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Chapter 5

Q. 5.9

Solve the problem given in Example 5.1 by E.R.R. method.

Step-by-Step

Verified Solution

Step 1: The cash flow diagram for the given problem is shown in Fig. 5.1.
Step 2: Since a specific value of e is not given take e = M.A.R.R. = 15\%.
Step 3: In this problem ₹5,00,000 is the only cash outflow and it occurs at period 0. Therefore, the result of this step is ₹5,00,000.
Step 4: All cash inflows (positive cash flows/receipts) are compounded to period n (the future) at e\% in the following manner:

₹2,00,000\left(F/A,15\%,5\right)+₹1,00,000

Step 5: The result of this step is given as:

₹5,00,000=₹2,00,000\left(F/A,15\%,5\right)+₹1,00,000

Step 6: The L.H.S. of the equation given at Step 5 represents present worth whereas, its R.H.S. represents future worth. Therefore, the two sides cannot be equal. Introduce the factor \left(F/P,i^{\prime}\% ,n\right) on to the L.H.S. to bring a balance between two sides of equation in the following manner:

₹5,00,000\left(F/P,i^{\prime}\% ,5\right)=₹2,00,000\left(F/A,15\%,5\right)+₹1,00,000

Step 7: The equation given at Step 6 is solved as:

₹5,00,000\left(F/P,i^{\prime}\% ,5\right)=₹2,00,000\left(6.7424\right)+₹1,00,000

 

₹5,00,000\left(F/P,i^{\prime}\% ,5\right)=₹14,48,480

 

\left(F/P,i^{\prime}\% ,5\right)=₹14,48,480/₹5,00,000

 

\left(F/P,i^{\prime}\% ,5\right)=2.897

 

\left(1+i^{\prime}\right)^{5}=2.897

 

1+i^{\prime}=1.237

 

i^{\prime}=1.237-1=0.237=23.7\%

Since the value of i^{\prime}\% = 23.7\% \gt M.A.R.R. = 15\%, the investment in the project is economically justified.