Pipeline Project
I am here to make the report for Pipeline Project.
a) i) Let me report the first problem. Here is the photo of the problem.
5m
Mountain
15m
Private ground
45m
Refinery
As we planned, cost for materials, labor and fees to run the pipeline across
BLM ground is $500,000 per mile. So if we plan to build the pipeline running west
for 5 miles, south for 15 miles and then east for 45 miles to the refinery, our cost
would be 5 m*$500,000 + 15 m*$500,000 + 45 m*$500,000 = $32,500,000.
ii) Let’s see the next problem. We can choose another way to reach the
refinery which is heading east through the mountain for 40 miles and the south for 15
miles to the refinery. Let’s see the picture.
40m
Mountain
15m
Private ground
Refinery
So the basic cost would be (45 m-5 m)*$500,000 + 15 m*$500,000 = $27,500,000.
But we have to pay the additional part. First, we have to pay for drilling the existing
mountain and that would be a one-time cost $3,000,000. And then we need to pay for
the environmental impact study before we are allowed to drill which is $420,000.
Also the study of impact will last for 6 mouths. If we delay the project by 6 mouths, it
would cost company another $120,000 per mouths. For this part, the payment will be
$3,000,000 + $420,000 + 6 months * $120,000 = $4,140,000. So the total cost of
this way will be $4,140,000 + $27,500,000 = $31,640,000.
b) i) In the third case, we will choose a closer way. We plan to build the pipeline
in the shortest distance crossing the private ground to the refinery. Here is the
sketch map. (The black line is the pipeline.)
Mountain
Private ground
15m
40m
According to the Pythagorean Theorem, the distance is 42.72 miles.
Refinery
15m
(152 + 402 ≈ 42.722 )
40m
Because we have to incur an additional $350,000 per mile right-of-way fees in the
private ground, plus the regular pipeline fees ($500,000 per mile), the cost will be
42.72 m *$350,000 + 42.72 m * $500,000 = $36,312,000.
ii) Now it is the fourth case. We have another plan which is straight across the
private ground for 15 miles, the staight to the refinery for 40 miles.
Mountain
15m
Private ground
40m
Refinery
As I mention before, company has to pay the additional $350,000 per miles in the
private ground. So the cost of the part in private ground will be
15 m * $350,000 + 15 m * $500,000 = $12,750,000. In addition, we need to pay
the regular payment when pipeline goes in the BLM ground ($500,000 per mile). For
this part, the cost will be 40 m * $500,000 = $20,000,000. So the total cost will be
$12,750,000 + $20,000,000 = $32,750,000.
c) The cases I listed above are some easy cases to work, but those
are not the most economical way to bulid the pipeline. So I have a
suggestion. And my suggestion has two ways to work. Here is the sketch
map.
Mountain
Private ground
②
①
Let me show you the idea ①.
Refinery
Mountain
Private ground
15m
X
40m
①
Refinery
I assume we are going to build X miles pipeline on the BLM ground (showed in the
sketch map). According to the Pythagorean Theorem, the length of pipeline which
will be built on the private ground will be √(1825 − 80x + 𝑥 2 ) miles.
15m
[152 + (40 − 𝑥)2 = 1825 − 80x + 𝑥 2 ]
(40 – x)m
So the cost function
C(x) = [√(𝟏𝟖𝟐𝟓 − 𝟖𝟎𝐱 + 𝒙𝟐 ) * (350,000 +
500,000)] + [x * 500,000] = 850,000√(𝟏𝟖𝟐𝟓 − 𝟖𝟎𝐱 + 𝒙𝟐 )
+ 500,000x.
According to the Fermat’s Theorem, we may find a
minimum Cost when C’(x) =0.
So C’(x) = [425,000/√(𝟏𝟖𝟐𝟓 − 𝟖𝟎𝐱 + 𝒙𝟐 )] *
(2x-80) + 500,000. If C’(x) =0, then [425,000/√(𝟏𝟖𝟐𝟓 −
𝟖𝟎𝐱 + 𝒙𝟐 )] * (2x-80) + 500,000 = 0. So x =
50.91089451179962 or x = 29.089105488200378. But x is not possible to be more
than 40. So x≈29 m. In conclusion, when we build 18.6 miles pipeline on the private
ground and 29 miles on the BLM ground as the sketch map showed, we get most
economical. The cost will be $30,310,914
The next one is idea ②
40m
Mountain
Private ground
15m
②
x
Refinery
As the same principle as idea ①, I assume we are going to build X miles pipeline on
the BLM ground (showed in the sketch map). According to the Pythagorean Theorem,
the length of pipeline which will be built on the private ground will be √(1825 −
30x + 𝑥 2 ) miles.
40m
(15-x)m
[402 + (15 − 𝑥)2 = 1825 − 30x + 𝑥 2 ]
We can get another cost function:
C(x) = [√(𝟏𝟖𝟐𝟓 − 𝟑𝟎𝐱 + 𝒙𝟐 ) * (350,000 + 500,000)] + [x * 500,000] =
850,000√(𝟏𝟖𝟐𝟓 − 𝟑𝟎𝐱 + 𝒙𝟐 ) + 500,000x.
According to the Fermat’s Theorem, we may find a minimum Cost when C’(x) =0.
So C’(x) = [425,000/√(𝟏𝟖𝟐𝟓 − 𝟑𝟎𝐱 + 𝒙𝟐 )] * (2x-30) + 500,000. If C’(x) =0,
then [425,000/√(𝟏𝟖𝟐𝟓 − 𝟑𝟎𝐱 + 𝒙𝟐 )] * (2x-30) + 500,000 = 0.
So x = 70.55555555555556 or x = -8.80952380952381.
According to the First Derivative Test,
-
C’(x)
+
-8.8
We know that when x greater than -8.8 C’(x) is
positive. It means when x greater than -8.8, the function is
increasing. But x is not possible to be more than 15 and less
than 0. So when x is smallest which 0 is, the cost will be the
most economical. But when x = 0, it will be the same
situation as b) i). So I combine this idea with b) i).
At last, let’s compare all of the cost:
a) i) $32,500,000
ii) $31,640,000
b) i) $36,312,000
ii) $32,750,000
c)
$30,310,914 (lowest)
So I highly recommend c) plan, it will help company save at least $1,330,000!
Reflection:
In the calculus class, I have learned many things about derivative and calculus. Now
I know Mean Value Theorem and l’Hospital’s Rule which are really powerful. After
taking this class, I learned how to analyse functions step by step, instead of just
finding their solution. Now I can draw a sketch graph of a more complex function. I
learned how to use calculus in the real life from this class. I have to admit, calculus is
a very useful tool in life. Just like what I have done above, I can use calculus to work
out what is the most economical solution of a project. And that can help people save a
lot of money. It is useful, isn’t it?
I am planning to study biology, I think how to get the critical number and draw a
graph of a function in calculus can help me in this area, because biology need a large
number of experiments to support a theory. Among the experiments, we have
statistical data and analyse the date. That is where I would use calculus.