UNIT – III

UNIT IIICASH FLOW

TYPES OF CASH FLOW DIAGRAMCOST DOMINATED CASH FLOW DIAGRAM

REVENUE DOMINATED CASH FLOW DIAGRAM

COST DOMINATED CASH FLOW DIAGRAMOut flow cash is assigned with positive sign.

The profit, revenue, salvage value is assigned with negative sign.

REVENUE DOMINATED CASH FLOW DIAGRAMOut flow cash is assigned with Negative sign.

The Profit, Revenue, Salvage Value is assigned with Positive sign.

REVENUE DOMINATED CASH FLOW DIAGRAM (PRESENT WORTH METHOD)P0123.jnRnSHere P initial investment. Rj net revenue at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

PW(i) = -P + R1{1/(1+i)1) + R2{1/(1+i)2)+. + Rj {1/(1+i)j } + Rn{1/(1+i)n } + S{1/(1+i)n)

R1R2R3 .Rj

COST DOMINATED CASH FLOW DIAGRAM (PRESENT WORTH METHOD)PHere P initial investment. Cj net cost of operation and maintenance at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

PW(i) = P + C1{1/(1+i)1) + C2{1/(1+i)2)+. + Cj {1/(1+i)j } + Cn{1/(1+i)n } - S{1/(1+i)n }0123.jnS C1C2C3.CjCn

DECISION PARAMETERAlternative with minimum cost, then the alternative with the least present worth is selected.

Alternative with maximum profit, then the alternative with the maximum present worth is selected.

Alpha industry is planning to expand its production operation. It has identified three different technologies for meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in Table 4.1 suggest the best technology which is to be implemented based on the present worth method of comparison assuming 20% interest rate, compounded annually.

ALTERNATIVE Initial outlay (Rs.) Annual revenue (Rs.) Life (years)Technology 1 12,00,000 4,00,000 10Technology 2 20,00,000 6,00,000 10Technology 3 18,00,000 5,00,000 10

PW(i%) = - P + A * (P/A, i%, n )PW(20%) = -1200000 + 4,00,000 (P/A,20%,10) = -1200000 + 4,00,000 * (4.195) = Rs.4,77,000.Technology I

Initial outlay, P = Rs.12,00,000.Annual Revenue, A = Rs. 4,00,000.Interest Rate, i = 20 %Life of this technology, n = 10 years.

0123.jn4,00,0004,00,0004,00,0004,00,000 .Rj12,00,000

PW(i%) = - P + A * (P/A, i%, n )PW(20%) = -20,00,000 + 6,00,000 (P/A,20%,10) = -20,00,000 + 6,00,000 * (4.195) = Rs.5,15,500.Technology II


0123.jn6,00,0006,00,0006,00,0006,00,000 .Rj20,00,000

PW(i%) = - P + A * (P/A, i%, n )PW(20%) = - 18,00,000 + 5,00,000 (P/A,20%,10) = - 18,00,000 + 5,00,000 * (4.195) = Rs.2,96,250.Technology III


0123.jn5,00,0005,00,0005,00,0005,00,000 .Rj18,00,000

Result : From the above calculations, it is clear that the present worth of technology 2 is the highest among all the technologies. Therefore, technology 2 is suggested for implementation to expand the production.

An engineer has two bids for an elevator to be installed in a new building. The details of the bids for the elevators are as follows:

bid Engineers s estimatesInitial cost(Rs.)ServiceLife(years)Annual operations&Maintenance cost (Rs).Alpha Elevator Inc. 4,50,0001527,000Beta Elevator Inc.5,40,0001528,500

BiD 1: Alpha Elevator Inc.Initial Cost, P = Rs. 4,50,000Annual operation and maintenance cost , A = Rs 27,000Life , n = 15 yearsInterest rate , i = 15%

012315 27,00027,00027,00027000450000PW(i%) = P + A * (P/A , i%, n ) = 4,50,000 + 27,000 * (P/A, 15%, 15) = 4,50,000 + 27,000 * 5.8474 = Rs.6,07,879.80

BiD 2: Beta Elevator Inc.Initial Cost, P = Rs. 5,40,000Annual operation and maintenance cost , A = Rs 28,500Life , n = 15 yearsInterest rate , i = 15%

012315 28,50028,50028,50028,5005,40,000PW(i%) = P + A * (P/A , i%, n ) = 5,40,000 + 28,500 * (P/A, 15%, 15) = 5,40,000 + 28,500 * 5.8474 = Rs.7,60,650.90

Result :The Total present worth cost of bid 1 is less than that of bid 2.

Hence bid 1 is selected for the implementation.

REVENUE DOMINATED CASH FLOW DIAGRAM (FUTURE WORTH METHOD)P0123.jnRnSHere P initial investment. Rj net revenue at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

FW(i) = -P (1+i)n + R1(1+i)n-1 + R2(1+i)n-2+. + Rj (1+i)n-j + Rn + S

R1R2R3 .Rj

COST DOMINATED CASH FLOW DIAGRAM (FUTURE WORTH METHOD)PHere P initial investment. Cj net cost of operation and maintenance at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

PW(i) = P (1+i)n + C1 (1+i)n-1 + C2 (1+i)n-2 +. + Cj (1+i)n-j +. + Cn - S0123.jnS C1C2C3.CjCn

DECISION PARAMETERAlternatives with the maximum future worth of NET REVENUE is selected.

Alternative with the minimum future worth of NET COST is selected.

Consider the following two mutually exclusive alternatives:At I = 18%, Select the best alternative based on future worth method of comparison.

AlternativeEnd of year01234A (Rs.)-50,00,00020,00,00020,00,00020,00,00020,00,000B (Rs.)-45,00,00018,00,00018,00,00018,00,00018,00,000

ALTERNATIVE IInitial investment, P = Rs. 50,00,000Annual equivalent investment, A = Rs. 20,00,000Interest rate, i = 18%Life of the alternative n = 4 years 0123420,00,00020,00,00020,00,00020,00,000 5000000FW (i%) = -P (F/P,i%,n) + A (F/A,i%,n)= -50,00,000 (F/P,18%,4) + 20,00,000 (F/A,18%,4)= -50,00,000 (1.939) + 20,00,000 (5.215)= Rs.7,35,000.

ALTERNATIVE IIInitial investment, P = Rs. 45,00,000Annual equivalent investment, A = Rs. 18,00,000Interest rate, i = 18%Life of the alternative n = 4 years 0123418,00,00018,00,00018,00,00018,00,000 45,00,000FW (i%) = -P (F/P,i%,n) + A (F/A,i%,n)= -45,00,000 (F/P,18%,4) + 18,00,000 (F/A,18%,4)= -45,00,000 (1.939) + 18,00,000 (5.215)= Rs.6,61,500.

RESULT :The future worth of alternative A is greater thanB ,hence A is to be selected.

M/S Krishna castings Ltd. is planning to replace its annealing furnace.It has received tenders from three different original manufacturers of annealing furnace. The details are as follows.Which is the best alternative based on future worth method at i = 20%?

Manufacturer123Initial cost (Rs.)80,00,00070,00,00090,00,000Life (years)121212Annual operation and maintenance cost (Rs.)8,00,0009,00,0008,50,000Salvage value after12 years5,00,0004,00,0007,00,000

Alternative IFirst cost , P= Rs. 80,00,000Life, n = 12 yearsAnnual operations and maintenance cost, A = Rs. 8,00,000Salvage value at the end of furnace life S = Rs. 5,00,0000123.125,00,000 8,00,0008,00,0008,00,000.8,00,00080,00,000FW (i%) = P (F/P,i%,n) + A (F/A,i%,n) S= 80,00,000 ((F/P,20%,12) + 8,00,000 (F/A,20%,12) 5,00,000 = 80,00,000 (8.916) + 8,00,000 (39.581) 5,00,000= Rs. 10,24,92,800.

Alternative IIFirst cost , P= Rs. 70,00,000Life, n = 12 yearsAnnual operations and maintenance cost, A = Rs. 9,00,000Salvage value at the end of furnace life S = Rs. 4,00,0000123.124,00,000 9,00,0009,00,0009,00,000.9,00,00070,00,000FW (i%) = P (F/P,i%,n) + A (F/A,i%,n) S= 70,00,000 ((F/P,20%,12) + 9,00,000 (F/A,20%,12) 4,00,000 = 70,00,000 (8.916) + 9,00,000 (39.581) 4,00,000= Rs. 9,76,34,900.

Alternative IIIFirst cost , P= Rs. 90,00,000Life, n = 12 yearsAnnual operations and maintenance cost, A = Rs. 8,50,000Salvage value at the end of furnace life S = Rs. 7,00,0000123.127,00,000 8,50,0008,50,0008,50,000.8,50,00090,00,000FW (i%) = P (F/P,i%,n) + A (F/A,i%,n) S= 90,00,000 ((F/P,20%,12) + 8,50,000 (F/A,20%,12) 7,00,000 = 90,00,000 (8.916) + 8,50,000 (39.581) 7,00,000= Rs. 11,31,87,850.

RESULT:The future worth cost of alternative 2 is less thanthat of the other two alternative. Hence they gofor selecting the second alternatives

REVENUE DOMINATED CASH FLOW DIAGRAM (ANNUAL EQUIVALENT METHOD)P0123.jnRnSHere P initial investment. Rj net revenue at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

FW(i) = -P (1+i)n + R1(1+i)1 + R2(1+i)2+. + Rj (1+i)j + Rn(1+i)n + S/(1+i)jn

R1R2R3 .Rj

COST DOMINATED CASH FLOW DIAGRAM (FUTURE WORTH METHOD)PHere P initial investment. Cj net cost of operation and maintenance at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

PW(i) = P (1+i)n + C1 (1+i)n-1 + C2 (1+i)n-2 +. + Cj (1+i)n-j +. + Cn - S0123.jnS C1C2C3.CjCn

DECISION PARAMETERAlternatives with the maximum future worth of NET REVENUE is selected.

Alternative with the minimum future worth of NET COST is selected.

A company is planning to purchase an advanced machine centre.Three original manufacturers have responded to its tender whose particulars are tabulated as followDetermine the best alternative based on the annual equivalent method by assuming i = 20%, compounded annually.

ManufacturerDown payment(Rs.)Yearly equal installment(Rs.)No. of Installments.15,00,0002,00,0001524,00,0003,00,0001536,00,0001,50,00015

ALTERNATIVE I

Down payment, P = Rs.5,00,000Yearly equal installment, A = Rs.2,00,000 n = 15 years i= 20%5,00,0000123.15 2,00,0002,00,0002,00,000.2,00,000AE (i%) = P(A/P,i%,n) + A = 5,00,000 (A/P,20%,15) + 2,00,000 = 5,00,000 (0.2139) + 2,00,000 = Rs. 3,06,950.

ALTERNATIVE II

Down payment, P = Rs.4,00,000Yearly equal installment, A = Rs.3,00,000 n = 15 years i= 20%4,00,0000123.15 3,00,0003,00,0003,00,000.3,00,000AE (i%) = P(A/P,i%,n) + A = 4,00,000 (A/P,20%,15) + 3,00,000 = 4,00,000 (0.2139) + 3,00,000 = Rs. 3,85,560.

ALTERNATIVE II

Down payment, P = Rs.6,00,000Yearly equal installment, A = Rs. 1,50,000 n = 15 years i= 20%6,00,0000123.15 1,50,0001,50,0001,50,000.1,50,000AE (i%) = P(A/P,i%,n) + A = 6,00,000 (A/P,20%,15) + 1,50,000 = 6,00,000 (0.2139) + 1,50,000 = Rs. 2,78,340.

A company invests in one of the two mutually exclusive alternatives.The life of both alternatives is estimated to be 5 yearsWith the following investments , annual return and salvage value.Determine the best alternative based on the annual equivalent method by assuming i = 25%, compounded annually.

Alternative A(Rs.)Alternative B(Rs.)Investment (Rs.)-1,50,000-1,75,000Annual equal return (Rs.)60,00070,000Salvage value (Rs.)15,00035,000

Alternative AInitial investment , P = Rs.1,50,000Annual equal return, A = Rs. 60,000Salvage value, S = Rs.15,000Life n = 5 yearsP01234560,000 +15,00060,00060,00060,00060,000AE (i%) = - P (A/P,i%,n) + A + S (A/F,i%,n) = -1,50,000 (A/P,25%,5) + 60,000 + 15,000 (A/F,25%,5) = -1,50,000 (0.3718) + 60,000 + 15,000 (0.1218) = Rs. 6,507.

Alternative BInitial investment , P = Rs.1,75,000Annual equal return, A = Rs. 70,000Salvage value, S = Rs.35,000Life n = 5 years1,75,00001234570,000 +35,00070,00070,00070,00070,000AE (i%) = - P (A/P,i%,n) + A + S (A/F,i%,n) = -1,75,000 (A/P,25%,5) + 70,000 + 35,000 (A/F,25%,5) = -1,75,000 (0.3718) + 70,000 + 35,000 (0.1218) = Rs. 9,198.

RESULT :The annual equivalent net return of alternative B is more than that ofalternative A. Thus the company should select alternative B.

RATE OF RETURN METHODThe expenditure is always assigned with negative sign.The revenue/inflow are assigned with positive sign.

RATE OF RETURN METHOD(CASH FLOW DIAGRAM)P0123.jnRnSHere P initial investment. Rj net revenue at the end of jth year. S the salvage value at the end of nth year. i the interest rate compounded annually.

PW(i) = -P + R1(1+i)1 + R2(1+i)2+. + Rj (1+i)j +..+ Rn(1+i)n + S/(1+i)n

R1R2R3 .Rj

A person is planning a new business.The initial outlay and cash flow for the new business are listed below.The expected life of the business is five years.Find the rate of return for the new business.

Period012345Cash flow (Rs.)-1,00,00030,00030,00030,00030,00030,000

Initial Investment , P = Rs. 1,00,000Annual equal revenue, A = Rs.30,000Life n = 5 years

0123.4530,00030,00030,00030,000 .30,0001,00,000The present worth of the business isPW(i) = - P + A (P/A,i%,n) = -1,00,000 + 30,000 (P/A,i%,5)

When i= 10%PW(i%) = -1,00,000 + 30,000 (P/A,i%,5) = - 1,00,000 + 30,000 (P/A,10%,5) = - 1,00,000 + 30,000 (3.7908) = Rs.13,724.

When i= 15%PW(i%) = -1,00,000 + 30,000 (P/A,i%,5) = - 1,00,000 + 30,000 (P/A,15%,5) = - 1,00,000 + 30,000 (3.3522) = Rs. 566.

When i= 18%PW(i%) = -1,00,000 + 30,000 (P/A,i%,5) = - 1,00,000 + 30,000 (P/A,18%,5) = - 1,00,000 + 30,000 (3.1272) = - Rs.6,184.

i = i% + PW(i%) positive O ------------------------------------- * (i%) PW(i%) PW(i%) negative

= 15% + 566 O ------------------------------------- * (3%) 566 (- 6184)

= 15% + 0.252 %

= 15.252 %

*****

UNIT – III

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