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Question 1 For the City of Hamilgaryton. Four road projects are given in Table 1. The current condition and replacement value of each asset is provided in Table 1. The condition levels are in a scale of 1 to 5, where 1 is considered best condition level and 5 is worst condition level. Table 1: Current condition and replacement value of each asset group Roads Current Conditon Current replacement value $400,000 $200,000 $300,000 $400,000 R1 R2 R3 R4 Four maintenance actions are considered: Replacement, Major repair, Minor repair, Preventive maintenance, Do nothing The cost associated with each maintenance action is provided in Table 2. Table 2: Cost table for the different maintenance actions (the cost is provided as a percentage of the current replacement value) Maintenance action Asset group Preventive Do nothing maintenance Minor repair Major repair Sewer 2% 15% 35% Road 0 10% 25% Water 0 2 % 15% 35% Bridges 0 2% 10% 25% Replacement 2% 100% 100% 100% 100% Each maintenance action improves condition level of the asset. For each condition level and maintenance action, and the corresponding improvement in the new condition level are provided in Table 3. If the values fall in between, you need to interpolate. For example, Table 3 shows that, if the asset is in condition state 5, replacement will bring it to condition level 1, whereas, minor repair will bring it to condition level 4. Table 3: Improvement in condition level due to the maintenance actions Maintenance action Condition level Preventive Do nothing Minor repair Major repair maintenance Replacement 11 2 1.5 2.5 3.5 4.5 21 3 1 4 For all assets R1 to R4 (Table 1), assume, inspection is done every five years, and if any of the maintenance actions are performed, the condition levels are improved as shown in Table 3. Assume that, the deterioration follows a linear function of time, Deterioration = 0.05t. Note, mathematically, the deterioration can go beyond condition level 5, however, it should be capped at 5. a. Compute the cost and new conditions for the five maintenance action (Do nothing, Preventive maintenance, Minor repair, Major repair, Replacement) decisions made at to, ts, and t1o'. b. By t1o, you should have about 125 decisions. To reach the 125 decisions in t10, you have taken different path (maintenance action and cost). Compute the lifecycle cost (LCC) for the 125 decision path for discount rate of 3%, 6% and 9%. When you plot the condition levels at year 10 and LCC, you should get a figure like this: $200,000 $400,000 $600 ODD $800,000 $1 OD DDD 1.200.000 $1,400.000 .. Proposed Condition. Condition Current Condition This plot should be combined for all roads, R1 to R4, combined in one graph. Discuss what a Pareto front. From the plot above, show the Pareto front. d. If condition level 3 is the desired LOS, determine the optimal decision action. If the LOS is set to conditions levels 2 and 3.5, which alternatives are the best options? For the optimal decision made for condition levels 3, 2 and 3.5, compute the cash flow diagram for the decisions taken at to, ts, and t1o. The condition given is in Table 1 is for to. This condition can be considered the do nothing action. Then you need take you can apply the Preventive maintenance, Minor repair, Major repair, Replacement. 2 Note, in principle, at any level the condition should not go below the desired LOS. However, since the inspection/assessment is done only on at to, ts, and t1o, it is possible it will go down. Do you your assessment only on to, ts, and t1o. Question 1 For the City of Hamilgaryton. Four road projects are given in Table 1. The current condition and replacement value of each asset is provided in Table 1. The condition levels are in a scale of 1 to 5, where 1 is considered best condition level and 5 is worst condition level. Table 1: Current condition and replacement value of each asset group Roads Current Conditon Current replacement value $400,000 $200,000 $300,000 $400,000 R1 R2 R3 R4 Four maintenance actions are considered: Replacement, Major repair, Minor repair, Preventive maintenance, Do nothing The cost associated with each maintenance action is provided in Table 2. Table 2: Cost table for the different maintenance actions (the cost is provided as a percentage of the current replacement value) Maintenance action Asset group Preventive Do nothing maintenance Minor repair Major repair Sewer 2% 15% 35% Road 0 10% 25% Water 0 2 % 15% 35% Bridges 0 2% 10% 25% Replacement 2% 100% 100% 100% 100% Each maintenance action improves condition level of the asset. For each condition level and maintenance action, and the corresponding improvement in the new condition level are provided in Table 3. If the values fall in between, you need to interpolate. For example, Table 3 shows that, if the asset is in condition state 5, replacement will bring it to condition level 1, whereas, minor repair will bring it to condition level 4. Table 3: Improvement in condition level due to the maintenance actions Maintenance action Condition level Preventive Do nothing Minor repair Major repair maintenance Replacement 11 2 1.5 2.5 3.5 4.5 21 3 1 4 For all assets R1 to R4 (Table 1), assume, inspection is done every five years, and if any of the maintenance actions are performed, the condition levels are improved as shown in Table 3. Assume that, the deterioration follows a linear function of time, Deterioration = 0.05t. Note, mathematically, the deterioration can go beyond condition level 5, however, it should be capped at 5. a. Compute the cost and new conditions for the five maintenance action (Do nothing, Preventive maintenance, Minor repair, Major repair, Replacement) decisions made at to, ts, and t1o'. b. By t1o, you should have about 125 decisions. To reach the 125 decisions in t10, you have taken different path (maintenance action and cost). Compute the lifecycle cost (LCC) for the 125 decision path for discount rate of 3%, 6% and 9%. When you plot the condition levels at year 10 and LCC, you should get a figure like this: $200,000 $400,000 $600 ODD $800,000 $1 OD DDD 1.200.000 $1,400.000 .. Proposed Condition. Condition Current Condition This plot should be combined for all roads, R1 to R4, combined in one graph. Discuss what a Pareto front. From the plot above, show the Pareto front. d. If condition level 3 is the desired LOS, determine the optimal decision action. If the LOS is set to conditions levels 2 and 3.5, which alternatives are the best options? For the optimal decision made for condition levels 3, 2 and 3.5, compute the cash flow diagram for the decisions taken at to, ts, and t1o. The condition given is in Table 1 is for to. This condition can be considered the do nothing action. Then you need take you can apply the Preventive maintenance, Minor repair, Major repair, Replacement. 2 Note, in principle, at any level the condition should not go below the desired LOS. However, since the inspection/assessment is done only on at to, ts, and t1o, it is possible it will go down. Do you your assessment only on to, ts, and t1o

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