W19_LUCKY_The Decision to propose Gas Engine Alternatives to a customer

W19_LUCKY_The Decision to propose Gas Engine Alternatives to a customer

1.      Problem Recognition

Our application engineer came to me seeking advice on a gas engine he wants to propose to a customer for purchase.  

2.      Problem Definition

What are some of the economics-based questions the engineer should answer as they evaluate the alternatives to propose (1) Gas Engine A or (2) Gas Engine B or (3) Gas Engine C?

a.      Assumptions

Key assumptions include:

·         There is no room for not proposing an alternative

·         The three alternatives are equally well supported by our company in Nigeria

·         Salvage value equals initial cost

·         Compare the alternatives using MARR = 12%

·         Year, n = 20

 

3.      Feasible Alternatives

Feasible alternatives include:

A.    Propose Gas Engine A

B.     Propose Gas Engine B

C.     Propose Gas Engine C

 

 

4.      Development of outcomes  for each alternative

For the three alternatives 

                                                            A                                 B                                 C

·         Initial cost                   $150                            $120                            $110

·         Maintenance Cost       $20                              $35                              $39

·         AB First                      35.5                             39.6                             39.6    

·         Useful life                   40                                45                                40

·         Rate of Return                        10%                             15%                             16.4%

 

 

 

 

D.    Selection Criteria

Compare using MARR = 12%

·         Future worth

·         Benefit cost

·         Payback period

 

Disregard alternative(s) with a return < MARR

 

E.     Analysis and Comparison of the alternatives

Future worth: option A

              

 

 

F =-110 -20(F/P, 12%, 4) – 20 (F/P, 12%, 10) -110(F/P, 12%, 20) + 35.5(F/A, 12%, 20)

F =-110 -20(1.5735) -20(3.1058)-110(9.6463) +35.5(72.0524)

F = -110 -31.47-62.12-1,061.09+2,557.78

F_A = $1,293.19

Future worth: option B

                  

 

 

F =-120 -35(F/P, 12%, 4) – 35 (F/P, 12%, 10) -120(F/P, 12%, 20) + 39.6(F/A, 12%, 20)

F =-120 -35(1.5735) -35(3.1058)-120(9.6463) +39.6(72.0524)

F = -120 -55.07-108.70-1,157.56+2,853.28

F_B = $1,411.95

Future worth: option C

                  

 

 

 

F =-110 -39(F/P, 12%, 4) – 39 (F/P, 12%, 10) -110(F/P, 12%, 20) + 39.6(F/A, 12%, 20)

F =-110 -39(1.5735) -39(3.1058)-110(9.6463) +39.6(72.0524)

F = -110 -61.37-121.13-1,061.09+2,853.28

F_C = $2,559.71

Future worth Analysis

                                                            A                                 B                                 C

·         Initial cost                   $150                            $120                            $110

·         Maintenance Cost       $20                              $35                              $39

·         AB First                      35.5                             39.6                             39.6    

·         Useful life                   40                                45                                40

·         Rate of Return                        10%                             15%                             16.4%

·         Future worth          $1,293.19                     $1,411.95                        $2,559.71

 

Benefit-cost Ratio Analysis

                        Year                            C                     A                                 C-A

             0                                 -110                 -150                             40

            1                                  39.6                 35.5                             4.10

            2                                  39.6                 35.5                             4.10

            3                                  39.6                 35.5                             4.10

Present worth of cost = 0

Present worth of benefit = 40+4.10(P/A, 12%, 40) + 150(P/F, 12%, 40) = 4.10(8.2438) + 150(0.0107) =75.41

B/C = infinity

Reject A

Benefit-cost Ratio Analysis

                        Year                            B                     C                                 B-C

             0                                 -120                 -110                             -10

            1                                  39.6                 39.6                             0

            2                                  39.6                 39.6                             0

            3                                  39.6                 39.6                             0

Present worth of cost = 10

Present worth of benefit = 0

B/C =0/10 < 1

Reject B

   Payback period

·         A  150/35.5 = 4.23 years

·         B  120/39.6 = 3.03 years

·         C  110/39.6 = 2.78 years

 

 

F.     Selection of preferred alternative

Summary

                                                            A                                 B                                 C

·         Initial cost                   $150                            $120                            $110

·         Maintenance Cost       $20                              $35                              $39

·         AB First                      35.5                             39.6                             39.6    

·         Useful life                   40                                45                                40

·         Rate of Return                        10%                             15%                             16.4%

·         Future worth          $1,293.19                     $1,411.95                        $2,559.71

·         Benefit Cost                C-A = infinity       B-C  = 0

·         Payback period           4.23                             3.03                             2.78    

Select Gas Engine C

I would advise our application engineer to propose Gas Engine type B.

G.    Performance monitoring and post evaluation of results

The focus on speed versus efficiency limitations of the payback period analysis as well as the assumptions for MARR, the salvage value and the period, n, used for the analysis can change and thus would present a new scenario for choosing an alternative given the new parameters.

A further review with lease or buy option will be considered in a subsequent blog.

 

 

 

Reference

1.      Sullivan, W., Wicks, E., Koelling, P., Kumar, p., & Kumar, N. (2012).Chapter 1 Introduction to Engineering Economy (pp. 29). Engineering economy (15^th edition). England: Pearson Education Limited.

2.      Sullivan, W., Wicks, E., Koelling, P., Kumar, p., & Kumar, N. (2012).Chapter 14 The Time Value of Money (pp. 131 -134). Engineering economy (15^th edition). England: Pearson Education Limited.

3.      Lindeburg, M. (1992).Chapter 13 Engineering Economic Analysis (pp.13-3).Engineer-In-Training Reference Manual (8^th edition).U.S.A: Professional Publications Inc.

4.       Giammalvo, P. (2012, October 22). Integrated portfolio (asset), program (operations) and project management methodology course (cost engineering) slides (An AACE methodology course). Lagos, Nigeria: Lonadek

5.       Lee, J. (2000).  Engineering Economy Review. Retrieved from http://tofudi.com/read-file/engineering-economy-review-main-concepts-suggestions-for-pdf-288876/