Present Worth Sample Problem


You are collecting bids for the installation and servicing of a new service elevator for your company's building, and need to select the cheapest offer as this is your first chance to impress a new boss. You are looking at a minimum attractive rate of return at 12% and a 15 year life span for all bids. Bid A will cost $15,000 to install and will have a $1000 per year servicing fee. Bid B has a $24,000 installation package and is so confident in their product that they will service the elevator free of charge. While, Bid C will cost only $11,000 to install but has a hefty $2,000 per year servicing fee. Choose the best bid.


First organize the information given:

Bid A Bid B Bid C
Initial Cost $15,000 $24,000 $11,000
Yearly Cost $1,000 $0 $2,000
MARR = i 0.12 Life = N 15

Second obtain the appropriate equation for Present Worth:

P = A ( ((1+i)N - 1) / (i (1+i)N))

Then solve present worth for each bid:

PWA = 15,000 + 1,000(P\A, i=.12, N=15)
PWA = 15,000 + 6,810.86 = $21810.86

PWB = $24,000

PWC = 11,000 + 2,000(P\A, i =.12, N=15)
PWC = 11,000 + 13,621.73 = $24,621.73

Choose the cheapest bid: Bid A


The idea of present worth is to take all the expenses that will accumulate over the life of the asset and move them to the present date. This allows us to evaluate the best option without biasing ourselves with nice service plans or a cheap initial installation. If you look at the given information Bid B and C look awfully tempting, but after the calculations we see that they will end up costing the company more money over the course of the plans.

To calculate the Present Worth of each bid, the initial cost, which was not manipulated because all three are in time period 0 (i.e. the present) and the yearly expense was added using the equation for Present Worth for a constant series. If you look at the equation the denominator includes a function to moved the value to time period 0 (i.e. the present).

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