Financial Analysis

March 23, 2018 | Author: MF Yousuf | Category: Internal Rate Of Return, Interest, Investing, Loans, Money



Accounting & Financial managementSession – Financial Analysis Prof. Dr. R.A Khan Introduction • Financial Analysis deals with the problems of investing money or capital. • Money has nominal value and real value • Nominal value: Legal value assigned to a particular unit. • Real value: It is defined as the purchasing power and with the passage of time real value reduce. • The Time therefore is an important element in investment decision Interest Factors for Discrete Compounding, Discrete Cash Flow • Single Payment compound amount (F/P, i%, N) = (1+i)n • Single payment present worth (Discount) (P/F, i%, N) = 1/(1+i)n • Sinking fund (A/F, i%, N) = i / [(1+i)n – 1] • Uniform series compounded amount (F/A, i%, N) = [(1+i)n – 1] / i Interest Factors for Discrete Compounding, Discrete Cash Flow • Annuity factor (P/A, i%, N) = [(1+i)n-1] / [i(1+i)n] • Capital recovery factor (A/P, i%, N) = [i(1+i)n] / [(1+i)n -1] • Uniform gradient P.W factor (P/G,i%, N) = (1/i) [ {(1+i)n -1/i(1+i)n} – {n/(1+i)n}] • Uniform gradient annual series (A/G, i%, N) = (1/i)- [n / (1+i)n – 1] Criteria of Financial Analysis 1. Net Present Value 2. Internal Rate of Return 3. Benefit Cost Ratio . 1 • A person buys a small piece of land for $5000 down and annual payment of $500 for next 6 years from now.Example . What is the present worth of the investment if the interest rate is 8% per year? . What is the present worth of the investment if the interest rate is 8% per year? .Example .2 • A person buys a small piece of land for $5000 down and deferred annual payments of $500 for 6 years starting 3 years from now. 8%.Solution • PW = 5000 + 500 (P/A. 8%.2) = $6981. 6) (P/F.6 . They estimate that the cost of computerizing such level crossing is expected to be Rs. If the life of the system is 15 years and rate of interest is 8% per year.Example • Pakistan Railway is considering a program of computerizing level crossings by installing automatic barriers. . determine the net present value of the project in zero year. resulting in cost saving 0.5 millions.4 million per year. If some improvements could be made in the factory to make it more suited for larger production.Exercise • Mixer machines are manufactured in a factory. the manufacturing cost of machine is $ 2000 from 1 to 10 months. these improvements would reduce the manufacturing cost of mixer machine to $1500 for first 10 months. $ 1600 in moth 11th and $1800 in 12th month. 1400 in 11th month and $1650 in 12th month. if market rate of interest is 2% per month . • Calculate NPV of the project in year zero. if no improvements are made in the factory. The improvement cost is expected to be $ 2500 in year zero. However extra maintenance cost on the automatic system is expected to be $450/ year. They estimate that they need 45 valves and timers at a cost of $85 per set. spend 25% of their time in watering. .Example • A University is considering installing electric valves with automatic timers on some of their sprinkler systems. The present cost of water is $2200 / year. The initial installation cost is expected to be $2000. Assume interest rate is 16% per year. each of whom earns $12000 per year. If the automatic system is installed the manpower cost for watering could be reduced by 80% and water charges by 35%. If the automatic system is expected to last for 8 years. which system should be used on the basis of present worth analysis. At the present time there are four employees who are incharg of maintaining lawns. (b)Use EUAC analysis Description Machine. using an interest rate of 15% per annum.A Machine -B First cost $ 11000 18000 Annual O&M cost $ 3500 3100 Salvage value $ 1000 2000 6 9 Life.Exercise • A plant superintendent is trying to decide between two excavating machines with the estimates presented below. years . (a)Determine which machine should be selected on the basis of PWC analysis. Exercise • An investment company is considering building a 25 unit’s apartment complex in a growing town.000 • Building investment cost $ 225. it is felt that the company could average 90% of full occupancy for the complex each year. what is the minimum monthly rent that should be charged if a 12% MARR/ year is desired? Use the annual worth method. • Land investment $ 50.000 • Study period 25years • Upkeep expense per month $ 35 • Property tax and insurance 10% of total initial investment . If the following items are reasonably accurate estimates. Because of the long term growth potential of the town. if i = 10% per year. Description Type A Type B First cost ($) 2500 3500 Annual operating cost ($) 900 700 Salvage value 200 350 Life years 5 5 .PW of alternative evaluation • Make a present worth comparison of the equal service machines for which the costs are shown in table below. Gradient Factor • Uniform increase or decrease in the amount is known as gradient amount. It is designated by letter “ G”. • Gradient factor is used for converting gradient amount into present worth or in annual worth . % Gradient • Amount increase or decrease by constant %.1} / ( r – i)} • Where r = % change in amount i = rate of interest n = Total period of change . • Amount {(1+ r / 1+ i)n . Example • Calculate equivalent present worth of $35000 now and annual series of $7000 per year for 5 years beginning 1 year from now. which starts to increase annually at 12% thereafter for the next 8 years. . Use interest rate 15% per year. 15)](P/F.15%. 4) + [7000(1.Solution • PW = 35000 + 7000 (P/A.0.15%.12/1.12 .4) • PW = $83232 .15)9 – 1/ (0. Capitalized Cost • Capitalized cost refer to the present worth value of a project that is assumed last forever or permanent or perpetual life project. . 4.Capitalized Cost Calculation 1. 3. Cash-flow diagram Find PW of nonrecurring amounts Find EUAW of all recurring amounts EUAW / i% to get the capitalized cost Add value obtained in step 2 to the value obtained in step 4 . 5. 2. .000 and an additional investment cost of $ 50.000 after 10 years. In addition there is expected to be major rework cost of $15000 every 13 years. The annual operating cost will be $5000 for the first four years and $8000 thereafter.Example • Calculate the capitalized cost of a project that has an initial cost of $150. Assume interest rate is 5% per year. . Calculate the capitalized cost of the park if the interest rate is 6% per year. CDGK expect to receive $11000 in income the first year.after that it will remain same.Example • CDGK has estimated the first cost of new amusement park to be $ 40. They expect to improve the park by adding new rides every year for the next 5 years at a cost of $6000 per year.000. 14000 the second and amounts increasing by $3000 per year until year 8 after which the income will remain constant. Annual operating cost are expected to be $12000the first year: These will increase by $2000per year until year 5. This cut the heating bill by $25/ month and the air conditioning cost by $20/ month. Assuming that the winter season is the first 6 months of the year and the summer season is the next 6 months. what was the equivalent amount of his savings after the first 3 years at an interest rate of 1% per month .Exercise • A businessman purchased a building and insulated the ceiling with 6 inches of foam. the company decided to purchase an additional unit for the machine which would make it fully automatic. what was its EUAC at interest rate of 9% per year. If the company used the machine for 13 years with no salvage value. The additional unit had a first cost of $9100. The cost for operating the machine in fully automatic condition was $1200/ yr. After 4 years from the initial purchase. Its annual maintenance and operation cost was $ 2700. .Exercise • A company purchased a machine for $18000. Internal Rate of Return • IRR is the rate of interest that equate present worth of benefit to present worth of cost. • The critical value of interest rate at which NPV = 0 . Trial & Error Method for ROR • IRR = A% + [a / (a-b)] (B% .A%) where • A% : low interest rate • B% : High interest rate • a : Positive NPV • b : Negative NPV . .Example • If $5000 is invested now in common stock that is expected to yield $100 per year for 10 years and $7000 at the end of10 years what is the rate of return. Solution • • • • Assume i= 5% NPV at 5% rate of interest = $69.46/ (69.16% .19 IRR = 5% + [ 69.5%) IRR = 5.46 +355.46 Assume i= 6% NPV at 6% rate of interest = -$355.19)](6% . it is said to be an “amortized” loan.Amortized Loan • If a loan is to be repaid in equal periodic amounts. . The loan is to be repaid in three equal payment at the end of each year.Example • A firm borrows $10000. from a bank at 6% per year rate of interest on balance amount that is outstanding at the beginning of each year. Determine the amount of installment and develop loan amortization table. . 6%. Interest Paid Balance Amount Amount Amount Principal Amount 1 $10000 $3741.6 3529.9 2 6850.3 3 3529.1 Amortization Table Year Beginning Inst.1 $600 $3141.3 0.1 211.3 3741.5 3329.1 $6850.8 3529.9 3741. 3) = $ 3741.00 .1 411.Solution • Installment Amount: 10000 (A/ P. assume interest rate is 10% per year.Exercise • Set up an amortizing schedule for a $25000 loan to be repaid in equal installments at the end of each of the next five years. . • B/C = AWB / AWC > 1 Desirable Modified Approach: • B/C = [AB – ADB – O&M] / CR .Benefit Cost Ratio • It is the ratio of equivalent worth of benefit to equivalent worth of cost. 000 per year Using an interest rate 7% to determine B/C of the dam.000. As a result of the dam flood damage will be reduced by an average of $120.Example • A small flood control dam is expected to have an initial cost of $2. In addition minor reconstruction will be required every five years at a cost of 100.8 million and an annual upkeep cost of $20000. (b) Assume the dam will be permanent. Assume the dam is expected to last 25 years. . 000 Annual maintenance Rs.100. years 20 25 .Example • Two independent sites are under consideration for a business.60000 Salvage value Rs.25000 Rs.30000 Life. Which site should be selected on the basis of B/C analysis using an interest rate 12 % per year.800 Rs.000 Rs.150.1000 Annual Income Rs. Assume that the financial details are as follows Initial Cost Site “A” (000) Site “B” (000) Rs. 20000 Rs. 1.Mutually Exclusive Alternatives • The steps for the incremental B/C analysis are summarized below. Alternatives arrange in ascending order according to their annual cost 2. Choose that alternative which requires heavy cost for which funds are available and for which incremental B/C is justified . Compare lower cost alternative to higher cost alternative only if that lower cost alternative is justified 3. Location A (000) B (000) C (000) D (000) Buildg Cost ($) 200 275 190 350 Annual cash-inflow 22 35 19. If the MARR is 10%. years 30 30 30 30 .5 42 Life.Example • Four different building locations have been suggested of which only one will be selected. use B/C analysis to select the most economically best location. Cost and annual cash –flow information are given in table below. $ (miln) A.Exercise The Corps of Engineer intends to construct an earthen dam on the river Indus. The construction cost and annual benefits are tabulated below. site Cost. Six different sites are suggested. If a MARR of 6% per year for public projects is used and dam life is long enough to be considered infinite for analysis purposes. select the best location from an economic perspective. Inc ($) A 6 350000 B 8 420000 C 3 125000 D 10 400000 E 5 350000 F 11 700000 . and the environmental impact have been approved. it would reduce the travel time and distance for local commuters.Example • Two road routes are under consideration for a new inter state highway segment. The northerly route N would be located about 5 km from central business district and would require longer travel distance and time by local commuter traffic.. . Assume that the costs for the routes are as follows…. and although its construction cost would be higher. The southerly route S would pass directly through the down town area . 000 55 200 • If the roads are assumed to last 30 years with no salvage value.000 35 450 15. Initial cost Maintenance cost per year Road user cost per year Route N.. which route should be accepted on the basis of B/C analysis using an interest rate of 5% per year? .Cont. $(000) 10. $(000) RouteS.
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