Engineering Economics Assignment: IRR and Cash Flow Analysis

This economics assignment delves into the principles of engineering economics, focusing on the application of the Internal Rate of Return (IRR) for project evaluation. The solution begins with a comparative analysis of two investment types, Fe and Al, using tabular methods to determine incremental cash flows across a four-year period. The IRR is calculated to assess the viability of each investment option, with detailed calculations and spreadsheet representations. The assignment further explores a scenario involving two projects, 2B and 4R, evaluating their costs, annual expenses, and applying a Minimum Acceptable Rate of Return (MARR) of 6% over a 20-year period. The IRR is calculated using a trial-and-error approach, demonstrating how to determine the rate at which the present value of cash inflows equals the present value of cash outflows, thus aiding in the decision-making process for project selection. The analysis concludes with a recommendation based on the calculated IRR compared to the MARR, guiding the selection of the optimal project.

Running head: ECONOMICS 1
Engineering Economics
Name
Institution

ECONOMICS 2
Question 1
a. By the use of tabulation, the incremental cash flow will be as follows;
Type Fe Type Al
Cash flow
Year Cash Flow, $ $ Cash flow, $ (AL-Fe)
0 -150,000 -280,000 -130,000
1 -92,000 -74,000 18,000
2 -212,000 -74,000 138,000
3 -92,000 -74,000 18,000
4 -62,000 -4,000 58,000
∆I* value 27.35%
Using the spreadsheet above, we calculate ∆ i as follows;
0 = -130+18 ( P
A , ∆i¿ , 4 ) + 120 ( P
F , ∆ i¿ ,2 )+40 ( P
F , ∆ i¿ , 4 )
From the above findings, we then enter the incremental cash flow where we apply the IRR to get;
∆ i=27.35% . Since 27.35 %> MARR=20 % we therefore select type Al in the spreadsheet.
b.
Type Fe Type Fe Type Al
Cash flow
Incrementa
l
Incrementa
l
Year
Cash Flow,
$
I and ∆I,
% PW, $ $ PW, $
Cash flow,
$ PW, $
0 -150,000 5%
-
560,390 -280,000 -484,811 -130,000 75,579
1 -92,000 10%
-
520,311 -74,000 -466,759 18,000 53,552
2 -212,000 15%
-
486,243 -74,000 -451,246 138,000 34,997
3 -92,000 20% - -74,000 -437,809 18,000 19,221

ECONOMICS 3
457,029
4 -62,000 25%
-
431,779 -4,000 -426,086 58,000 5,693
30%
-
409,796 -415,793 -5,997
35%
-
390,531 -406,699 -16,169
∆I*
value 27.35%
From the graphs above in (b), since 27.35 %> MARR=20 % we therefore select type Al in the
spreadsheet.
c.

ECONOMICS 4
From (c) graph above, since 27.35 %> MARR=20 % we therefore select type Al in the
spreadsheet. It therefore shows that the answer is as in (a) above.
Question 2
From the question;
The cost of designing 2B = %3,000,000
The annual Cost for designing 2B = $135,000
The cost for 4R = $3,500,000
Annual Cost for 4R = $70,000
MARR = 6%
Time Period is 20 years
Incremental rate of return (IRR) is calculated using the following formula;

ECONOMICS 5
0= ( Cost Design 2 B−Cost Design 4 R )+ ( Annual Cost 2 B− Annual Cost 4 R ) ( ( 1+ IRR )Time−1
IRR (1+ IRR )Time )
If we substitute IRR to be 7% thus using trial analysis, we substitute the value of IRR in the
above formula to calculate the internal Rate of Return as follows;
0= ( 3,000,000−3,500,000 ) + ( 132,000−70,0000 ) ( ( 1+0.07 ) 20−1
0.07 ( 1+0.07 ) 20 )
0= (−500,000 ) + ( 62,000 ) ( 3.8697−1
0.07∗3.8697 )
0= (−500,000 ) + ( 62,000 ) ( 2.8697
0.2709 )
0= ( −500,000 ) + ( 62,000 ) ( 10.5932 )
0= (−500,000 ) + ( 656,778.40 )
0<156,778.40
From the above calculation, if we decide to substitute with the higher IRR of 7%, the will be
greater than zero. As a result, we can decide to decrease the value to approximately 6.8% which
is within the MARR given i.e.
0= ( 3,000,000−3,500,000 )+ (132,000−70,0000 ) ( ( 1+0.068 )20 −1
0.07 ( 1+0.068 )20 )
0= (−500,000 ) + ( 62,000 ) ( 3.7276−1
0.068∗3.7276 )
0= ( −500,000 ) + ( 62,000 ) ( 2.7276
0.2535 )

ECONOMICS 6
0= (−500,000 ) + ( 62,000 ) ( 10.76 )
0= (−500,000 ) + ( 667,120 )
0<167,120
Comparing the two values, we therefore conclude that the incremental rate of return is 6.8%
since IRR is greater than MARR which is 6% thus we select project for 4R.