Estimation of financial needs and condition of infrastructure assets
2024-04-29
Intervention strategies
Intervention programs
The construction of a maintenance intervention program can be:
Example agency rules
If less than 5% of the surface area of a bridge requires an intervention according to the intervention strategy to be followed, no intervention is performed (to ensure interventions are not performed on very small parts of bridges).
If more than 5% of the surface area of a structure requires an intervention according to the intervention strategy to be followed, than interventions for all parts of the bridge that are in states 3, 4 and 5 are performed (to ensure a sufficient return period between interventions)
Step | Description |
---|---|
1 | Determine the time intervals and time period in which maintenance interventions are to be executed. |
2 | Estimate the percentage of each item in each state at \(t=0\). |
3 | Determine the probable maintenance interventions to be executed at t in which a maintenance intervention is to be executed without consideration of other items (based on actual or expected state of items). |
4 | Determine the maintenance interventions to be executed at each t in which a maintenance intervention is to be executed with consideration of other items (based on actual or expected state of items and agency rules). (Attention: Not covered in Excel spread sheet explicitly) |
5 | Estimate probable costs of maintenance interventions to be executed at time \(t\). |
6 | Estimate the probable state of each item at \(t= t+1\). |
7 | Repeat steps 2 to 7 (with t= t+ 1) until the probable states of all items, the maintenance interventions to be executed, and maintenance intervention costs are determined over the specified time period (a first intervention program exists). |
8 | Verify all constraints at time t, e.g., budget limit in each time interval. |
9 | Determine the maintenance interventions to be included in the maintenance intervention program so that all constraints are satisfied. |
Bridge identifier | Intervention program | intervention program |
1 | 1 | |
B1 | 7,29 | 22 |
B2 | 1,24 | 10 |
B3 | 1,23 | 9 |
\(\cdots\) | \(\cdots\) | \(\cdots\) |
\(\cdots\) | \(\cdots\) | \(\cdots\) |
Intervention Program 1
Intervention Program 2
Compare costs and effects on service overtime
Intervention Program 1
Intervention Program 2
Compare the condition evolution overtime
\[ B/C = \frac{(C_{IS-R}^l-C_{IS-O}^l)+C_t}{C_{IS-O}^l} \]
\[ B/C = \frac{(C_{IS-R}^l-C_{IS-O}^l)+C_t}{C_{IS-O}^l} \]
The benefit could be taken as the difference between the long term costs if the reference maintenance intervention strategy was followed and the long term costs if the optimal maintenance intervention strategy was followed
This would lead to all objects of the same type in the same state having the same B/C ratio.
It would be nice, perhaps, to have the larger objects prioritised before the smaller objects.
This can be done by adding the cost of the intervention on the object to the benefit. I agree it is not strictly a benefit but it is a way of prioritizing the large objects that will bring us the most benefit over the small objects – There are many other ways.
I challenge you to use a better one!
The cost could be taken as the long term costs if the optimal maintenance intervention strategy was followed
\(C_t\) : Cost in year \(t\)
\(C_{IS-O}^l\) : long term costs if selected maintenance intervention strategy followed
Bridge identifier | Intervention program 1 | Maintenance cost | B/C |
---|---|---|---|
B1 | 7, 29 | 100, 100 | 2.8, 2.8 |
B2 | 1, 24 | 100, 100 | 3.1, 3.1 |
B3 | 1, 23 | 100, 100 | 2.7, 2.7 |
Budget limit = 150 CHF per year
Means we can’t do B2 and B3 in year 1
The B/C tells us it is more beneficial to us to do B2, so we postpone B3
Bridge identifier | Intervention program 1 | Maintenance cost | B/C |
---|---|---|---|
B1 | 7, 29 | 100, 100 | 2.8, 2.8 |
B2 | 1, 24 | 100, 100 | 3.1, 3.1 |
B3 | 1, 23 | 100, 100 | 2.7, 2.7 |
Bridge identifier | Intervention program 1 | Maintenance cost | B/C |
---|---|---|---|
B1 | 7, 29 | 100, 100 | 2.8, 2.8 |
B2 | 1, 24 | 100, 100 | 3.1, 3.1 |
B3 | 2, 23 | 105, 100 | 2.8, 2.7 |
Budget limit = 150 CHF per year
Means we can’t do B2 and B3 in year 1
The B/C tells us it is more beneficial to us to do B2, so we postpone B3
Without constraint
With constraint
Compare the costs and effects on service overtime
Without constraint
With constraint
Compare the condition evolution overtime
Questions?
Hamed Mehranfar
Doctoral student
hmehranfar@ethz.ch
ETH Zürich
Institute of Construction and Infrastructure Management (IBI)
Chair of Infrastructure Management
HIL G 32.2, Stefano-Franscini-Platz 5
8093 Zürich