2. A German company is planning to replace the old machine for a new one. The new machine will cost Euro 4 million, the old one can not be sold. The new machine will be used for 20 years, then it can be sold for 25% of the purchase price. Each year the company will save Euro 450,000 on the production costs, but the maintenance costs will be higher too: Euro 40,000 per year. In year 10, a special maintenance has to be performed which will cost Euro 600,000.
The company wants you to calculate the MIRR, with given that recovered funds will be reinvested at the Euro short term interest rate, which is 0.5% monthly. For the external financing rate, the company uses the MARR, which is 7.5% annual nominal, compounded monthly.
Cost Savings per year = Savings - maintenance cost = 450000 - 40000 = 410000
Cost of machine = 4000000
Future value of cost savings = Future value of annuity = cost savings * [ ( 1 + int rate )^ no of years - 1 ] / int rate
= 410000 * [ 1.0617^20 - 1 ] / 0.0617
= 15351439.79 (a)
future value of special maintenance cost = 600000 * 1.0617^10 = 1091638.04 (b)
sale value of machine = 4*0.25 = 1 million (c)
Future value of cash flows = a +c - b = 15259802
MIRR =( Ending value / beginning value ) 1 / no of years
=( 15259802 / 4000000 ) ^ 1/20 - 1
= 6.92%
B. No the company should not accept the project as the MIRR is lesser than the MARR
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