Example 361

The local highway department advertised for bids for a concrete highway bridge. The requirements specified that the concrete must exceed 25 MPa in compressive strength and must have a factor of safety of at least 2.0 with respect to a standard HS-20 truck crossing the bridge at all locations. Company A submitted a bid for $70,000, and Company B submitted a bid for $100,000. Both designs met the specifications, and both bidders were qualified to do the work. Which company should get the job?


Company A, based on the lower cost. EXAMPLE 36.2

The local highway department expects the useful life of the highway bridge to be 60 years. The highway department engineer investigates the two proposals in Example 36.1 further and determines that the design submitted by Company A is less expensive because the concrete is not as durable. This engineer estimates that the Company A bridge will require a major rehabilitation every 20 years for a cost of $30,000 and will require a major inspection every 5 years at a cost of $5,000 each time. The Company B bridge will require a major rehabilitation every 30 years for a cost of only $20,000. The inspection of this bridge will also cost $5,000, but it only needs to be done every 10 years. The engineer uses a discount rate of money (which accounts for inflation) of 3% and determines that the preventive maintenance and failure costs of both designs are about equal and, therefore, irrelevant to the decision. Both options require a major rehabilitation at year 60, so there is no difference in salvage value. Which company should get the job?


The analysis requires that the costs associated with both projects be converted to present value for a valid comparison. Even though the Company A bid appears to now cost $185,000 (i.e., undiscounted total cumulative cost = $70,000 + 2 x $30,000 + 11 x $5,000), over the life of the structure compared to $145,000 (i.e., undiscounted total cumulative cost = $100,000 + $20,000 + 5 x $5,000) for Company B, the timing of the payments is important. The equation for converting a future value (FV) to a present value (PV) is

0 0

Post a comment