an engineering company is working in coal mining operation has installed an in-s

Question

an engineering company is working in coal mining operation has installed an in-shaft monitoring system for oxygen tank and gear readiness for emergencies. Based on Mai taints patterns for previous systems, cost are minimal for the first few years, increase for a time period, and then Direction questions carefully. Please draw the cash flow diagrams and explain the steps h to solve the problems then solve the problem. Show the details in S. Pease write legible hand writing, otherwise there is a deduction from your Flow Diagramn is deductible points equal to 20% of the total Please read the qu eions cto solve the problems then solve thc ere i Please read the that you are going to approac solving the problems. I grade 10 points. Missing Cash points for each question leasev Question (D-20 points in-shaft n engineering company is working in coal mining operation has installed an onitoring system for oxygen tank and gear readiness for emergencies. Based on maintenance patterns for previous systems, costs are minimal for the first few years, increase for a time period, and then level off. Maintenance costs are expected to be $180,000 in year 3, and increase y 6% per year through year 6 and remain constant thereafter for the expected 10-year life of this system. If similar systems will replace the current one, determine the perpetual equivalent annual maintenance cost at i-10% per year. Question (2)-(20 point A woman wishes to make a quarterly deposit into her savings account so that at the end of 10 years the account balance will be $10,000. If the account earns 8% annual interest, compounded quarterly, how much should she deposi each quarter? quarterly, how much should she deposit each quarter? Question (3)-20 points) The maintenance on a machine is expected to be $180 at the end of the first year, then increasing y $35 each year for the next 7 years. What sum of money would need to be set aside now to e year period? Assume 8% interest. Additionally, we wish now to

Explanation / Answer

1.

Maintenance cost in year 3 = $180,000

Maintenance cost in year 4 = 180000*1.06 = $190800

Maintenance cost in year 5 = 180000*1.06^2 = $202248

Maintenance cost in year 6 = 180000*1.06^3 = $214382.88

Maintenance cost from year 7 to 10 = $214382.88 per year

Now,

Interest rate (I) = 10%

Present value of Maintenance cost = 180000/ (1+I)^3 + 190800/(1+I)^4 + 202248/(1+I)^5 + 214382.88/(1+I)^6 + 214382.88/(1+I)^7 +214382.88/(1+I)^8 +214382.88/(1+I)^9 +214382.88/(1+I)^10

Present value of Maintenance cost = 180000/ (1+.1)^3 + 190800/(1+.1)^4 + 202248/(1+.1)^5 + 214382.88/(1+.1)^6 + 214382.88/(1+.1)^7 +214382.88/(1+.1)^8 +214382.88/(1+.1)^9 +214382.88/(1+.1)^10

Present value of Maintenance cost =$895745.93

Thus,

Perpetual equivalent Annual Maintenance Cost = Present value of Maintenance cost*interest rate

Perpetual equivalent Annual Maintenance Cost = 895745.93*.1 = $89574.59

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