[128shot]

Alright.


How do you plug the numbers in to all these calculations to get these estimated numbers on production?

[gt1370a]

If it were as simple as "plugging some numbers into some equations" I doubt there would be so much controversy about this.

The way I have read it, start with one oil-producing country. Plot their historical production data. Make an educated guess about their total Ultimate Recoverable Reserves URR (lot of controversy in this, take proved reserves, add "possible" reserves for some unknown future oil price). Use historical data and URR to fit a curve (some controversy here too about what kind of curve is appropriate - gaussian, etc.). Then, repeat for every other oil-producing country in the world. Then figure out a way to build in tar sands LNG, etc. Sum it all up and you have a big curve.

[pup55]

As a public service, I will walk through one of these calculations so you can see how to do it.

Here are the “best fit” variables for the 2005 verhulst curve that we did awhile back:

n 0.149454128
qinf 2,461.82
t(1/2) 106.8467633
k 0.039628337


q-inf is the estimate of the ultimate remaining reserves URR.

T-1/2 is the best fit estimate of the year in which 50% of the URR will be gone (close to the peak, but not exactly). Year 1 is 1900, so therefore year 106.8 is somewhere toward then end of 2006.

N and K are the fudge factors that make the system work.

The equation we put up computes the remaining reserves for each year. Let’s do it for year 1 first:

1. Compute t minus t-1/2: t is 1, so t minus t(1/2) is -105.84
2. Compute tau. Tau equals 1/k, so since k= .0396… per the above, tau is always going to equal 25.23447
3. Compute t minus t-1/2 divided by tau. We take the value from step above, and divide it by the value from step 2. The result for the first year is 4.1945. The result from the second year will be different because t will be equal to 2.
4. compute 2 to the nth power. N is the variable above, two to the nth power is 1.091. It is going to be the same every year.
5. Compute 2 to the nth power minus 1. It is .091, and it is also going to be the same every year.
6. Compute (exp) of (t-t(1/2 divided by tau). We computed t-t(1/2) divided by tau in step (3) above. Excel will compute the exponential function. (exp). It will be different every year. For the first year, it is .01507881
7. Multiply 2 to the Nth minus 1 (which we computed in 4 above) by exp (t-t(1/2) divided by tau) which we computed in 6 above. For the first year, this will be .0016457 It will change every year, because t changes every year.
8. add one to the result of step 7. For the first year, this will be 1.0016457
9. Take the result of step 8, and raise it to the power of one over K. K is in the table above. The excel function for this is: power( x, 1/k) where x is the cell where you have the result of step 8, and k is the cell where you have k.
10. Next, take q-inf from the table above, and divide it by the result of step 9. This gives you the remaining reserves after the first year. The result of step 9 is the fraction of the reserves that were used during the first year. The result for the first year is 2434.882. This is how many gb of oil were left in the ground at the end of the first year.


So, now let’s figure out how much oil was left after the second year: t is now equal to 2:

1. t munus t-1/2 = -104.847. t is now equal to 2, so the result is 1 higher.
2. 25.23447 same as last year.
3. T minus t-1/2 divided by tau: 4.1549, a little different because t=2.
4. 2 to the nth: 1.091
5. 2 to the nth minus one: 0.091 just like last year.
6. Exp (t-t1/2 divided by tau): 0.015687315
7. 2 to the nth minus one, times exp (t-t1/2 divided by tau): 0.0017123
8. the result of step 7 plus one: 1.00171
9. The result of step 8 raised to the 1/k power: 1.01151
10. The remaining reserves at the end of year 2, Q-inf divided by the result of step 9: 2433.800

So for the production for year 2, easy enough to subtract the amount of reserves at the end of year 1 from the amount of reserves at the end of year 2, it’s 1.08 gb. That’s the production value for year 2. It’s the amount that had to have been removed from the ground during year 2.

So now you do this for all of the years, which is relatively easy using excel provided you use the $ for absolute references so when you copy the equations you divide by the same constants every year, and you get the points that are used to draw these curves.

What you should see is that as t, the year, gets bigger and bigger, the first term gets less and less negative, and the production fraction, step 9, gets bigger and bigger every year. When t finally gets past 106.8, the production fraction gets smaller and smaller, which is the phenomenon of shrinking production after the peak is reached.

When you get clear out into the future, the function approaches zero (but never gets to zero).

The way that n, k, q-inf and t-1/2 are arrived at is that we took the historical production for all of the years between 1900 and 2004 and used the “solver” function to do trial and error until the software found the values for the variables that when put into the equation, gave the closest fit with the historical data. A lot of the conversation in this forum is to try to figure out which equation does the best job of fitting the data points for the historical production, and more importantly, whether or not any of these methods is particularly useful at trying to predict the future.

Anyway this is the calculation. Once you have the production for all of the years, it is easy to produce a graph. If you have any issues with using excel, or another spreadsheet program, print this post out, and ask your prof or teacher or a local nerd to give you a hand.

The way my spreadsheet works is that I have it set up so that you can change the variables and that will change the appearance of the curve, so I can experiment with different variables and do play time with the shape of the curve.