What is the solution to this problem? Thanks! The College of Engineering (COM) a
ID: 2506562 • Letter: W
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What is the solution to this problem? Thanks!
The College of Engineering (COM) at Valpo is purchasing a 3-D visualization software which costs dollar 5,000. COE would also like to purchase a maintenance plan, for which there are two alternatives. A 5-year maintenance plan (Plan A) costs dollar 500 in year 1. and decreases by dollar 100 each year. A 10-year maintenance plan (Plan B) costs dollar 200 annually, hut it comes with free maintenance for the first two years. Interest rate is 8% per year C umpire the two "purchase and maintenance" alternatives using the present worth analysis and determine the preferred alternative. I se the least common multiple (LCM) approach to compare the alternatixes for the same number of years.Explanation / Answer
Given Interest Rate = 8%
Cost of visualization software = $5,000.
Following is the analysis of alternative 1 maintenace plan as per given conditions:
Cash outflow at the end of year 1 : $500
Cash outflow at the end of year 2 : $400
Cash outflow at the end of year 3 : $300
Cash outflow at the end of year 4 : $200
Cash outflow at the end of year 5 : $100
Since, the maintenance cost decreases by $100 as per given conditions.
Therefore, present value of this maintenace plan is given by: [discounting cash flows to present using given interest rate)
Present Value = $500/(1+0.08) + $400/(1+0.08)^2 + $300/(1+0.08)^3 + $200/(1+0.08)^4 + $100/(1+0.08)^5
= 462.96 + 342.94 + 238.15 + 147.01 + 68.06
= $1259.12
Therefore, the total purchase and maintentance plan cost will be = $5000 + $1259.12 = $6259.12 ----(1)
Following is the analysis of alternative 2 maintenace plan as per given conditions:
Cash outflow at the end of year 1 : $0
Cash outflow at the end of year 2 : $0
Cash outflow at the end of year 3 : $200
Cash outflow at the end of year 4 : $200
Cash outflow at the end of year 5 : $200
Cash outflow at the end of year 6 : $200
Cash outflow at the end of year 7 : $200
Cash outflow at the end of year 8 : $200
Cash outflow at the end of year 9 : $200
Cash outflow at the end of year 10 : $200
Therefore, present value of this maintenace plan is given by: [discounting cash flows to present using given interest rate)
where,
C = Cash flow per period (Here C is $200)
i = Interest rate (Here C is 8%)
n = Number of payments (Here n is 8 i.e from year 3 to year 10)
First discount all equal cash flows of $200 using present value of annunity factor which is given by:
(Here $200 is annutiy).. So, we find out present value of maintenance plan at the end of year 2
i.e. beginning of year 3 and then discount that cost to present value in the next step
Value of Cost of maintenance plan at the end of year 2 = 200*[(1 - (1.08)^-8)/0.08] = $1149.33
Therefore, present value of cost of maintenance = $1149.33/(1.08^2) = $985.37
Therefore, the total purchase and maintentance plan cost will be = $5000 + $985.37 = $5985.37 ----(2)
From (1) and (2), we can conclude that,
Second alternative is the preferred alternative as its cost is less and maintenance cover is more.
Comparison of both programs using LCM approach:
LCM of 5 and 10 is 10 years, therefore, we calculate the cost of alternative 1 for 10years i.e renewed after 5 years
New cost of alternative = Cost of alternative 1 as calculated in earlier + Present value of renewed cost of maintenance
= $1259.12 + $1259.12/(1.08^5)
.....[ Since, value of renewed maintenance at the end of 5th year will be same as present value of a 5 year maintenance
Therefore, new cost of alternative = $1259.12+ $856.94 = $2116.06
Therefore , new total cost = $5000 + $2116.06 = $7116.06 -----(3)
From, (2) and (3) we can conclude that,
Still second alternative is the preferred alternative as its cost is less and maintenance cover is more.
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