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## Q. 5.10

Solve the problem given in Example 5.2 by E.R.R. method.

## Verified Solution

Step 1: The cash flow diagram for the given problem is shown in Fig. 5.2.
Step 2: Since a specific value of e is not given take $e = M.A.R.R. = 10\%$.
Step 3: All cash outflows (negative cash flows/expenditures) are discounted to period 0 (the present) at $e\%$ in the following manner:

$₹1,05,815.4+₹30,000\left(P/A,10\%,5\right)$

$=₹1,05,815.4+₹30,000\left(3.7908\right)$

$=₹2,19,539.4$
Step 4: All cash inflows (positive cash flows/receipts) are compounded to period n (the future) at $e\%$ in the following manner:

$₹53,000\left(F/A,10\%,5\right)+₹30,000$

$=₹53,000\left(6.1051\right)+₹30,000$

$=₹3,53,570.3$

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

$₹2,19,539.4=₹3,53,570.3$

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:

$₹2,19,539.4\left(F/P,i^{\prime}\% ,5\right)=₹3,53,570.3$

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

$₹2,19,539.4\left(F/P,i^{\prime}\% ,5\right)=₹3,53,570.3$

$\left(F/P,i^{\prime}\% ,5\right)=₹3,53,570.3/₹2,19,539.4$

$\left(F/P,i^{\prime}\% ,5\right)=1.61$

$\left(1+i^{\prime}\right)^{5}=1.61$

$1+i^{\prime}=1.0999$

$i^{\prime}=1.0999-1=0.0999=9.99\%$ which is almost equal to $10\%$.

Since the value of $i^{\prime}\% = 9.99\% = M.A.R.R. = 10\%$, the investment in the project is economically barely justified.