Solve the problem given in Example 5.2 by the I.R.R. method.
Chapter 5
Q. 5.8
Step-by-Step
Verified Solution
Step 1: The cash flow diagram for the given problem is shown in Fig. 5.2.
Step 2: The P.W. of the net receipts at an interest rate of i^{\prime } is calculated as:
Step 3: The P.W. of the net expenditures at an interest rate of i^{\prime } is calculated as:
P.W.=₹1,05,815.4+₹30,000\left(P/A,i^{\prime}\% ,5\right)Step 4: The net present worth N.P.W. is obtained as:
N.P.W.=₹53,000\left(P/A,i^{\prime}\% ,5\right)+₹30,000\left(P/F,i^{\prime}\% ,5\right)-₹1,05,815.4-₹30,000\left(P/A,i^{\prime}\% ,5\right)Step 5: 0 =₹53,000\left(P/A,i^{\prime}\% ,5\right)+₹30,000\left(P/F,i^{\prime}\% ,5\right)-₹1,05,815.4-₹30,000\left(P/A,i^{\prime}\% ,5\right)
The equation given at Step 5 normally involves trial-and-error calculations until the i^{\prime }\% is found. However, since we do not know the exact value of i^{\prime }\%, we will probably try a relatively low i^{\prime }\%, such as 5\%, and a relatively high i^{\prime }\%, such as 12\%.
At i^{\prime }\% = 5\%:
₹53,000\left(P/A,5\%,5\right)+₹30,000\left(P/F,5\%,5\right)-₹1,05,815.4-₹30,000\left(P/A,5\% ,5\right)₹53,000\left(4.3295\right)+₹30,000\left(0.7835\right)-₹1,05,815.4-₹30,000\left(4.3295\right)=+₹17,268.1
At i^{\prime }\% = 12\%:
₹53,000\left(P/A,25\%,5\right)+₹30,000\left(P/F,25\%,5\right)-₹1,05,815.4-₹30,000\left(P/A,25\% ,5\right)₹53,000\left(3.6048\right)+₹30,000\left(0.5674\right)-₹1,05,815.4-₹30,000\left(3.6048\right)=-₹5,883
Since we have both a positive and a negative P.W. of net cash flows, linear interpolation can be used as given below to find an approximate value of i^{\prime }\%
\frac{12\%-5\%}{₹17,268.1-\left(-₹5,883\right)}=\frac{i^{\prime }\%-5\%}{₹17,268.1-₹0}i^{\prime}\%=5\%+\frac{₹17,268.1}{₹17,268.1-\left(-₹5,883\right)}\left(12\%-5\%\right)
i^{\prime }\% = 10.22\%, which is approximately equal to 10\%.
Step 6: Since the value of i^{\prime }\% = M.A.R.R., the investment in the project is economically barely justified.
Note: Let us check whether the value of N.P.W. at i^{\prime } = 10\% is 0.
N.P.W.=₹53,000\left(P/A,i^{\prime}\%,5\right)+₹30,000\left(P/F,i^{\prime}\%,5\right)-₹1,05,815.4-₹30,000\left(P/A,i^{\prime}\%,5\right)At i^{\prime }\% = 10\%:
N.P.W.=₹53,000\left(P/A,10\%,5\right)+₹30,000\left(P/F,10\%,5\right)-₹1,05,815.4-₹30,000\left(P/A,10\%,5\right)=₹53,000\left(3.7908\right)+₹30,000\left(0.6209\right)-₹1,05,815.4-₹30,000\left(3.7908\right)
N.P.W.=0
Thus i^{\prime } = 10\% which is equal to the given M.A.R.R. and therefore, the investment in the project is economically barely justified.
