an account earning 3% per year such that $11,000 can be paid

out every two years indefinitely (starting two years from now).

2. The PW of the incremental investment is:

(8%) = −$400 + (

15,000

19

−

15,000

24ℎ

) (

$3.50

) (/, 8%, 10)

= −$400 + (164.5 gal/yr) ($3.50/gal) (6.7101)

= $3,463

This is a very attractive investment in the Ford truck ($3,463 > 0).

3. The bond pays (0.05)($5,000) = $250 once per year. The yield, or effective annual interest rate, can be

found as follows:

0 = −$5,500 + $250 (P/A, i’%, 10) + $5,000 (P/F, i’%, 10)

i’% = RATE(NPER,PMT,PV,FV) = RATE(10,250,-5500,5000) = 3.78%

$5,000

A = $250 / year

Buyer’s

Viewpoint: 0 1 2 3 4 5 6 7 8 9 10

End of year

2

4. P = $150 (P/A, 2%/qtr., 60 qtr.) + $10,000 (P/F, 2%/qtr., 60 qtr.)

P = $150(34.7609) + $10,000(0.3048)

P = $5,214 + $3,048

P = $8,262

5. FW(15%) = −$10,000 (F/P,15%,5) + ($8,000 − $4000)(F/A,15%,5) − $1,000

= −$10,000 (2.0114) + $4000(6.7424) − $1,000

= $5,855.60

Since FW(15%) ≥ 0, the project is acceptable.

6. Let X = annual savings required.

AW(20%) = 0 = X – [$3,000,000(A/P, 20%, 7) − $300,000(A/F, 20%, 7)]

X = $808,980 / year

Net savings per pallet = ($808,980/year) / (1,000,000 pallets/year) = $0.81 per pallet

$255

7.

$3,000

$3,000 = $255 (P/A, i′, 15)

i′ = RATE(NPER,PMT,PV) = RATE(15,255,-3000) = 3.2% per month

r = 12 × 3.2% = 38.4% compounded monthly (APR)

ieff = (1 + ⁄ )

− 1 = (1 +

0.384

12

)

12

– 1 = 0.459 or 45.9% per year

0

1 2 3 14 15

3

$375

8.

$350

The CFD is from the lender’s viewpoint. The net cash flow will have a $350 outflow at end of month 0,

uniform series of inflows of $25 from at end of months 1-11, and finally a $375 outflow at end of month

12. Using IRR function, we can then find the interest rate as shown in the table below:

EOM

Net Cash

Flow

0 ($350)

1 $25

2 $25

3 $25

4 $25

5 $25

6 $25

7 $25

8 $25

9 $25

10 $25

11 $25

12 $375

IRR 7.14%

The monthly interest rate (rmonthly) = 7.14%;

The nominal annual rate (r) = 12 × 7.14% = 85.68%;

The effective annual interest rate is

ieff = (1 + ⁄ )

− 1 = (1 +

0.8568

12

)

12

– 1 = (1.0714)12 – 1 = 1.288 (128.8%).

Jess’s wife is correct in her worry.

0 1 2 3 11 12

学霸联盟