THE CHANGING SCENARIO OF AN ANNUAL VALUE ANALYSIS

Harry, owner of a battery distributor in Atlanta, Georgia, did an economic analysis 3 years ago when he decided to put in-line voltage protectors on all the important parts of his test equipment. The estimates used and the annual value analysis at MARR = 15% are summarized here. Two protectors from two different manufacturers are compared.

The Excel spreadsheet shown below was used to make the decision. The Lloyd company was the clear choice due to its substantially higher AW value. Lloyd's protectors were installed.

During a quick review this past year (year 3 of operation), it was obvious that maintenance costs and repair savings had not been in line with estimates (includes quarterly inspection) will increase from $300 to $1200 next year and from there will increase 10% per year for the next 10 were $35,000, $32,000, and $28,000. Harry believes savings could decrease by $2,000 per year thereafter. Finally, these 3 year old protectors are worthless in today's market. This makes the residual value in 7 years zero. 3 years ago. In fact, the maintenance contract (which years. Also repair savings for the last 3 years instead of $3,000. Requirements Build the spreadsheet in Excel according to the image shown below.

a. Make the graph to project the estimated maintenance costs and repair savings for Lloyd Company, assuming the protectors last 7 more years. For this consider the old estimates (as shown in the Excel file)

* This is a money flow graph where we use arrows and to identify the series we join them with a line from the beginning to the end of the series.

b. With the new estimates of the problem, replace the values of the estimates in the Excel file. What is the recalculated AW for Lloyd's protectors? Would Lloyd be the economic choice if we go by these new estimates? Why or why not?

c. How has the capital recovery amount for Lloyd's protectors changed with these new estimates? You must set the difference in AW (annual value).

MARR =15% \begin{tabular}{lrr} & PowerUp(\$) & Lloyd(\$) \\ \hline Installation cost & 22,000 & 36,000 \\ \hline Annual maintenance cost & -600 & -300 \\ Residual value & 2,000 & 3,000 \\ Equipment repair savings & 25,000 & 35,000 \\ Useful life, years & 6 & 10 \\ \hline \end{tabular} \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 4 & A & B & C & D & \multicolumn{2}{|c|}{E} & F & G & H & & \multirow[t]{2}{*}{ K } & \multirow[t]{2}{*}{ L } \\ \hline 1 & & & & & & & & & & & & \\ \hline 2 & & & & \multirow{2}{*}{\multicolumn{3}{|c|}{=PMT(C3,6,C10,C16,0)}} & \multirow{2}{*}{\multicolumn{3}{|c|}{=PMT(C3,6,NPV(C3,E11:E:16),0,0)}} & & & \\ \hline 3 & & MARR = & 15% & & & & & & & & \multirow{2}{*}{\multicolumn{2}{|c|}{=SUM(C8:E8)}} \\ \hline 4 & & & & & & & & & & & & \\ \hline 5 & & & & Powrup & & & 7 & Lloyd's & & & & \\ \hline 6 & & & Investment & Annual & Rep & pair & Investment & Annual & Repair & AW PowrUp & & \\ \hline 7 & & YEAR & and salvage & maint & savi & ings. & and salvage & maint & savings & 518,815.26 & & \\ \hline 8 & & AW values & (155,584.74) & +(15600.00) & $25, & .000 .00 & ($57,025} & ($300.00) & $35,000 & & & \\ \hline 9 & & & & 4 & & & & & & AW Lloyd's & & \\ \hline 10 & & 0 & $(22,000.00) & & & & (36,000) & & & $27,675 & & \\ \hline 11 & & 1 & & (600) & s & 25,000 & & (300) & 35,000 & & & \\ \hline 12 & & 2 & & (600) & s & 25,000 & & (300) & 35,000 & & & \\ \hline 13 & & 3 & & (600) & s & 25,000 & & (300) & 35,000 & & & \\ \hline 14 & & 4 & & (600) & s & 25,000 & & (300) & 35,000 & & & \\ \hline 15 & & 5 & & (600) & 5 & 25,000 & & (300) & 35,000 & & & \\ \hline 16 & & 6 & 52,000.00 & (600) & s & 25,000 & & (300) & 35,000 & & & \\ \hline 17 & & 7 & & & & & & (300) & 35,000 & & & \\ \hline 18 & & 8 & & & & & & (300) & 35,000 & & & \\ \hline 19 & & 9 & =PMT(C3, & 6,NPV(C3,D11:D: & 16),0,0) & & & (300) & 35,000 & & & \\ \hline 20 & & 10 & & & & & 3,000 & (300) & 35,000 & & & \\ \hline 21 & & & & & & & & & & & & \\ \hline \end{tabular}