Scheduling a Backlog of Oilwell Workovers
Authors: Brennan J.J. , Barnes J.W. , Knapp R.M. 
Published: December 2007

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There are few wells with excess productive capacity today. To maximize production from each well, a greater number of wells have been analyzed to ascertain if workovers are economically justifiable. Frequently, such decisions lead to a backlog of wells awaiting attention from workover rigs. Each day a well waits for workover means potentially immediate production loss. This amounts to potentially immediate production loss. This amounts to the difference between the current production level and the production level after workover completion. The decision has been made that the workovers will be performed. The only question remaining is the order in performed. The only question remaining is the order in which they will be performed.

When the number of wells with planned workovers is greater than the number of available rigs, the order or sequence in which the workovers are performed can have a significant impact on the total amount of oil produced from those wells during the workover period.

Currently, workovers are scheduled either by intuition or by following some accepted "rule of thumb." Such rules of thumb probably will result in a schedule that fails to maximize total production from all wells during the workover period. This situation is unfortunate, because efficient techniques for obtaining the optimum schedule already exist. nose techniques use only information that normally would be required for economical justification of the workover. They are subject to the following assumptions.Setup and workover times are independent of the order of the workover schedule. The setup times are included in the workover times. (Travel times between wells are assumed to be insignificant or independent of the schedule order.)Each rig begins operation at time zero and operates without idle time on one well at a time until all wells assigned to it are completed.Once started, each workover is performed to completion (no cancellations).Rigs essentially are identical in their capabilities.

The example below is a backlog of oilwell workovers that occurred in a field operated by a major oil company.

Well A B C D E F G H I

ti 5 6 16 5 7 11 5 1 5 pi 51 112 106 102 51 108 88 57 40 pi 51 112 106 102 51 108 88 57 40 Each well in the backlog example is designated by a capital letter and has two parameters associated with it. ti is the number of days needed to perform the well workover and pi is the expected increase in productivity in barrels per day resulting from the workover.

Because, during the workover period, we want to minimize the "loss" of potential oil production or to maximize the production of the nine wells, we should recognize first that there is a certain "unavoidable" loss. That unavoidable loss is the total barrels of potential production lost when the number of available rigs equals production lost when the number of available rigs equals the number of wells that need workover. In the example, nine rigs will yield a lost of Sigma p t = 5*51 + 6*112 + 16*106 + . . . + 1*57 + 5*40 = 5,375 bbl. Clearly, avoidable and unavoidable losses. 5,375 bbl. Further, the "total loss" is the sum of the avoidable and unavoidable losses.

Suppose only two rigs are available. How can we select an optimal schedule out of the 3,628,800 schedules or sequences that might be formed?

One scheduling strategy is to arrange the workovers by descending values of pi. This would be expected to increase the daily production quickly. Another strategy would be to schedule the wells by ascending values of ti. This results in performing the quickest jobs first. (In both rules ties are broken arbitrarily.) The wells, when ordered by descending pi, appear as B, F, C, D, G, H, A, E, and I with A arbitrarily placed before E, and when ordered by ascending t, appear as H, A, I, G, D, B, E, F, and C with A, I, G, and D ordered arbitrarily.

P. 1651