A13.10 - Example of risk analysis
A13.10 - Example of risk analysis
Introduction
The following example illustrates the application of these risk analysis.
In this example, a minor bridge structure has been assessed to have a limited residual life and has been tentatively programmed for replacement after 5 years. However, the design of the bridge pre-dates modern earthquake design codes and the bridge would be damaged to an extent requiring replacement in an earthquake of return period of 200 years or more.
Calculating probability of risk
The annual probability of the bridge being destroyed by earthquake in any one year, denoted as p, is 1/200 = 0.005. The probability of the bridge surviving for 5 years and then being replaced as programmed, is calculated as follows:
- The probability of an earthquake in the first year = p = 1/200 = 0.005.
- The probability of the bridge surviving for one year is therefore (1 − p) = 0.995.
- The probability of the bridge being destroyed in year 2 is the probability of it surviving through year 1 multiplied by the probability of an earthquake in year 2 = p(1 − p) = 0.005 × 0.995 = 0.004975 and so on for five years.
In the general case, the probabilities of the bridge being destroyed in each year are:
year 1 p
year 2 p(1 − p)
year 3 p(1 − p)2
…year n p(1 − p) n − 1
and the probability of the bridge surviving to n years and then being replaced is therefore:
1 − p − p(1 − p) − p(1 − p)2 − … − p(1 − p)(n − 1) = (1 − p)n
The probability of survival to the end of year 5 is therefore:
(1 − 0.005)5 = 0.97525
In the event of earthquake damage, a temporary Bailey Bridge would have to be erected while a new permanent structure was being built. This would impose an additional cost on the road controlling authority which would not occur in the case of a planned replacement. There would also be disruption to traffic at the time of the earthquake.
Calculating costs if risk occurs
Assume that the bridge replacement cost is $2.5 million over 2 years. Making the assumption that an earthquake, if it occurred, would on average occur mid-year, it is then assumed that these costs are distributed $1.5 million in the first year, and $1.0 million in the next year.
Assume that the cost of erecting a temporary Bailey Bridge is $0.2 million spread over six months, the disruption cost during planned replacement of the bridge is zero (the old bridge remains open), and the disruption cost of unplanned delays while the Bailey is being constructed is $0.5 million and disruption during Bailey use (during the 2 years it takes to construct the new bridge) is $0.2 million per year.
If the bridge is destroyed before planned replacement, then the costs at the start of the year in which the earthquake occurs are:
| Roading costs: | $million | ||
| Bailey bridge | $0.1 × 0.9535 | (SPPWF yr 0.5) | |
| $0.1 × 0.9091 | (SPPWF yr 1.0) | ||
| Permanent replacement bridge | $1.5 × 0.9091 | (SPPWF yr 1.0) | |
| $1.0 × 0.8264 | (SPPWF yr 2.0) | ||
| total | $2.376 million | ||
| Road user costs: | |||
| Initial disruption costs | $0.5 × 0.9535 | (SPPWF yr 0.5) | |
| $0.2 × 0.5 × 0.9091 | (SPPWF yr 1.0) | ||
| Ongoing disruption costs | $0.2 × 0.8668 | (SPPWF yr 1.5) | |
| $0.2 × 0.5 × 0.8264 | (SPPWF yr 2.0) | ||
| total | $0.83 million | ||
where: SPPWF is the single payment present worth factor.
Calculating expected values
The probability of the bridge being destroyed by an earthquake in each of years 1, 2, 3 and 4 are then multiplied by the above costs and benefits to give expected values in each year. The same is done in year 5 for the costs of planned replacement of the bridge. The expected values of costs and benefits in each year are then as follows:
| Year | Probability | Costs | Benefits | Expected value (costs) | Expected value (benefits) |
|---|---|---|---|---|---|
| 1 | 0.005000 | 2,376,000 | −830,000 | 11,880 | −4,150 |
| 2 | 0.004975 | 2,376,000 | −830,000 | 11,821 | −4,129 |
| 3 | 0.004950 | 2,376,000 | −830,000 | 11,761 | −4,109 |
| 4 | 0.004925 | 2,376,000 | −830,000 | 11,702 | −4,088 |
| 5 | 0.004901 | 2,376,000 | −830,000 | 11,645 | −4,068 |
| Year 5 replacement | 0.975250 | 2,190,000 | 2,136,000 |
Remaining calculations
The above costs and benefits are effectively discounted to the start of each year and each must be further discounted by the SPPWF factor for (year - 1).
The example does not take account of any benefits that may arise from bridge replacement such as a reduction in annual maintenance costs, road user benefits from improved alignment or reduction in bridge loading restrictions. These should be dealt with in a similar way, by discounting future costs and benefits to the start of each year 1 to 5 and then multiplying by the probability of loss of earthquake occurrence to give expected values, which should then be further discounted to time zero.
