The document discusses a presentation on solving a rate of return (RoR) problem related to an investment made by a contractor in a warehouse building costing $350,000. The contractor is expected to save $7,000 annually for 10 years, leading to a net return of $497,000 in year 10, and the estimated rate of return is calculated to be 5.30%. The analysis includes cash flow diagrams and interpolation methods to determine the final rate of return.

1. My Presentation Topic is :-
Solve this problem on the solution of this
ROR ( Rate of Return ) problem
Course Name: Project Planning and Management
Course Code : CE 403
Final Term
Submitted By
Mohammad Khasrul Alam
ID No. : 003-18-23
Department Of Civil Engineering
Southern University Bangladesh.
Submitted To
Mohammad Ayanul Huq Chowdhury
Lecturer,
Department Of Civil Engineering
Southern University Bangladesh.

2. Rate of Return :
Rate of return (ROR) is the rate paid on the unpaid balance of
borrowed money, or the rate earned on the unrecovered
balance of an investment, so that the final payment or receipt
brings the balance to exactly zero with interest considered.
 PWD= costs or disbursements
 PWR= incomes or receipts
𝑃𝑊𝐷=𝑃𝑊𝑅
0=−𝑃𝑊𝐷+𝑃𝑊𝑅
𝐴𝑊𝐷=𝐴𝑊𝑅
0=−𝐴𝑊𝐷+𝐴𝑊𝑅
 If i* ≥ MARR, accept the alternative as economically viable.
 If i* < MARR, the alternative is not economically viable.

3. Problem :
A small industrial contractor purchased a
warehouse building for storing equipment and
materials that are not immediately needed at
construction job sites. The cost of the building
was $A=350,000. He was expecting a saving of
$B=7000 per year for 10 years and a net return
of $C=497,000 in year 10. What rate of return
did the contractor make on this investment?
 Given Data,
$A=350,000 $B=7000 $C=497,000

4. Solution :
 Draw Cash Flow Diagram :

5. Continue…
 Use equation :
0=−𝑃𝑊𝐷+𝑃𝑊𝑅
0=−350,000+7,000𝑃/𝐴,𝑖∗,10+497,000(𝑃/𝐹,𝑖∗,10)
P = $350,000,
n = 10, and
F = 10(7,000) + 497,000 = $567,000
» 350, 000 = 567,000 (P/F, i, 10)
» (P/F, i, 10) = .620
The roughly estimated i is between 4% and 5%

6. Continue…
 Use the estimation procedure to determine I :
Use 5% as the first trial because approximate rate for
the P/F factor will be lower than the true value when
the time value of money is considered …
i =5%
0=−350,000+7,000(P/A,5%,10)+497,000(P/F,5%,10)
» 0<$ 9167.03
Again, i= 6%
0=−350,000+7,000(P/A,6%,10)+497,000(P/F,6%,10)
» 0>$−20957. 18

7. Continue…
We get,
y1 = 5.0 x1 = 9167.03 y= ?
y2 = 6.0 x2 = -20957.18 x =0
 Since the interest rate of 6% is too high, linearly
interpolate between 5% and 6%
𝑖∗=5.0+ 0-9167.03 (6.0−5.0)
(-20957.18)-9167.03
=5.0+0.30%=5.30%
5.30% rate of return did the contractor make on this
investment.
» i = 5.30%
Answer

8. Thank You
See you again any other class and
any other presentation.
End