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Case 1)Determine the capitalized cost of a research laboratory which requires Php 5,000,000 for original construction Php 100,000 at the end of every year for the first 6 years and then Php 120,000 each year thereafter for operating expenses, and Php 500,000 every 5 years for replacement of equipment with interest at 12% per annum

Finance Oct 13, 2020

Case 1)Determine the capitalized cost of a research laboratory which requires Php 5,000,000 for original construction Php 100,000 at the end of every year for the first 6 years and then Php 120,000 each year thereafter for operating expenses, and Php 500,000 every 5 years for replacement of equipment with interest at 12% per annum.

Case 2)A man purchased a foreclosed property for Php 425,000. In the first month that he owned the house, he spent Php 75,000 for repairs and remodelling. Immediately after the house was remodeled, he was offered Php 545,000to sell the house. After some consideration, he decided to keep the property and have it rented for Php 4,500 per month starting two months after the purchase. He collected rent for 15 months and then sold the property for Php 600,000. If the interest rate was 1.5% per month, how much extra earnings did he make or lose by not selling the house immediately after it was remodeled.

Expert Solution

Case 1-

Original Construction      5,000,000.00
Operating expense(Year 1 to 6)         100,000.00
Operating expense(Year 7 to perpetual)         120,000.00
Replacement cost for every 5 years         500,000.00

Present value of Operating expense(year 1 to 6)-

Year Cash Outflow AF@ 12% Present value
1_6 100000 4.1114 411140

The present value of operating expense from year 1 to 6 is 411,100.

Present value of Operating expense(year 7 to perpetual)-

Value at year 6= Operating cost/Interest

=120000/12%

=1,000,000

Present Value is -

Year Cash Outflow AF@ 12% Present value
6 1000000 0.5066 506600

The present value of operating expense from year 6 to perpetual is 506,600.

Present value of replacement cost-

Present value= S/(1+i)k-1

Where,

S= Replacement cost

i= Interest Rate

k=Replacement period.

Present value of replacement cost= 500000/(1.12)5-1

=500000/0.7623

=655,910.

Calculation of amount to be capitalised-

Original Construction 5000000
Present value of Operating expense(Year 1 to 6) 411140
Present value of Operating expense(Year 7 to perpetual) 506600
Replacement cost for every 5 years 655910
Amount to be capitalised 6573650

The capitalised cost of research laboratory is 6,573,650.

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