Econ Lesson 4

Chapter 5 Present Worth Analysis

What we will cover We will apply the P/F/A/G factors to project analysis Ch. 5 is Present Worth Analysis Ch. 6 is Annual Cash Flow Analysis Ch. 7 is Rate of Return Analysis Ch. 8 Incremental Analysis • • • •

Chapter 5 Outline Assumptions in Solving Economic Analysis Problems

•

Economic Criteria

•

Applying Present Worth Techniques Techniques ◦ Useful Lives Equal the Analysis Period ◦ Useful Lives Different from the Analysis Period ◦ Infinite Analysis Period

•

in Solving Economic Analysis Problems End-of-Year Convention:

0

F

 A

 A

A

1

2

n-1

 A n

P •

Viewpoint Viewpo int of Economic Analysis Studies  –

Generally, the point of view from the total firm is Generally, taken 4

Economic Criteria Si t u at i o n Cr i t er i o n Neither input nor output Maximize (Output Input) fixed Fixed input Maximize output Fixed output Minimize input  –

Methods: Present worth  Annual cash flow Rate of return • • •

5

to Present Worth Techniques Situation

Criterion

Neither input nor output fixed: Typical cases

Maximize Net Present Worth (present worth of benefits minus present worth of costs)

Fixed input: amount of money or other input resources are fixed Fixed output: fixed task, benefit, or other outputs

Maximize present worth of benefits or other outputs Minimize present worth of costs or other inputs

Applying Present Worth when Useful Lives are Equal Device B

Device A

4500 5000 4000 3000 3500

 A=3000 0

1 P=10,000

2

3

4

5

0

2

1

3

4

5

P=13,500

 =  10,000 + 3000(, 7%, 5) = 10,000 + 3000   4.1 4.100 00 = $2 $230 300 0  =  13,500 + 3000(, 7%, 5) + 500(,7%,5) =  13,500 + 3000 4.1 4.100 + 500 7.6 7.647 = $2624

PWProject = PW of Benefits

 –

PW of Costs

Applying Present Worth when Useful Lives are Equal Build a full-sized facility for $400 million mil lion now, now, or build a reduced-size facility now for $300 million and expand it i t 25 years hence for an additional $350 million, mi llion, at 6% interest?

For the single-stage construction     = $4 $400 00     

For the two-stage construction      = $300 $300      + 35 350 0    ( (, 6%, 25) = $3 $300 00      + 81. 81.6 6      = $3 $381 81.6 .6     

Applying Present Worth when Useful Lives are Equal Manufacturer

Cost

Useful Life

End-of-Useful-Life Salvage Value

Speedy

$1500

5 years

$200

 Allied

$1600

5 years

$325

Speedy     = 1500 1500  200 200((,7%,5) = 1500  200 0.7130 = $1357

 Allied     = 1600 1600  325 325((,7%,5) = 1600  325 0.7130 = $1368 9

Applying Present Worth when Useful Lives are Equal  Alternatives

Cost

Uniform Annual Benefit

End-of-Useful-Life Salvage Value

 Atlas

$2000

$450

$100

Tom Thumb

$3000

$600

$700

 Atlas  = 450(, 8%, 6) + 100(, 8%, 6)  2000 = 450 4.6 .623 23 + 10 100 0 0. 0.6 6302  2000 = $143 Tom Thumb

 = 600(, 8%, 6) + 700(, 8%, 6)  3000 = 600 4.6 .623 23 + 70 700 0 0. 0.6 6302  30 300 00 = $215

Present Worth when Useful Lives are Different Diff erent from Analysis Analysis Period •

Least Common Multiple of Useful Lives from Various Alternativ Alternatives es • •

•

 Assuming service will be needed needed indefinitely Repeating same cash flows in each cycle for each alternative

 Analysis Period •

Need to estimate the terminal values for all alternatives at the end of the analysis period

Applying Presen Presentt Worth Worth using using Analysis Period  All t er  A ern n at atii v es

Alt. 1

Alt. 2

Initial Cost

$50,000

$75,000

Estimated salvage value at end of useful life

$10,000

$12,000

Useful Life

7 years

13 years

Estimated market value, end of 10-year

$20,000

$15,000

.  = 50,0 50,000 00 + (10,0 (10,000 00  50,00 50,000)( 0)(, 8%, 7) + 20,000(,8%,10) = 50,00 50,000 0  40, 40,00 000 0 0.5 0.5835 835 + 20, 20,00 000 0 0.4 0.4632 632 = $64,076 .  = 75,000 + 15,000(,8%,10) = 75,000 + 15,000 0.5835 = $68,052

Present Worth with Infinite Analysis Period (Capitalized Cost)  

  = 

Capitalized Cost How much should one set aside to pay $50 per year for maintenance on a gravesite if interest is assumed to be 4%?

 

  50 = =  = $1250  0.04

Capitalized Cost 0 $8 million 0

70

140 $8 million

$8 million 70 =

∞

$8 million

n=70  A

$8 million

  = 8 ( , 7%, 70) = $4960          

  4960  = 8     + =  = $8,071,000  0.07

Capitalized Cost (Alternate Solution 1) 0 $8 million 0

70

140 $8 million

$8 million 70 =

∞

$8 million

n=70  A

$8 million

  = 8 ( , 7%, 70) = $565,000        

  565000 = = = $8,071,00 $8,071,000 0  0.07

Capitalized Cost (Alternate (Alternat e Solution 2) 0 $8 million

70

140

$8 million

$8 million

∞

$8 million

  (1 + 7%) = 113.989    = (1         

  8       = 8     + =  = $8,071,000  113.989

Multiple Alternatives Pipe Size (in.)

2 Initial Cost

4

5

$22,000 $23,000 $25,000 $30,000

Cost of pumping ($/hr)

  "   "   "   "

3

$1.20

$0.65

$0.50

= 22, 22,000 + 1.20 × 20 2000( , 7%, 5) = = 23, 23,000 + 0.65 × 20 2000( , 7%, 5) = = 25, 25,000 + 0.50 × 20 2000( , 7%, 5) = = 30, 30,000 + 0.40 × 20 2000( , 7%, 5) =

$0.40

$3 $ 31,840 $28,330 $2 $29,100 $2 $33,280 $3

Multiple Alternatives Total Investment

Uniform Net  Annual Benefit

Terminal Value

$0

$0

$0

B: Vegetable market

$50,000

$5,100

$30,000

C: Gas Station

$95,000

$10,500

$30,000

D: Small motel

$350,000

$36,000

$150,000

 Alternatives  A: Do Nothing

 = $0   = 50,000 + 5,100( , 10% 0%,, 20) + 30 30,0 ,00 00(, 10% 0%,, 20 20)) = $2 $2,1 ,120 20   = 95 95,00 ,000 0 + 10,5 10,500( 00(, 10% 0%,, 20) + 30, 0,00 000( 0(, 10 10%, %, 20 20)) = $1 $1,1 ,140 40   = 350,000 + 36,000( , 10 10%, %, 20 20)) + 15 150, 0,00 000( 0(, 10 10%, %, 20 20)) = $2 $21, 1,21 210 0

Year 0 1-10 10

Cash Flow -$610 +200 per year   -1500

 = 6 610 + 200(, 10%, 10)  1500( , 10%, 10) = 610 + 200 6.145  1500 0.3855 = $41

Example Ex ample 5-10 5-10 Year 0 1 2-5 6 7 8

Alt. B -$1500 +700 +300 +400 +500 +600

700 300 300 0

1

2

3

300

300

4

5

400 6

500 600 7

8

1500

 = 1500 + 300(, 8%, 8) + 400(, 8%, 1) + 100(,8%,4)(,8%,4) = 15 1500 00 + 30 300 0 5. 5.74 747 7 + 40 400 0 0. 0.92 9259 59 + 10 100 0 4. 4.65 650 0 (0 (0.7 .735 350) 0) = $936.24

Example 5-11 Year 0 1 2 3 4 5-8

Alt. A -$2000 +1000 +850 +700 +550 +400

1000 850

0

1

2

700 550 3

4

400 400 5

6

400 400 7

8

2000

  = 2000 2000 + 40 400( 0(, 8%, 8) + 600(, 8%, 4)  150(,8%,4) = 200 2000 0 + 40 400 0 5. 5.74 747 7 + 60 600( 0(3. 3.31 312) 2)  15 150 0 4. 4.65 650 0 = $1588. $1588.50 50

Example 5-13 A 15-yr municipal bond was issued 5 yrs ago. Its coupon interest rate is 8%, interest interest payments are are made semi-annually, semi-annually, with with face value of $1000. What should be the bond’s price if market rate is 12.36%. 1000 40 0 1 2 3 4 5

6 7 8

9 10 11 12 13 14 15 16 17 18 19 20

( 1 + − ) = 1 + 0.1236 − = 6%  = 40  , 6%, 20 + 1000  , 6%, 20 = $770.60

Problem 5-2 1. What variables do we have? 2. What do we have a time = 0?

Problem 5-2 Identify variables Do we have an A? Do we have a P? Do we have an G? When do they start? • • • •

What are the possible solutions?