Economics Chapter 5 From the annual cash flow and balance table

Chapter 5: Present-Worth Analysis

Identifying Cash Inflows and Outflows

5.1

Revenue =

$45 (0.40)(5,000) 5 $450, 000××=

(No. of engineer:5)

n

Inflow

Outflow

Net flow

0

$500,000

-$500,000

5.2

(a) Cash inflows: (1) savings in labor, $45,000 per year, (2) salvage value,

$3,000 at year 5.

(b) Cash outflows: (1) capital expenditure = $30,000 at year 0, (2) operating

costs = $5,000 per year.

(c) Estimating project cash flows:

n

Inflow

Outflow

Net flow

0

$30,000

-$30,000

1

$45,000

5,000

40,000

2

45,000

3

5,000

40,000

4

45,000

5

5,000

43,000

Payback Period

5.3

(a) Conventional payback period: $100,000/$25,000 = 4 years

(b) Discounted payback period at i = 15%:

n

Net Cash Flow

Cost of Funds (15%)

Cumulative Cash

Flow

0

-$100,000

1

$25,000

-$100,000(0.15) =-$15,000

-$90,000

2

7

1

$450,000

175,000

2

175,000

8

175,000

5.4

(a) Conventional payback period: 0.75 years

5.5

•Compute the required annual net cash profit to pay off the

investment and interest.

5.6

Project

(a) Payback

Period

(b) Discounted Payback

Period

A

3.40

3.70

3.18

3.66

NPW Criterion

5.7

PW(12%) $100,000 $30,000( / ,12%,5) $8,143.3PA=−+ =

5.8

-$30,000

5.9

(a)

(b) Draw the graphs using Excel or other software (not provided).

5.10

(a)

(b)

PW(20%) $250,000 $40,000( / ,20%,35) $50,000( /,20%,35)

$50, 254

PA PF=−+ +

=−

(c)

5.11

Given: Estimated remaining service life = 25 years, current rental income =

$250,000 per year,

O&M costs = $85,000

for the first year increasing by

$5,000 thereafter, salvage value

=$50,000,

and

MARR 12%=

.

5.12

5.13

PW(15%) $500,000 2,200,00( / ,15%,1) $2,700,000( / ,15%,8)

$1,500,000( / ,15%,8)

PF PA

PF

=− − +

+

= $10,193,077

5.14

Future Worth and Project Balance

5.16

5.17

(a)

FW(15%) $12,500( / ,15%,3) $6,400( / ,15%,2)

$14,400( / ,15%,1) $7,200

$13,213

FW(15%) $12,500( / ,15%,3) $3,000( / ,15%,2)

$23,000( / ,15%,1) $13,000

$16,472

=−+

++

=

=− −

++

=

A

B

FP FP

FP

FP FP

FP

 All projects are acceptable.

(b)

Sample calculation for Project A:

5.18

(a)

i

a

=12.68% : $5,000(F/P,12.68%,20) =$54,438.58

5.19

(a)

(b) The original cash flows of the project are as follows.

n

A

Project Balance

0

-$10,000

2

(c) No, the project is not acceptable.

5.20

(a)

•Without software updates,

(b)

Note that without software updates, you expect $2,457,281. So, this is

the minimum price that you may be willing to accept.

5.21

(a) First find the interest rate that is used in calculating the project balances.

We can set up the following project balance equations:

11

PB( ) $10,000(1 ) $11,000

iiA

=−+ + =−

n

A

PB( )n

i

0

-$10,000

2

3

5

(b) Conventional payback period

=3

years ( or 2.5 years with continuous cash

flows)

5.22

(a) From the project balance diagram, note that

1

PW(24%) 0=

for project 1

and

2

PW(23%) 0=

for project 2.

5.23

(a) The original cash flows of the project are as follows

n

A

Project Balance

0

-$3,000

1

2

-$1,500

3

4

5

5.24

(a) No graphs are given.

Period

(n)

Project Balance, PB(i)n

A

B

C

D

0

-$2,500.00

-$3,500.00

-$2,800.00

-$2,300.00

1

-2,750.00

-2,250.00

-4,880.00

-3,530.00

2

3

4

(b)

Project

A

B

C

D

5.25

(a)

Period

(n)

Project Balance, PB(i)n

A

B

C

D

0

-$1,000.00

1

2

3

(b) Discounted payback period:

(c) & (d)

Period

(n)

Area of Negative Project Balance, PB(i)n

A

B

C

D

0

-$1,000.00

1

2

5.26

n

Cash flow

Project Balance

0

-$152,000

1

$42,400

-$132,400

2

$53,400

3

4

$62,500

5

$33,000

At the end of year 3, you still need to recover $61,189 from your previous

investment. And you expect two more cash inflows from the project if you

5.27

From the annual cash flow and balance table, shown below, it will take 6.54

years approximately.

n

Cash flow

Balance

0

-$20,000.00

1

$5,000.00

-$18,000.00

2

$5,000.00

-$15,700.00

3

$5,000.00

-$13,055.00

4

$5,000.00

-$10,013.25

5

$5,000.00

6

$5,000.00

7

$5,000.00

$2,133.60

8

$5,000.00

$7,453.64

9

$5,000.00

5.28

! The present worth of the expenses already spent on the project:

expense

PW(20%) $10M( / , 20%,5) $74.416M=−=−FA

5.29

A

$800( / ,10%,3) $150( / ,10%,2)

FW( ) ( / ,15%,3)

$150( / ,10%,1) $350

FP FP

iFP

FP

−+

⎛⎞

=⎜⎟

++

⎝⎠

5.30

(a) Since a project’s terminal project balance is equivalent to its future worth,

we can easily find the equivalent present worth for each project by

$401.88

=−

(b)

00

PB(10%) = =-$1,000

A

(c)

5

FW(20%) PB(20%) $1, 000==−

(Note that the terminal project balance is the future worth of the project.)

5.31

(a)

A

B

C

PW(0%) = 0

PW(18%) = $575( / ,18%,5)=$251.34

PW(12%) = 0

PF

(c) The net cash flows for each project are as follows:

Net Cash Flow

n

A

B

C

0

-$1,000

2

5

(d)

Capitalized Equivalent Worth

5.32

(a)

PW(5%) $1, 000,000( / ,5%,5) $1,300,000( / ,5%,5)( / ,5%,5)

PA PA PF

=+

5.33

! Capitalized equivalent amount:

5.34

! Find the equivalent annual series for the first cycle:

! Capitalized equivalent amount:

5.35

Given: r =5% compounded monthly, maintenance cost = $80,000 per year

5.36

Given: Construction cost = $15,000,000, renovation cost = $3,000,000

every 15 years, annual

&OM

costs = $1,000,000 and

5%i=

per year

(a)

(b)

(c)

•15-year cycle with 10% of interest:

•20-year cycle with 10% of interest:

5.37

Given: Cost to design and build

=$830,000

, rework cost

=

$120,000

every 10 years, new type of gear

=$80,000

at the end of 5th year, annual

operating costs

=$70,000

for the first 15 years and

$100,000

thereafter