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02-1 Time Value of Money MIN E 408: Mining Enterprise Economics Review of Economi c Indicators Resources:  An y ec onomics book  A cl ass ic: Rung e, 1998. Minin g Ec onomics an d St rategy. 295 p. Slides modified from Dr. Oy Leuangthong Time Value Of Money Basis: mone y rec eived today is wor th more t han th e same amount in future. Impact of in teres t on 2 al terna te pro jects ( strea ms of c ash flo w) is used to select one of them. Inter est = money paid for t he use of someone e lse’s/ bank’s money = cost of borrowed money. Inter est rate = {int erest payable / amo unt borrowed}*1 00%

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• Time Value of Money

MIN E 408: Mining Enterprise Economics

Review of Economic Indicators

Resources: Any economics book

 A class ic: Runge, 1998. Mining Economics and Strategy. 295 p.

Slides modified from Dr. Oy Leuangthong

Time Value Of Money

• Basis: money received today is worth more than the same amount infuture.

• Impact of interest on 2 alternate projects (streams of cash flow) isused to select one of them.

• Interest = money paid for the use of someone else’s/bank’s money

= cost of borrowed money.

• Interest rate = {interest payable / amount borrowed}*100%

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Simple Interest, I

•  Annualized percentage amount borrowed paid for the use ofmoney

(A) if n=1; i=10%; P=$1000.00

(B) if n=2 years; i=10%; P=$1000.00

* * I P n i

2 $1000.00*0.1*2 $200.00 I   

1 $1000.00*0.1*1 $100.00 I  

Compound Interest

• Interest plus principal or remaining balance forms basis for subsequentyear’s interest

• Interest is applied to each period’s basis

Basis = principal + interest

• Future value of present sum

• Present value of future sum

(1 )nF P i

(1 )   nP F i

 

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Example

Estimate the value of $1,000.00 today @ 8% interest per annum at theend of 5 years.

Or the present value of this future amount is given by

469,1$)08.01(1000 5 F 

000,1$)08.01(1469 5   P

 Another ExampleOne period on ly:

• You have a mining property for sale

•  ABC Company offers you $1.3 mil lion and is ready to wire thecash into your savings account today!

• XYZ Ltd of fers you $1.5 million but will pay th is next year 

• You trust that each company wil l deliver on its promise

• Your Bank, offers you 12% interest rate on both your savingsand credit account.

•  Assume( i) You don’t really need cash

(ii) You need cash badly

What wil l you do?

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 Another Example cont’di ) You don’t real ly need cash

i i) You need cash badly

 Another Example cont’d

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Future Value Of Annualized Amount

(1 ) 1ni

F Ai

 

E.g. Find the future sum at the end of 5 years of an annualizedseries of $250.46 at the end of each year for the next 5 years at aninterest of 8% p.a.

5(1 0.08) 1250.46* $1,469

0.08F 

Thus the annualized amount is given

For previous example, the annualized amount can be calculated giventhat F = $1469:

(1 ) 1n

i A F 

i

 

Future Value Of Annualized Amount

5

0.081469* $250.46

1 0.08) 1 A

 

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Present Sum Of Annualized Amount

• Present Sum of Annualized Equivalent Series

(1 ) 1

(1 )

n

n

iP A

i i

 

(1 )

(1 ) 1

n

n

i i A P

i

 

Estimate the present sum of annualized equivalent series of $250.46 per year@ 8% interest p.a.

5

5(1.08) 1250.46 $1,0000.08(1 )

Pi

Gradient Series

Future worth of a Gradient Series

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Gradient Series

Present Sum of A Gradient Series

Relation between Uniform and Gradient Series

2

1 (1 )(1 )   nni i

P Gi

 

1

(1 ) 1n

n A G

i i

Example• The maintenance costs on a certain equipment are estimated to be

$500 in year 1 and to increase by $100/year. What is A if n=5 years.and i=8% per annum?

• Solution

5

1

(1 ) 1

( / , %, )

1 5500 100* $685

0.08 (1 0.08) 1

n

n

 A B G i i

 B G A G i n

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1 (1 ) (1 )

1

n nk i

C i k i k 

P

nC i k 

i

   

     

Geometric Series

1 (1 ) /(1 )(1 )

( )

(1 )(1 )

n n

n

n

k iC i i k  

i k F 

n iC i k 

   

 

 

Example

Find the present equivalent of this cash flow stream;

C = $10,000.00; i = 15%; k = 10%; and n = 5 years.

5 5

5 5

1 (1 ) (1 )

1 (1.10) (1.15)$10,000

0.15 0.10

$39,858

k iP C 

i k 

 

 

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Problem Solving Method - Recommended

Use the following method to solve interest problems;

1. Draw a cash flow diagram. Include magnitude, timing and direction ofcash flows (Cash inflow or income is +ve; and cash outflow or cost is

 –ve). Include interest rate in diagram.

2. Using the various notations for the factors, write an equation torepresent the answer.

3. Put the values of the factors from the tables into the equation.

4. Calculate the answer 

5. Check for accuracy.

Example A piece of machinery requires an overhaul today which will cost$1,000.00. It will require another overhaul 3 years hence which will cost$1,500.00. Annual operating costs are $300/year. Find the present-worth of these costs at 6% if the machine has a life of 7 years.

n F|P,i,n P|F,i,n F|A,i,n A|F,i,n P|A,i,n A|P,i,n A|G,i,n F|G,i,n P|G,i,n

1 1.06 0.9434 1 1 0.9434 1.06 0 0 0

2 1.1236 0.89 2.06 0.48544 1.83339 0.54544 0.48544 1 0.89

3 1.19102 0.83962 3.1836 0.31411 2.67301 0.37411 0.96118 3.06 2.56924

4 1. 26248 0. 79209 4. 37462 0. 22859 3. 46511 0. 28859 1. 42723 6. 2436 4. 94552

5 1. 33823 0. 74726 5. 63709 0. 1774 4. 21236 0. 2374 1. 88363 10. 61822 7. 93455

6 1. 41852 0. 70496 6. 97532 0. 14336 4. 91732 0. 20336 2. 3304 16. 25531 11. 45935

7 1 .50363 0.66506 8.39384 0.11914 5.58238 0.17914 2 .76758 23.23063 15 .44969

8 1 .59385 0.62741 9.89747 0.10104 6.20979 0.16104 3 .19521 31.62447 19 .84158

9 1 .68948 0 .5919 11.49132 0.08702 6.80169 0.14702 3 .61333 41.52193 24 .57677

10 1 .79085 0.55839 13.18079 0.07587 7.36009 0.13587 4 .02201 53.01325 29 .6023211 1.8983 0.52679 14.97164 0.06679 7.88687 0.12679 4 .42129 66.19404 34.8702

12 2.0122 0.49697 16.86994 0.05928 8.38384 0.11928 4 .81126 81.16569 40 .33686

13 2 .13293 0.46884 18.88214 0.05296 8.85268 0.11296 5 .19198 98.03563 45 .96293

14 2.2609 0 .4423 21.01507 0.04758 9.29498 0.10758 5 .56352 116.9178 51 .71284

15 2 .39656 0.41727 23.27597 0.04296 9.71225 0.10296 5 .92598 137.9328 57 .55455

16 2 .54035 0.39365 25.67253 0.03895 10 .1059 0.09895 6 .27943 161.2088 63 .45925

17 2.69277 0.37136 28.21288 0.03544 10.47726 0.09544 6.62397 186.8813 69.40108

18 2 .85434 0.35034 30.90565 0.03236 10 .8276 0.09236 6.9597 215.0942 75 .35692

19 3.0256 0.33051 33.75999 0.02962 11.15812 0.08962 7.28673 245.9999 81.30615

20 3 .20714 0 .3118 36.78559 0.02718 11.46992 0.08718 7 .60515 279.7599 87 .23044

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Solution (1)1. Construct cash flow diagram. Only costs (outflows) are involved so

arrows point downwards

2. Write an equation to represent it

3. Put in values from table

4. Calculate the answer 5. Check the accuracy

)7%,6,|()3%,6,|(0   AP AF PF PP  

14.934,3$P

58238.5300)83962.0(15001000   P

Problem Solving Method - Alternate

Use the followi ng method to solve interest problems;

1. Draw a cash flow diagram. Include magnitude, timing and direction ofcash flows (Cash inflow or income is +ve or ; and cash outflow or costis -ve or . Include interest rate in diagram.

2. Using the discrete interest rate formulae table, write a mathematicalequation to represent the answer.

3. Put the actual values from the diagram (question) into the equation.

4. Calculate the answer 

5. Check for accuracy.

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Solution (2)

1. Construct cash flow diagram. Since only costs (outflows) are involved, thearrows will point downward

2. Write down mathematical equation

3. Put in values from diagram

4. Calculate the answer 

5. Check the accuracy

 

  

 

 

7

73

0)1(

1)1()1(

ii

ii AiF PP

14.934,3$P

 

 

 

 

 

7

73

)06.01(06.0

1)06.01(300)06.01(15001000P

Review Of Main Points

• Cash Flow Diagrams

• Time Value of Money:

 – Present Value

 – Future Value

 – Annualized Value – constant amount

 – Gradient Series – constant amount of change

 – Geometric series – constant rate of change