Question

To resolve the issue of Coronavirus testing, a city decided to set up a plant to produce low cost testing kits. This facility will operate for 12 months and then it will be dismantled. It will cost the city $P to buy the main machine. In addition, the city will spend $45,000 as planning cost before the work commences. The monthly operating and maintenance cost to run the facility will be $52,500. The city also expects to lose additional $43,000 every month for the duration of the facility. It is estimated that, this plant will save taxpayers who will use the testing facility about $15 per usage. The city expects 0.5% of its 2 Million citizens to use the facility every month for 12 months. The facility will be upgraded at a cost of $40,000 at the end of month 5, $75,000 at the end of month 10, and will then be dismantled at the end of month 12 for $100,000. After dismantling, the city will sell the used machine at it salvage value of $72,000. Using benefit-cost ratio analysis with an interest rate of 5%, what is the maximum value of P that the city can spend to buy the main machine?

Answer 1

Facility Cost =$P today at t=0

Planning cost =$45000 at t=0

Monthly costs =$52500+$43000 =$95500 t=1 to t=12

Additional Cost = $40000 at t=5, $75000 at t=10, $100000 at t=12

Interest rate per month = 5%/12 =0.41667%

Present value of costs

=P+45000 + 95500/0.0041667*(1-1/1.0041667^12) +40000/1.0041667^5+75000/1.0041667^10+ 100000/1.0041667^12

= $P+1366811.95

Savings every month = 0.5% of 2 million * $15 = $150000

Present value of Benefits = 150000/.0041667 *(1-1/1.0041667^12)+72000/1.0041667^12 =$1820678.93

For maximum value of P, Benefit cost ratio =1

=> PV of Benefits/PV of Cost= 1

=>1820678.93/(P+1366811.95) =1

=> P = 453866.99 or $453867 which is the maximum value of P