Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 1

UNIT - 2

ENERGY ECONOMIC ANALYSIS: The time value of money concept, developing cash flow models,

payback analysis, depreciation, taxes and tax credit numerical problems. 7 Hours

Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 2

Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 3

Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 4

Types of interest: Simple interest:

Total interest, I = (P x n x R) / 100 = P x n x r

P principal amount

n number of periods

r rate of interest

Compound interest:

Amount payable at the end of 1st year, F1 = P (1+r)

Amount payable at the end of 2nd year, F2 = [P (1+r)] (1+r) = P (1+r)2

Amount payable at the end of 3rd year, F3 = [P (1+r)2] (1+r) = P (1+r)3 so on

Amount payable (or accumulated) at the end of nth year, Fn = P (1+r)n

Effective interest:

e = (1 +

)

1

re = effective rate of interest

r = nominal rate of interest

c = number of compounding in a year

Amount payable (accumulated) at the end of nth year, F = P (1+re)n = P (1 +

)

True or continuous compounding:

re = er -1 and F = P (.) ; e = 2.71828

CASH FLOW MODELS

SPCA (Single payment compound amount) F = P x (SPCA) SPCA = (1 + )

SPPW (Single payment present worth) P = F x (SPPW) SPPW = 1

(1+)

USCA (Uniform series compound amount) F = D x (USCA) USCA =

(1+)1

SFP (Sinking fund payment) D = F x (SFP) SFP =

(1+)1

USPW (Uniform series present worth) P = D x (USPW) USPW =

(1+)1

(1+)

CR (Capital recovery) D = P x (CR) CR =

(1+)

(1+)1

GPW (Gradient present worth) P = D x (GPW)

GPW =

1+

1+[1(

1+

1+)

]

11+

1+

Where P = Present worth, F = Future worth, D = Uniform series deposits, r = rate of interest, e = escalation rate

Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 5

Methods of depreciation:

1. Straight line method:

Depreciation amount, D = ()

2. Sum of years digits method:

Sum of years digits, =(+1)

2

Where n = life of equipment (number of years)

Depreciation amount for 1st year, 1 =

( )

Depreciation amount for 2nd year, 2 =1

( )

Depreciation amount for 3rd year, 3 =2

( ) so on..

Depreciation amount for nth year, =1

( )

3. Declining balance method: (Reducing balance method or Diminishing balance method)

Annual depreciation rate, = 1 (

)

1

= 1

Book value at the end of nth year, BVn = P (1-d)n

4. Sinking fund method:

Total amount to be deposited, (P - L) = (1+)1

Annual amount to be deposited, D = ( )

(1+)1

IMPORTANT QUESTIONS:

1. Write a short note on time value of money concept. (05M Dec. 2011, 05M Jan. 2010)

2. Write a brief note on

a. Pay back analysis

b. Causes of depreciation. (06M - June/July 2011)

3. What is time value of money concept? What are the different cash flow models? (08M Dec. 2010)

4. Explain pay back analysis. (04M Dec. 2010)

5. Explain pay back analysis. Mention its advantages and disadvantages. (05M - June/July 2014, 06M

Jan. 2010)

6. Explain cash flow model. (10M - June/July 2014)

7. Explain the concept of time value of money. (06M - June 2012)

8. What is depreciation? Explain i) Sinking fund, ii) Declining balance methods of calculating

depreciation. (08M - June 2012)

9. What is depreciation? What are the causes of depreciation? (05M Dec. 2010)

10. Write a short note on causes of depreciation.

11. Develop cash flow model for Uniform series compound amount factor. (06M Jan. 2010)

Unit-2: ENERGY ECONOMIC ANALYSIS 10EE842

Savyasachi.G.K EEE, VVCE, Mysuru P a g e | 6

12. What is meant by life cycle cost analysis? Develop single payment compound amount cash flow

model. (06M June/July 2011)

13. What do you understand by depreciation? Explain how depreciation reserve is calculated using: a)

Sum of years digits method, b) Straight line depreciation method. (06M June/July 2011)

14. Discuss the different classification of costs of electric energy generated. (10M Dec. 2011)

15. A distribution transformer costs Rs. 2,00,00,000 and has a salvage value of Rs. 10,00,000. The life

of the transformer is 20 years. If the rate of annual compound interest is 8%, calculate the amount

to be saved annually for replacement of the transformer at the end of 20 years by sinking fund

method. (06M - June 2012) [D = Rs.415192]

16. The equipment in a power station costs Rs. 15,00,000/- and has a salvage value of Rs. 60,000/- at

the end of 25 years. Determine the depreciation value of the equipment at the end of 20 years by

the following methods:

a. Straight line method

b. Diminishing value method

c. Sinking fund method at 5% compounded annually. (10M Dec. 2011) [D=57600,

D20=1152000, d=0.12081, D20=1385779, D=30171.54, D20=997651]

17. (2.13.13 pg.no. 55) Calculate the depreciation rate using the

a. Straight line

b. Sum of years digits and

c. Declining balance methods, for the data given below:

Salvage value Rs. 0. Life of the equipment = 5 years. Initial expenditure P = Rs. 1,50,000.

For declining balance use a 200% rate. (08M Dec. 2010)

18. You have accumulated Rs. 5000 in credit card debit. The credit card company charges 18%

nominal annual interest compounded monthly. You can only afford to pay only Rs. 100 per month.

How many months will it take you to pay off debit and how much money will you have to pay as

interest? (08M Jan. 2010)

19. A motor drive consumes 40,000 units per annum. By upgrading to high efficiency spare parts the

consumption can be reduced by 5%. The additional cost of up-gradation is Rs. 35,000. Assume

energy charge of Rs. 5 per unit and life of motors 15 years. Is the charge justified? Take rate of

interest/return = 20%. Use annual cost method. (08M June/July 2011)

20. (2.13.19 pg.no. 60) A plant cost Rs. 7,56,000 and it is estimated that after 25 years it will have to

be replaced by a new one. At that instant its salvage value will be Rs. 1,56,000. Calculate: i) Annual

deposit to be made in order to replace the plant after 25 years; ii) the value of the plant after 10

years on the following basis:

a. Straight line depreciation method

b. Reducing balance method

c. Sinking fund method at 8% annual compound interest. (05M - June/July 2014)

21. Develop cash flow model for Sinking Fund Payment (SFP). [or Develop cash flow model for Uniform

series deposits of (D) with Present worth (P) whose Future worth is (F) with compound interest (r)

for (n) number of periods]

22. Develop cash flow model for Uniform series present worth. [or Develop cash flow model for Present

worth (P) whose Future worth is (F) for Uniform series deposits of (D) with compound interest (r) for

(n) number of periods].

23. Develop cash flow model for Capital Recovery (CR). [or Develop cash flow model for Uniform series

deposits of (D) with Present worth (P) whose Future worth is (F) with compound interest (r) for (n)

number of periods].