Drilling time and cost

Quick refresher

On a days versus depth curve, the sloping stretches show the bottom of the hole getting deeper. The horizontal stretches are time without deepening: trips (pulling the drill string out and running it in), casing, logging, pressure tests, and also stops caused by complications (problems in the hole) and repairs. Even without complications the bit is on bottom for only about half the time. In well A-2 it is 53.1% of the productive time and 45.5% of the calendar time.
We divide the calendar time of drilling into productive time and non-productive time (NPT: repairs, complications, accidents and standby, which means waiting). Invisible lost time is slow work inside productive operations. We can see it only by comparing with the best performance or with the technical limit.
Commercial speed is the metres drilled per rig-month (720 h), counted over the whole calendar time of drilling and casing. Technical speed is the same, counted over the productive time only. Their ratio equals the share of productive time.
The cost estimate has three parts. Time-dependent costs are the spread rate (the daily cost of the rig and the services) multiplied by the days. Depth- and material-dependent costs are pipe, bits, mud and cement. Lump-sum costs are the site, mobilisation (moving the rig and crew in), logging and completion. Speeding up the work reduces only the first part.
The cost of a well is skewed to the right (it has a long tail of high costs). A well rarely costs less than planned, and then only a little. It often costs more, and sometimes by a lot. So the mean is higher than the median. We calculate the contingency from a risk register (a list of the events that may happen) and we state its level. The P90 of cost is the sum that will not be exceeded with 90% probability.
On the learning curve, each new well of a series cuts the excess over the technical limit by the same factor. We decide whether to complete a well or abandon it by the extra cost that this choice adds. The money already spent on drilling is not part of it.

Key relationships

v_\text{com} = \dfrac{720\,H}{T_\text{cal}}, \qquad v_\text{tech} = \dfrac{720\,H}{T_\text{prod}}

commercial and technical speed, m/rig-month; footage in metres, time in hours

C = C_\text{lump} + r\,T + \sum_j c_j L_j

the estimate: lump-sum costs, the spread rate times the number of days, the unit price times the quantity

\mathrm{E}[\Delta C] = \sum_k p_k\,\Delta C_k

the mean contingency from a register of independent risks

t_n = C_1\,\e^{C_2(1-n)} + C_3

the learning curve: the duration of the $n$ -th well of a series

Common pitfalls

Explaining the high cost of a well only by slow drilling. A large part of the money goes on time without deepening and on complications.
Counting only accidents as non-productive time. It includes repairs, complications and waiting too.
Adding a “10% for everything” contingency without knowing which risks it covers.
Confusing the P90 of cost with the P90 of hydrocarbon volumes. The P90 of cost is a high estimate: with 90% probability it will not be exceeded. The P90 of volumes is a low estimate: with 90% probability it will be exceeded.
Comparing wells by cost per metre without correcting for depth, casing programme and the amount of evaluation.
Counting the money already spent on drilling in the decision to complete a well.

After this lesson you can

draw and read a days versus depth curve: the stretches where the hole gets deeper, the time without deepening, and the stops caused by complications
divide the time of a well into productive time, non-productive time and invisible lost time, and calculate the mechanical, technical and commercial drilling speed
estimate how long a new well will take from offset wells (nearby wells already drilled), correcting for depth and allowing for non-productive time
prepare a well cost estimate with time-dependent, depth- and material-dependent and lump-sum costs, split into tangible and intangible; find the cost per metre
estimate a contingency from a risk register and read the P50 and P90 estimates of cost
describe the learning curve in a series of wells, and decide “complete or abandon” by the extra cost that the decision adds

The days versus depth curve

The clearest picture of well construction is the days versus depth curve. The horizontal axis shows the days from the start of drilling (spud). The vertical axis shows the depth of the bottom of the hole, measured along the hole. Depth increases downward, as in a geological section. The plan is drawn from the drilling programme, and the actual record from the daily reports. Both lines are kept on one sheet, so any divergence is seen at once.

The curve has two kinds of stretches. On sloping stretches the bottom of the hole gets deeper. This is mechanical drilling, with connections (adding a new stand of pipe to the string). The steeper the stretch downward, the faster the drilling. Horizontal stretches are time without deepening, called flat time in English-language practice. They include trips to change the bit, circulating (pumping mud to clean the hole), well logging, running and cementing casing, waiting on cement (WOC), and installing and pressure-testing the blowout preventers (BOP). The longest “shelves” are at the casing setting depths, because the casing work and the preparation for the next section happen there.

Shelves also appear in the middle of a section, where the plan has none. This is time that was not planned: repairing equipment, dealing with lost circulation (mud flowing away into the rock) or stuck pipe, or waiting on materials. We separate it from the planned time and analyse it on its own, in the next section.

CalculatedThe days versus depth curve of the offset well A-2, 2800 m deep. Its casing programme is the one in the lesson on well construction. The solid line is the actual record. The dashed line is the same sequence of work without non-productive time. The numbers mark the stops that were not in the plan, explained in the frame of the chart. The last shelf is completion and testing. The data in this example are assumed for illustration.

Worked exampleSlope and shelves of the curve of well A-2

Given (assumed data)

Casing completed: surface casing — on day 3.12 (600 m), intermediate casing — on day 13.90 (1900 m)
Interval 600–1900 m: mechanical drilling 142 h and connections 3.8 h; besides these there were trips, logging, casing, a BOP pressure test and a stop to repair the top drive (16 h)

Solution

Time of the interval from the curve: $13.90 - 3.12 = 10.78$ d, or 258.7 h.
Average rate of deepening over the interval: $1300/10.78 = 120.6$ m/d.
Rate on the sloping stretches: $1300/(142 + 3.8)$ $= 1300/145.8 = 8.92$ m/h, or 214 m/d.
Share of the interval's time when the bottom was getting deeper: $145.8/258.7 = 56.4$ %; the rest is shelves: trips, logging, casing, BOP and repair.

Answer: On average the bottom got deeper by 120.6 m/d, although the rate while drilling was 214 m/d. For 43.6% of the time in the interval the well was not getting deeper.

What it means. Most of the shelves are necessary work: a well cannot be built without trips, casing and pressure tests. But this is often where a reserve of time hides. Running the casing, testing the BOP and circulating can be done faster or slower, and the report counts all of this as productive time. Doubling the drilling speed is a job for bits and drilling parameters (Bit performance and cost per metre). Cutting the shelves by a third is often easier. The data in this example are assumed for illustration.

Where the time goes: productive, non-productive, invisible losses

To see where the time went, we break it down by type of work. In Russian-language practice this is the calendar time balance of drilling. It lists the time of mechanical drilling, trips, casing and auxiliary work, and, separately, the time of repairs, complications, accidents and standby. Preparatory work, rig-up and the testing of the well are not part of the drilling balance. They are counted in the full cycle of well construction.

Productive time: Work without which a well cannot be built: drilling with connections, trips, casing, logging, pressure tests. It is part of the plan.
Non-productive time (NPT): Time that was not in the plan and did not advance construction. It includes equipment repair. It includes dealing with complications: lost circulation, hole collapse, and bit balling (sticky clay packing around the bit). It includes dealing with accidents: stuck pipe (pipe that cannot be moved), tool breakage and tools left in the hole. It also includes waiting on materials, transport, weather or a decision.
Invisible lost time (ILT): Time that could be saved inside productive operations. Stands are pulled more slowly than they could be, a connection takes 8 minutes instead of 4, or a pressure test lasts longer than necessary. The report counts this time as productive, so we can see it only by comparing with the best performance. It is also called hidden NPT, although it is not part of the balance of non-productive time.
Technical limit and best composite well: The technical limit is the shortest time in which a well can be built if every operation is executed flawlessly with the available means. Specialists estimate it by analysing the operations. Closer to practice is the best composite well: the sum of the best times of each section or operation in a series of similar wells.

Mechanical, technical and commercial drilling speed

v_\text{mech} = \frac{H}{t_\text{bot}}, \qquad v_\text{tech} = \frac{720\,H}{T_\text{prod}}, \qquad v_\text{com} = \frac{720\,H}{T_\text{cal}}

where:

$H$ footage: the metres drilled by the well (or by a group of wells), m
$t_\text{bot}$ time of mechanical drilling, with the bit on the bottom, h
$T_\text{prod}$ productive time of drilling and casing, h
$T_\text{cal}$ calendar time of drilling and casing: productive plus non-productive, h
$720$ hours in a rig-month: 30 days of work of one rig, 24 h each, h
$v_\text{mech}$ mechanical speed, the rate of penetration (in the lesson Bit performance and cost per metre, the same quantity for one bit run is $\bar R$ ), m/h
$v_\text{tech},\ v_\text{com}$ technical and commercial speed, m/rig-month

Units: metres and hours; speeds in m/h and metres per rig-month (m/rig-month)

Holds when

$T_\text{prod}$ and $T_\text{cal}$ do not include preparatory work, rig-up or testing. A speed that includes them is called the cycle speed.
For a group of wells, use the total footage and the total time.
The trip-averaged rate of penetration describes one bit. It is the footage of a bit run divided by the drilling and tripping time of that run (measures of bit performance).

The mechanical speed tells us about the bit and the drilling parameters. The technical speed tells us how well the productive work is organised. The commercial speed covers everything, including accidents and standby. The ratio of the commercial speed to the technical speed is $T_\text{prod}/T_\text{cal}$ , the share of productive time. So the gap between the two speeds is the price of NPT.

Time balance of well A-2: drilling and casing to 2800 m

Type of work: Mechanical drilling (including coring)
Hours: 325.0
Share, %: 45.5

Type of work: Connections
Hours: 8.2
Share, %: 1.2

Type of work: Trips, circulation, wiper trips
Hours: 75.0
Share, %: 10.5

Type of work: Casing: running the strings, cementing, WOC
Hours: 120.0
Share, %: 16.8

Type of work: Auxiliary: BOP, drilling out, logging
Hours: 84.0
Share, %: 11.8

Type of work: Productive time
Hours: 612.2
Share, %: 85.7

Type of work: Repair of the top drive
Hours: 16.0
Share, %: 2.2

Type of work: Lost circulation (complication)
Hours: 44.0
Share, %: 6.2

Type of work: Stuck pipe (accident)
Hours: 30.0
Share, %: 4.2

Type of work: Waiting on casing pipe (standby)
Hours: 12.0
Share, %: 1.7

Type of work: Non-productive time
Hours: 102.0
Share, %: 14.3

Type of work: Calendar time
Hours: 714.2
Share, %: 100

Type of work	Hours	Share, %
Mechanical drilling (including coring)	325.0	45.5
Connections	8.2	1.2
Trips, circulation, wiper trips	75.0	10.5
Casing: running the strings, cementing, WOC	120.0	16.8
Auxiliary: BOP, drilling out, logging	84.0	11.8
Productive time	612.2	85.7
Repair of the top drive	16.0	2.2
Lost circulation (complication)	44.0	6.2
Stuck pipe (accident)	30.0	4.2
Waiting on casing pipe (standby)	12.0	1.7
Non-productive time	102.0	14.3
Calendar time	714.2	100

Completion and testing (96 h) are not part of the drilling balance. Connections are taken at 5 min per 28.35 m stand, as in the lesson Bit performance and cost per metre. The shares are rounded, so their sums may differ by 0.1. The data in this example are assumed for illustration.

Worked exampleThe time balance, speeds and invisible losses of well A-2

Given (assumed data)

Balance: mechanical drilling 325 h, connections 8.2 h, trips 75 h, casing 120 h, auxiliary work 84 h; NPT 102 h (table above)
Footage: 2800 m
Best composite well of the area: 23.45 d of productive time against 25.51 d for A-2 (the table of offset wells in the next section)
Spread rate of the rig and services: US$45 thousand/d (section on the cost estimate)

Solution

Productive time: $325 + 8.2 + 75 + 120 + 84 = 612.2$ h. With NPT, the calendar time of drilling and casing is $612.2 + 102 = 714.2$ h, or 29.76 d.
Shares: mechanical drilling — $325/714.2 = 45.5$ %, NPT — $102/714.2 = 14.3$ %.
Mechanical speed: $2800/325 = 8.62$ m/h. Technical: $720\cdot2800/612.2 = 3293$ m/rig-month. Commercial: $720\cdot2800/714.2 = 2823$ m/rig-month.
Check: $2823/3293 = 0.857$ — the share of productive time, $1 - 0.143$ .
Invisible losses against the best composite well: $25.51 - 23.45 = 2.06$ d. Together with NPT ( $102/24 = 4.25$ d) that is 6.31 d, or $6.31\cdot45 \approx 284$ US$ thousand.

Answer: The commercial speed of well A-2 is 2823 m/rig-month and the technical speed is 3293. Mechanical drilling took 45.5% of the time. Against the best performance of the area, the well lost about 6.31 d, or about US$284 thousand.

What it means. NPT is a clear loss, but not the only one. Another 2.06 d went on work that counts as productive time. The best composite well is a benchmark, not a norm. Different crews drilled its sections in different conditions, and it is hard to repeat all the best results in a single well. We can compare speeds only for wells of similar depth and design. In shallow wells the commercial speed is higher simply because trips and casing take a different share of the time. The data in this example are assumed for illustration.

Estimating a new well from offset wells

We almost always estimate a new well from wells already drilled nearby. These are offset wells, with the same casing programme, a similar rock section and, if possible, the same rig. The closer the offsets, the fewer the corrections and the more reliable the estimate. For a new area with no offsets, we build the estimate operation by operation. We take the drilling time of each section from the expected mechanical speed. We take the time of trips, casing and evaluation from norms and from experience in similar conditions.

The order of an estimate from offsets

1
Select the offsets
Choose wells in the same area or neighbouring ones, with the same casing programme and setting depths and a similar rig. Record the differences: a different mud, bits, contractor or amount of evaluation.
2
Break down the time
Break the time down by sections and operations, and separate the non-productive time. Carry the productive time over to the new well. Estimate the NPT separately, as a risk.
3
Adjust to the new well
Recalculate the drilling time for the metres of the new well. Recalculate the time of trips, casing and evaluation for its strings and programme. Allow for the depths, coring and the amount of logging.
4
Choose the level
For the cost estimate, take the mean or the median of the offsets. For the target set for the crew, take the best composite well or the technical limit.
5
Add the risks
Add the NPT, taken from its share in the offsets or from a risk register (see the section on contingency). Then convert days into money using the cost estimate.

Offset wells: productive time by section and NPT, d

Well: A-1
0–600 m: 3.50
600–1900 m: 9.40
1900–2800 m: 10.90
NPT: 2.10

Well: A-2
0–600 m: 3.95
600–1900 m: 9.78
1900–2800 m: 11.78
NPT: 4.25

Well: A-3
0–600 m: 3.75
600–1900 m: 9.05
1900–2800 m: 11.30
NPT: 2.30

Well	0–600 m	600–1900 m	1900–2800 m	NPT
A-1	3.50	9.40	10.90	2.10
A-2	3.95	9.78	11.78	4.25
A-3	3.75	9.05	11.30	2.30

All three have the same casing programme, core was cut in each, and two suites of logs were run. The values for A-2 are from its time balance. The completion time (4.0 d) is not included in the table. The data in this example are assumed for illustration.

Worked exampleDuration of well A-4, 2950 m deep

Given (assumed data)

New well: the same casing programme, planned depth 2950 m: the last section is 150 m longer
Addition for 150 m: drilling at the average speed of the last metres of the offsets, 5.0 m/h including connections; another 6 h for longer trips, wiper trips and casing running
Completion: 4.0 d, as in the offsets

Solution

Averages by section: $(3.50 + 3.95 + 3.75)/3 = 3.73$ ; $(9.40 + 9.78 + 9.05)/3 = 9.41$ ; $(10.90 + 11.78 + 11.30)/3 = 11.33$ d.
Addition for 150 m: $150/5.0 + 6 = 36$ h, or 1.50 d; the last section is $11.33 + 1.50 = 12.83$ d.
Productive time of drilling and casing: $3.73 + 9.41 + 12.83 = 25.97$ d; with completion, 29.97 d.
NPT of the offsets: on average $(2.10 + 4.25 + 2.30)/3 = 2.88$ d per $(23.80 + 25.51 + 24.10)/3 = 24.47$ d of productive time, that is 11.8%. For A-4 the allowance is $0.118\cdot25.97 = 3.06$ d. The expected duration is $29.97 + 3.06 = 33.03$ d.
Best composite well: $3.50 + 9.05 + 10.90 = 23.45$ d; with the same addition, 24.95 d of productive time of drilling and casing.

Answer: The base estimate for A-4 is 25.97 d of drilling and casing without NPT (29.97 d with completion). Allowing for the NPT of the offsets, it is about 33.03 d. The target for the crew is 24.95 d.

What it means. The base estimate goes into the cost estimate, and the NPT goes into the contingency. This shows how much the work itself costs and how much the risks cost (see the section on contingency). Three offsets are a small sample. One unlucky well shifts the mean noticeably, so we also examine the causes of the NPT and ask whether they will happen again. The target from the best composite well is 1.02 d shorter than the mean estimate. Better organisation of the work can give that much without any new technology. The data in this example are assumed for illustration.

The cost estimate of a well: what depends on time, what on depth

It helps to divide the costs of a well by what makes them grow. Time-dependent costs are charged for each day while the rig works on the well. They include the rate of the drilling contractor for the rig with its crew, and everything else paid by the day: drilling fluids service, the mud logging unit, tool rental, the operator's supervisors, transport, the camp (housing for the crew) and fuel. Their sum is the spread rate of work on the well. Depth- and material-dependent costs grow with the metres of hole and casing: casing pipe, bits, mud materials, cement. Lump-sum costs depend neither on days nor on metres: the site and road, mobilisation and rig-up, wellhead equipment, and service operations paid per job, such as cementing, logging and perforating.

Cost of a well by type of cost

C = C_\text{lump} + r\,T + \sum_j c_j\,L_j

where:

$C$ cost of the well according to the estimate, US$ thousand
$C_\text{lump}$ lump-sum costs: the site, mobilisation, wellhead equipment, service operations paid per job, US$ thousand
$r$ spread rate: the rig with its crew and all the services paid by the day, US$ thousand/d
$T$ days of work on the well according to the time estimate, d
$c_j$ unit price of the $j$ -th material: a metre of casing, a bit, a tonne of cement, US$ thousand per unit
$L_j$ quantity of the $j$ -th material — metres of casing, pieces, tonnes, units

Units: any currency; in the example, thousand US dollars and days

Holds when

$T$ is the number of days from the time estimate. The drilling contract decides how the time of repairing the rig is paid.
The boundary between the types is not sharp: bits depend on both metres and time, and mud consumption depends on the volume of the hole and on losses.
Prices are those of the date of the estimate. For a programme lasting several years, we allow for rising prices (escalation).

Speeding up the work reduces only $rT$ . Changing the casing programme or the depth mainly affects $\sum c_j L_j$ . The set of jobs decides the lump-sum costs, and they hardly depend on how quickly the jobs are done.

The second way to divide the costs is into tangible and intangible costs. Tangible costs create property that stays in the well and keeps its value: casing pipe, the casing head, the Christmas tree (the valves on the wellhead), tubing and the packer. Intangible costs are work and materials that are used up: rig time, services, mud, cement and site preparation. The division comes from accounting. Depreciation (writing off the cost over time) and tax deductions depend on it, and the rules differ from country to country. We do not need it to estimate time and risk, but the estimate usually shows it.

In international practice the approved cost estimate of a well is called the AFE (authority for expenditure). In joint projects the partners approve it in proportion to their working interests (their shares in the project). The operator may exceed it only by an agreed percentage. For a larger overspend it prepares a supplementary AFE. In post-Soviet practice the estimate in the well construction project sets the cost of a well. In essence it is the same calculation of time, materials and lump-sum jobs.

Spread rate on well A-4

Item: drilling rig with crew
US$ thousand/d: 28.0

Item: drilling fluids service
US$ thousand/d: 3.0

Item: mud logging unit
US$ thousand/d: 1.5

Item: rental of tools and equipment
US$ thousand/d: 2.5

Item: operator's supervisors
US$ thousand/d: 1.5

Item: transport, camp, communications
US$ thousand/d: 5.0

Item: fuel
US$ thousand/d: 3.5

Item: Total
US$ thousand/d: 45.0

Item	US$ thousand/d
drilling rig with crew	28.0
drilling fluids service	3.0
mud logging unit	1.5
rental of tools and equipment	2.5
operator's supervisors	1.5
transport, camp, communications	5.0
fuel	3.5
Total	45.0

That is US$1875 per hour. The rates are assumptions of the example, not data on prices in Kazakhstan.

Cost estimate of the completed well A-4, US$ thousand

Item: Drilling to the planned depth and logging
Amount
Type of cost

Item: rig and service time
Amount: 1063.80
Type of cost: time

Item: site, road, pits
Amount: 250.00
Type of cost: lump sum

Item: mobilisation, rig-up and rig-down
Amount: 380.00
Type of cost: lump sum

Item: casing head
Amount: 70.00
Type of cost: lump sum, tangible

Item: cementing of three strings (service)
Amount: 90.00
Type of cost: lump sum

Item: logging, two suites
Amount: 190.00
Type of cost: lump sum

Item: coring and core analysis
Amount: 90.00
Type of cost: lump sum

Item: casing pipe for three strings
Amount: 235.20
Type of cost: depth and materials, tangible

Item: bits
Amount: 190.00
Type of cost: depth and materials

Item: drilling fluid materials
Amount: 300.00
Type of cost: depth and materials

Item: cement and additives for three strings
Amount: 70.00
Type of cost: depth and materials

Item: Production casing and completion
Amount
Type of cost

Item: rig and service time
Amount: 284.85
Type of cost: time

Item: 168 mm casing pipe
Amount: 177.00
Type of cost: depth and materials, tangible

Item: cementing (service)
Amount: 40.00
Type of cost: lump sum

Item: cement and additives
Amount: 40.00
Type of cost: depth and materials

Item: Christmas tree
Amount: 70.00
Type of cost: lump sum, tangible

Item: perforating, bringing in and testing
Amount: 150.00
Type of cost: lump sum

Item: tubing and packer
Amount: 110.00
Type of cost: depth and materials, tangible

Item: Total
Amount: 3800.85
Type of cost

Item	Amount	Type of cost
Drilling to the planned depth and logging
rig and service time	1063.80	time
site, road, pits	250.00	lump sum
mobilisation, rig-up and rig-down	380.00	lump sum
casing head	70.00	lump sum, tangible
cementing of three strings (service)	90.00	lump sum
logging, two suites	190.00	lump sum
coring and core analysis	90.00	lump sum
casing pipe for three strings	235.20	depth and materials, tangible
bits	190.00	depth and materials
drilling fluid materials	300.00	depth and materials
cement and additives for three strings	70.00	depth and materials
Production casing and completion
rig and service time	284.85	time
168 mm casing pipe	177.00	depth and materials, tangible
cementing (service)	40.00	lump sum
cement and additives	40.00	depth and materials
Christmas tree	70.00	lump sum, tangible
perforating, bringing in and testing	150.00	lump sum
tubing and packer	110.00	depth and materials, tangible
Total	3800.85

Casing pipe by the length of the strings: conductor 426 mm — 30 m at US$140/m; surface casing 324 mm — 600 m at US$100/m; intermediate 245 mm — 1900 m at US$90/m; production 168 mm — 2950 m at US$60/m. The prices and rates are assumptions of the example.

Worked exampleCost estimate of well A-4

Given (assumed data)

Time: 25.97 d of drilling and casing without NPT and 4.0 d of completion (the example above); running and cementing the production casing take 2.33 d
Spread rate: US$45 thousand/d (table above)
Other items: from the table of the estimate

Solution

Days by phase: drilling to the planned depth with logging — $25.97 - 2.33 = 23.64$ d; production casing and completion — $2.33 + 4.0 = 6.33$ d.
Time-dependent costs: $45\cdot23.64 = 1063.8$ and $45\cdot6.33 = 284.85$ US$ thousand.
Pipe: $30\cdot0.14 + 600\cdot0.10$ $+ 1900\cdot0.09 = 235.2$ US$ thousand for the three strings and $2950\cdot0.06 = 177$ US$ thousand for the production casing.
Phases: drilling with logging — 2929.0 US$ thousand, casing and completion — 871.85; the completed well — 3800.85 US$ thousand.
By type: time — 1348.65 (35.5%), lump-sum — 1330 (35.0%), depth and materials — 1122.2 US$ thousand (29.5%). Tangible — 662.2 US$ thousand (17.4%).
Cost per metre of the well: $3800.85/2950 = 1.288$ US$ thousand/m, or US$1288/m (US$393 per foot).

Answer: The base estimate of the completed well A-4 is about US$3.80 million. More than a third of it is rig and service time.

What it means. A day of NPT costs US$45 thousand for time alone. So we count the contingency for risks (the next section) first of all in days. The cost per metre, US$1288/m, is good for comparing similar wells, but not different ones. In a shallow well the lump-sum costs are divided among fewer metres, and a complex well has more strings and more evaluation. Do not confuse it with the cost per metre drilled in a bit run. That figure answers a different question: which bit to use and when to change it (cost per metre drilled). The data in this example are assumed for illustration.

Contingency for risks: the risk register, P50 and P90

The base estimate assumes that everything goes without complications. This does not always happen, and the estimate is wrong in a lopsided way. A well cannot be drilled much faster than planned, because the technical limit bounds the time from below. But lost circulation, stuck pipe or a long wait can add a day or a week. So the distribution of the cost of a well is skewed to the right. A well rarely costs less than planned, and then only a little. It often costs more, sometimes much more. The mean is therefore greater than the median.

The simplest contingency (a reserve for unplanned costs) is a percentage of the base, the same for all wells. It says nothing about where an overspend comes from or how to reduce it. A more exact way is to compile a risk register. It lists the events that may happen, their probabilities and their consequences in days and money. We take the probabilities from the history of the area, from the same analysis of the NPT of the offsets.

Cost of an event and the mean contingency from a risk register

\Delta C_k = r\,\Delta t_k + c_k, \qquad \mathrm{E}[\Delta C] = \sum_k p_k\,\Delta C_k

where:

$\Delta C_k$ extra cost if the $k$ -th event happens, US$ thousand
$r$ spread rate, US$ thousand/d
$\Delta t_k$ time lost, d
$c_k$ direct cost of the event: materials, fishing (recovering stuck or lost tools), tools left in the hole, US$ thousand
$p_k$ probability of the event on this well, fraction
$\mathrm{E}[\Delta C]$ mean (expected) overspend, US$ thousand

Units: thousand US dollars and days

Holds when

The events are independent: each happens at most once, and its consequences do not depend on the others. Linked events are combined, or they are given by a conditional probability. An example is lost circulation, after which stuck pipe is more likely.
The register is incomplete, so we add a contingency for what it does not cover.
Means add up whatever the relation between the events, but P90 values do not. We find the P90 of a sum from the whole distribution, by listing the combinations or by simulation (why P90 values are not added).

The mean contingency is the sum of the contributions of the events. It shows which event costs the most on average, so measures to reduce risk start with that event. But a mean overspend does not protect against a large one. For that we look at the upper estimates of the distribution.

Risk register of well A-4

Event: lost circulation in the 2250–2450 m interval
Probability: 0.30
Days: 3.5
Direct, US$ thousand: 80
If it happens, US$ thousand: 237.5

Event: stuck pipe in the 215.9 mm hole
Probability: 0.12
Days: 6.0
Direct, US$ thousand: 300
If it happens, US$ thousand: 570.0

Event: caving and hole tightening: reaming (opening the hole up again)
Probability: 0.30
Days: 1.5
Direct, US$ thousand: 0
If it happens, US$ thousand: 67.5

Event: waiting on materials and transport
Probability: 0.25
Days: 2.0
Direct, US$ thousand: 0
If it happens, US$ thousand: 90.0

Event	Probability	Days	Direct, US$ thousand	If it happens, US$ thousand
lost circulation in the 2250–2450 m interval	0.30	3.5	80	237.5
stuck pipe in the 215.9 mm hole	0.12	6.0	300	570.0
caving and hole tightening: reaming (opening the hole up again)	0.30	1.5	0	67.5
waiting on materials and transport	0.25	2.0	0	90.0

The last column is the days at US$45 thousand per day plus the direct costs. The probabilities and consequences are assumptions of the example, consistent with the NPT of the offsets.

Worked exampleContingency for well A-4

Given (assumed data)

Base estimate: US$3800.85 thousand (example above)
Risks: four independent events from the table above

Solution

Cost of the events: lost circulation $3.5\cdot45 + 80 = 237.5$ ; stuck pipe $6\cdot45 + 300 = 570$ ; reaming $1.5\cdot45 = 67.5$ ; waiting $2\cdot45 = 90$ US$ thousand.
Mean overspend: $0.3\cdot237.5 + 0.12\cdot570$ $+ 0.3\cdot67.5 + 0.25\cdot90$ $= 182.4$ US$ thousand. On average this is 2.72 d of NPT, close to the 2.88 d in the offsets.
Distribution: there are 16 combinations of events, and the most probable is none at all ( $0.7\cdot0.88\cdot0.7\cdot0.75 = 0.3234$ ). The cumulative probability (the chance that the overspend is this amount or less) reaches 0.570 at US$90 thousand; this is P50. It reaches 0.814 at 305 (P80), and 0.924 at US$570 thousand, where the pipe sticks (P90).
Estimate to P90: $3800.85 + 570 = 4370.85$ US$ thousand. The contingency is 15.0% of the base. The “10% for everything” rule would give US$380 thousand and would cover only 86% of outcomes.

Answer: The mean overspend is US$182.4 thousand, P50 is US$90 thousand and P90 is US$570 thousand. The estimate that will not be exceeded with 90% probability is about US$4.37 million.

What it means. The mean (182.4) is twice the median (90). In most outcomes the overspend is small, but a rare stuck pipe costs US$570 thousand and pulls the mean up. Its probability is only 0.12, yet it accounts for 38% of the mean overspend. It alone decides the P90: without stuck pipe the overspend would not exceed US$395 thousand. So it pays better to lower the probability of stuck pipe than to fight frequent but small delays. We can do this by choosing the mud, the drilling parameters and the assembly. The “10%” rule does not cover P90 here, and the risk register shows why. Company policy decides which level to approve: P50, P80 or P90. The higher the level, the less often supplementary AFEs are needed, but the more money is frozen in the contingency. The data in this example are assumed for illustration.

CalculatedThe distribution of the overspend from the risk register of well A-4. The curve shows the probability that the overspend will not exceed a given sum. P50 (US$90 thousand), the mean (182.4) and P90 (570) are marked. The step at zero is the probability that none of the events happens (32%).

The learning curve in a series of wells

When the same rig drills a series of similar wells in one area, each new well is usually built faster. The crew learns the rock section and its dangerous intervals. The drilling programme is refined with experience. The best practices pass from shift to shift, and the services and supply adjust to the rhythm. The gain is large at first and then gradually fades. The time approaches a limit set by the operations themselves.

The learning curve (Brett–Millheim model)

t_n = C_1\,\e^{C_2(1-n)} + C_3

where:

$t_n$ duration of the $n$ -th well of the series, d
$n$ number of the well in the series: 1, 2, 3 …, —
$C_1$ by how much the first well exceeds the limit: $t_1 = C_1 + C_3$ , d
$C_2$ rate of learning, —
$C_3$ the limit that the time approaches with full learning, d

Units: days

Holds when

The wells are of one type in one area, with one rig and one contractor. A break in the work, a change of crew or a change of casing programme sets the series back.
We fit the constants to the actual wells of the series. It is more reliable to estimate $C_3$ separately, from the technical limit or the best composite well (fitting the curve).
Non-productive time improves less with learning than productive time, so it is often analysed separately.

The formula gives $(t_{n+1} - C_3)/(t_n - C_3) = \e^{-C_2}$ . So each well reduces the excess over the limit by the same factor. For $C_2 = 0.5$ the excess falls by 39% per well, and for $C_2 = 0.2$ by 18%.

Worked exampleA series of eight production wells from one pad

Given (assumed data)

Series: after the discovery by well A-4, 8 production wells about 2800 m deep are drilled from one pad (a site with several wellheads) with the same rig. The time is counted from spud to the move to the next wellhead
First four wells: 22.4; 18.2; 16.0 and 14.9 d
Technical limit: $C_3 = 13$ d — an estimate by specialists from the operations for this casing programme

Solution

Logarithms of the excess over the limit $y_n = \ln(t_n - 13)$ : $\ln 9.4 = 2.241$ ; $\ln 5.2 = 1.649$ ; $\ln 3.0 = 1.099$ ; $\ln 1.9 = 0.642$ . The model says they lie on a straight line $y = \ln C_1 - C_2 (n - 1)$ .
We fit the line by least squares (the line closest to all the points), with $x = n - 1 = 0, 1, 2, 3$ : $\sum x = 6$ , $\sum x^2 = 14$ , $\sum y = 5.631$ , $\sum xy = 5.773$ . The slope is $(4\cdot5.773 - 6\cdot5.631)/(4\cdot14 - 36)$ $= -0.5347$ , whence $C_2 = 0.535$ .
The intercept: $\ln C_1 = (5.631 + 0.535\cdot6)/4 = 2.210$ , $C_1 = \e^{2.210} = 9.12$ d.
Forecast: $t_5 = 13 + 9.12\cdot\e^{-0.535\cdot4}$ $= 13 + 9.12\cdot0.1177 = 14.07$ d; likewise $t_6 = 13.63$ , $t_7 = 13.37$ , $t_8 = 13.22$ d.
The whole series: $22.4 + 18.2 + 16.0 + 14.9$ $+ 14.07 + 13.63 + 13.37 + 13.22$ $= 125.79$ d against $8\cdot22.4 = 179.2$ d if all the wells were drilled like the first.

Answer: The remaining four wells will take about 54.29 d. The whole series takes 125.79 d, which is 53.41 d less than without learning. That is about US$2.40 million at a spread rate of US$45 thousand/d.

What it means. Each well of the series removes $1 - \e^{-0.535} = 41$ % of the remaining excess over the limit. So after four wells there is almost nothing left to gain: the eighth is only 0.22 d longer than the limit. The forecast is only as good as the limit $C_3$ is right. With another limit, the same four points give other values of $C_1$ and $C_2$ , and another forecast (see the advanced section). The comparison with “drilling like the first well” shows the price of learning, but it is not a target. We make the estimate for the series from the curve, not from the first well. The data in this example are assumed for illustration.

CalculatedThe learning curve of the series: the points are the first four wells, the line is the Brett–Millheim curve with

C_1 = 9.12

C_2 = 0.535

and a limit of

C_3 = 13

d; the dashed lines are the limit and the time of the first well. The data in this example are assumed for illustration.

Dry hole and completed well

It is useful to divide the cost estimate of an exploration or appraisal well into phases, by the decision that is made at the bottom. Drilling to the planned depth, with logging, is paid for in any outcome. Then comes the decision. If the formation has met expectations, the crew runs the production casing and completes the well. If not, it abandons the well: it sets cement plugs and restores the site. The dry hole cost is drilling plus abandonment. The completed well cost is drilling plus casing and completion.

Complete or abandon

V_\text{well} > C_\text{compl} - C_\text{aband}

where:

$V_\text{well}$ the value of the well's production: the NPV (net present value, the value today) of its future cash flows, leaving out the costs already spent and the cost of completion, US$ thousand
$C_\text{compl}$ the cost of the production casing and completion, US$ thousand
$C_\text{aband}$ the cost of abandonment that will be avoided if the well is completed, US$ thousand

Units: thousand US dollars at the date of the decision

Holds when

The cost of drilling to the planned depth has already been spent and does not depend on the decision, so it is not part of the comparison (sunk cost).
The abandonment of a completed well at the end of its life is included in $V_\text{well}$ . It comes years later, so it weighs less after discounting (converting future money into today's value).
If the well is also needed for other purposes, such as injection or observation, add their value to $V_\text{well}$ .

We decide on the increment. Completion pays if the production of the well is worth more than the extra cost of completing it compared with abandoning it.

Abandonment of the dry hole A-4, US$ thousand

Item: rig and service time
Amount: 90.00
Type of cost: time

Item: cement plugs (service and materials)
Amount: 60.00
Type of cost: lump sum

Item: site restoration
Amount: 60.00
Type of cost: lump sum

Item	Amount	Type of cost
rig and service time	90.00	time
cement plugs (service and materials)	60.00	lump sum
site restoration	60.00	lump sum

2.0 d of rig work on cement plugs; the rates are assumptions of the example.

Worked exampleWell A-4 at total depth: complete or not

Given (assumed data)

Estimate by phase: drilling with logging 2929.0, casing and completion 871.85, abandonment 210 US$ thousand
Log result: the formation is thinner than expected; the NPV of the well's production without the cost of completion is US$1200 thousand (an assumption of the example)

Solution

The completed well: $2929.0 + 871.85 = 3800.85$ US$ thousand. The dry hole: $2929.0 + 210 = 3139.0$ US$ thousand, or 82.6% of the completed one.
The extra cost of completion over abandonment: $871.85 - 210 = 661.85$ US$ thousand.
Comparison: $1200 > 661.85$ , so we complete the well. The gain from the decision is $1200 - 661.85 = 538.15$ US$ thousand.

Answer: The well should be completed: the production, US$1200 thousand, pays for the increase in cost of US$661.85 thousand, although the whole well cost US$3800.85 thousand.

What it means. It is a mistake to compare US$1200 thousand with the full cost of US$3800.85 thousand. US$2929.0 thousand has already been spent, and abandoning the well does not bring it back. The well project as a whole will make a loss, but the decision to complete reduces the loss by US$538.15 thousand. The same logic applies in the economics of exploration. In a failure the company loses the cost of a dry hole, US$3139.0 thousand. The cost of completion belongs to the “success” outcome, as part of the development costs (EMV of an exploration well). The data in this example are assumed for illustration.

Terms and symbols

Days versus depth curve: The depth of the bottom of the hole against the days from the start of drilling. Sloping stretches are deepening; horizontal ones are time without deepening.
Time without deepening (flat time): Trips, casing, WOC, logging, pressure tests and completion: everything that happens while the bottom of the hole is not getting deeper.
Non-productive time (NPT): Repairs, complications, accidents and standby: time that was not in the plan.
Invisible lost time (ILT): Slow execution of productive operations, compared with the best performance or the technical limit.
Technical limit, best composite well: The technical limit is the shortest time with flawless execution. The best composite well is the sum of the best times of the sections in a series of similar wells.
Technical and commercial speed: Metres per rig-month (720 h), counted over the productive time and over the whole calendar time of drilling and casing.
Spread rate: The sum of all the costs charged for a day of work on the well: the rig with its crew and the services paid by the day.
AFE, supplementary AFE: The approved budget of a well, and the approved increase of it in case of an overspend.
Tangible and intangible costs: Tangible costs buy property that stays in the well (pipe, wellhead equipment). Intangible costs pay for work and consumable materials.
Contingency, P50 and P90 of cost: An addition to the base estimate for events that it does not account for. P50 and P90 are the sums that will not be exceeded with a probability of 50% and 90%.
Learning curve: The reduction of time from well to well in a series. The Brett–Millheim model is $t_n = C_1 \e^{C_2(1-n)} + C_3$ .
Dry hole cost: Drilling to the planned depth with evaluation, plus abandonment.

Common mistakes and misconceptions

Wrong

“The well is expensive because it was drilled slowly — we need better bits.”

In fact

Mechanical drilling takes less than half of the time (in well A-2, 45.5%). Drilling twice as fast would have shortened the well by 6.8 d out of 29.8. The shelves and the non-productive time offer just as large a reserve of time.

Wrong

“Non-productive time means accidents.”

In fact

NPT includes equipment repairs, complications and waiting for materials or a decision. In well A-2 only the stuck pipe was an accident, and it was less than a third of the time lost.

Wrong

“We put in a 10% contingency, as always.”

In fact

We calculate the contingency from the risk register. It shows which risk is the most expensive and lets us choose the level: P50, P80 or P90. A percentage “for everything” may match the calculation by chance, but it will not show how to reduce the overspend.

Wrong

“The P90 of an estimate is an optimistic figure, like the P90 of hydrocarbon volumes.”

In fact

For costs P90 is the sum that will not be exceeded with 90% probability: a high, cautious estimate. The convention for volumes is the opposite.

Wrong

“The production of the well is worth less than we have spent on it — let us abandon it.”

In fact

The money spent on drilling cannot be recovered whatever we decide. We compare the value of the production with the increment of cost: completion minus abandonment.

In depth: advanced detail

How the learning curve is fitted and why the limit is set separately

If the limit $C_3$ is known, the model becomes a straight line: $\ln(t_n - C_3) = \ln C_1 - C_2 (n - 1)$ . We find the two remaining constants by least squares, as in the example of the lesson. It is unreliable to fit all three constants at once to four or five wells. Curves that stay close to the data can be obtained with very different limits, and the forecast for the distant wells depends on the limit most of all.

Worked exampleThe same forecast with a limit of 11 days

Given (assumed data)

Wells of the series: 22.4; 18.2; 16.0 and 14.9 d, as in the example of the lesson
Limit: $C_3 = 11$ d instead of 13

Solution

$y_n = \ln(t_n - 11)$ : 2.434; 1.974; 1.609; 1.361. $\sum y = 7.378$ , $\sum xy = 9.275$ .
The slope is $(4\cdot9.275 - 6\cdot7.378)/20$ $= -0.3584$ , $C_2 = 0.358$ ; $\ln C_1 = (7.378 + 0.358\cdot6)/4 = 2.382$ , $C_1 = 10.83$ d.
Forecast: $t_5 = 13.59$ , $t_6 = 12.81$ , $t_7 = 12.26$ , $t_8 = 11.88$ d. The four wells together take 50.54 d against 54.29 with a limit of 13.

Answer: With a limit of 11 d the curve learns more slowly (

C_2 = 0.358

) but settles lower. The last four wells will take 50.54 d instead of 54.29.

What it means. Both curves pass close to the four wells. The largest deviation is 0.28 d for a limit of 13 and 0.57 d for a limit of 11, so four points cannot tell us which limit is right. Yet the forecasts differ by 3.75 d, about US$169 thousand. So we estimate the limit separately, by analysing the operations or from the best wells of the neighbours. We refine the forecast as the series continues. The data in this example are assumed for illustration.

Probabilistic estimation of time

The risk register describes rare events. But ordinary operations also vary in duration: the mechanical speed changes from well to well, and running a casing string goes now faster, now slower. A full probabilistic estimate gives each operation a distribution of duration. The duration is never shorter than the technical limit, and the distribution has a long tail to the right. The estimate adds the events from the risk register and combines everything by Monte Carlo simulation (many random trials; Probabilistic volumetrics by Monte Carlo simulation). The result is a distribution of time and cost, with P10, P50, P90 and the mean.

The links between operations matter. If a crew works slowly, all its operations are slow. If the rock is harder than expected, all the sections are drilled more slowly. Such links widen the distribution of the total: the P90 of a well with linked operations is higher than with independent ones. A model without links gives a deceptively narrow range.

Connections to other topics

How a well is drilled: the process and the people involved — the stages of well construction, the drilling programme and the types of drilling contract.
Bit performance and cost per metre — mechanical and trip-averaged speed, the cost per metre of a bit run and the moment to pull the bit.
Drilling rig: power and hoisting systems — tripping operations and the work of the hoisting system.
Primary cementing — what makes up the time of cementing and of waiting on cement.
Expected monetary value and preference theory — the cost of a dry hole in the evaluation of exploration.
Unit costs per barrel — how well costs enter the capital expenditure per barrel.
Risk reduction and diversification — how the risk of an overspend is passed to the contractor or an insurer and shared with partners.

Check yourself

1
On the actual days versus depth curve, a horizontal shelf of one and a half days has appeared in the middle of a section, far from a casing setting depth. It was not in the plan. What does it most likely mean?
1. A.The rock is soft at this point, and drilling was fast
2. B.Running the next casing string
3. C.An error in the depth measurement
4. D.Time without deepening that was not planned: a repair, a complication or waiting
Show the answer
Correct answer: D. Time without deepening that was not planned: a repair, a complication or waiting
A horizontal stretch is time when the bottom of the hole was not getting deeper. Planned shelves are at casing setting depths and at places where the bit is changed. A long shelf in the middle of a section that is not in the plan is almost always non-productive time. We look for its cause in the daily report.
2
A well 3400 m deep was drilled and cased in 42 calendar days, of which 5 days were non-productive time. Find the commercial and technical drilling speed.
Show the answer
The calendar time is $42\cdot24 = 1008$ h, and the productive time is $37\cdot24 = 888$ h. The commercial speed is $720\cdot3400/1008 \approx 2429$ m/rig-month. The technical speed is $720\cdot3400/888 \approx 2757$ m/rig-month. Their ratio $37/42 \approx 0.88$ is the share of productive time.
3
Which costs in a well estimate grow primarily with the number of days of work?
1. A.The rate of the rig with its crew and the services paid by the day
2. B.Mobilisation and rig-up
3. C.The Christmas tree
4. D.Casing pipe
Show the answer
Correct answer: A. The rate of the rig with its crew and the services paid by the day
The spread rate is charged for each day. It covers the rig with its crew, drilling fluids service, mud logging, tool rental, supervisors, transport and fuel. Pipe depends on the length of the strings. The tree and mobilisation are lump-sum costs.
4
A risk register includes an event with a probability of 0.2: it adds 4 d at a spread rate of US$50 thousand per day and US$100 thousand of direct costs. What is its contribution to the mean contingency? Why is the estimate to P90 usually larger than the base plus the mean contingency?
Show the answer
If the event happens, it costs $4\cdot50 + 100 = 300$ US$ thousand. Its contribution to the mean contingency is $0.2\cdot300 = 60$ US$ thousand. The mean contingency is the average over all outcomes. P90 is the sum that is exceeded in only one case out of ten. The distribution of the overspend is skewed to the right. Its upper part is made up of combinations of several events, or of one large event. So the P90 of the overspend is greater than the mean.
5
A learning curve has been fitted to a series of wells: $C_1 = 10$ d, $C_2 = 0.4$ , $C_3 = 15$ d. How long will the third well take, and what does the time tend to?
Show the answer
$t_3 = 15 + 10\,\e^{0.4\cdot(1 - 3)}$ $= 15 + 10\,\e^{-0.8}$ $\approx 15 + 4.49 = 19.5$ d. As the well number grows, the exponential tends to zero, and the time tends to the limit $C_3 = 15$ d. The first well takes $C_1 + C_3 = 25$ d.
6
A well has been drilled to the planned depth for US$2.6 million. Completion will cost another US$0.7 million and abandonment US$0.2 million. The NPV of the well's production without the cost of completion is US$0.6 million. What should be done?
1. A.Complete: 0.6 million is more than 0.7 − 0.2 = 0.5 million
2. B.Abandon: 0.6 million is less than the cost of completion, 0.7 million
3. C.Abandon: 0.6 million is less than 2.6 + 0.7 million
4. D.It makes no difference: the project is loss-making whichever is decided
Show the answer
Correct answer: A. Complete: 0.6 million is more than 0.7 − 0.2 = 0.5 million
The 2.6 million is already spent and does not depend on the decision. Completion adds $0.7 - 0.2 = 0.5$ US$ million to the costs compared with abandonment and brings 0.6 million — a benefit of US$0.1 million. The well as a whole makes a loss, but completion reduces the loss.

Updated: 28 October 2026