Jack Porter Jarrod Phillips Scott Simon Weon Kim.

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Project 2: Petroleum Industry Management Jack Porter Jarrod Phillips Scott Simon Weon Kim

Transcript of Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Page 1: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Project 2: Petroleum Industry Management

Jack PorterJarrod PhillipsScott SimonWeon Kim

Page 2: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Our group provides consulting services to a petroleum company. We are to advise them on how to meet the demands of their customers for motor oil, diesel oil, and gasoline. The company has three plants and has decided not to store any excess production.

From a barrel of crude oil, factory #1 can produce 20 gallons of motor oil, 10 gallons of diesel oil, and 5 gallons of gasoline. Factory #2 can produce 4 gallons, 5 gallons, and 12 gallons, respectively, of motor oil, diesel oil, and gasoline. Factory #3 can produce 4 gallons, 5 gallons, and 12 gallons of motor oil, diesel oil, and gasoline.

Introduction

Page 3: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

From the given specifications, we can set up a system of equations to represent the output of the factories:

Motor Oil: 20f1 + 4f2 + 4f3 = x1

Diesel Oil: 10f1 + 14f2 + 5f3 = x2

Gasoline: 5f1 + 5f2 + 12f3 = x3

Where f1, f2 , and f3 are the number of barrels of oil provided to each factory, and x1 , x2 , and x3 are the outputs of each product, respectively. We can then use these equations to find the amount of oil to give to each factory to meet a given demand.

Setting up a System of Equations

Page 4: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

The first demand to meet is 5000 gallons of motor oil, 8500 gallons of diesel oil, and 10000 gallons of gasoline. Solving the system of equations gives us the input for each factory in barrels of oil, as shown in the table below:

Situation 1: First Demand

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 5000

Diesel Oil 10 14 5 8500

Gasoline 5 5 12 10000

Barrels of Oil (exact) 48.75 331.25 675.00 1055

Barrels of Oil (approximate) 49 332 675 1056

Page 5: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Next, we are given the situation where the demand has doubled. The solution for the new system of equations is shown below:

It is observed that if demand doubles, then the barrels of oil doubles as well.

Situation 2: Doubling the First Demand

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 10000

Diesel Oil 10 14 5 17000

Gasoline 5 5 12 20000

Barrels of Oil (exact) 97.5 662.5 1350 2110

Barrels of Oil (approximate) 98 663 1350 2111

Page 6: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

We are now given a different set of demands to meet: 2000 gallons of motor oil, 4000 gallons of diesel oil, and 4000 gallons of gasoline.

Situation 3: Second Demand

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 2000

Diesel Oil 10 14 5 4000

Gasoline 5 5 12 4000

Barrels of Oil (exact) 12.5 187.5 250 450

Barrels of Oil (approximate) 13 188 250 451

Page 7: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

We now add the first and second demand to get 7000 gallons of motor oil, 12500 gallons of diesel oil, and 14000 gallons of gasoline.

The solution to the added demands is simply the solution of both demands added together. Each of these solutions is said to be unique because there is only one combination of inputs that allows the company to meet the required demand.

Situation 4: Adding the First and Second Demand

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 7000

Diesel Oil 10 14 5 12500

Gasoline 5 5 12 14000

Barrels of Oil (exact) 61.25 518.75 925 1505

Barrels of Oil (approximate) 62 519 925 1506

Page 8: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Sensitivity analysis is performed by varying coefficients in the production equations and noting how this affects the solution.

We performed sensitivity analysis on our factories by varying the output of the first factory by ±3% individually for each product and recalculating the output for each variation.

A system that does not change much for modest changes in parameters is said to be robust.

Sensitivity Analysis

Page 9: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20.6 4 4 5000

Diesel Oil 10 14 5 8500

Gasoline 5 5 12 10000

Barrels of Oil (exact) 47.06734 332.36785 675.23534 1054.67053

Barrels of Oil (approximate)

48 333 676 1057

Percentage Change -0.03%

Sensitivity Analysis

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil19.4 4 4

5000

Diesel Oil10 14 5

8500

Gasoline5 5 12

10000

Barrels of Oil (exact)50.55743 330.04926 674.74721 1055.3539

Barrels of Oil (approximate)

51 331 675 1057

Percentage Change0.03%

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Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 5000

Diesel Oil 10.3 14 5 8500

Gasoline 5 5 12 10000

Barrels of Oil (exact) 48.92122 329.90467 675.48921 1054.3151

Barrels of Oil (approximate)

49 330 676 1055

Percentage Change -0.06%

Sensitivity Analysis

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 5000

Diesel Oil 9.7 14 5 8500

Gasoline 5 5 12 10000

Barrels of Oil (exact) 48.57997 332.58595 674.51420 1055.68012

Barrels of Oil (approximate)

49 333 675 1057

Percentage Change 0.06%

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Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 5000

Diesel Oil 10 14 5 8500

Gasoline 5.15 5 12 10000

Barrels of Oil (exact) 48.8599 331.4332 674.2671 1054.5602

Barrels of Oil (approximate)

49 336 675 1060

Percentage Change -0.04%

Sensitivity Analysis

Factory 1 Factory 2 Factory 3 Demand/Total Barrels

Motor Oil 20 4 4 5000

Diesel Oil 10 14 5 8500

Gasoline 4.85 5 12 10000

Barrels of Oil (exact) 48.6406 331.0676 675.7296 1055.4377

Barrels of Oil (approximate)

49 336 675 1060

Percentage Change 0.04%

Page 12: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Plant Three Shut DownIn our next situation, factory #3 is shut down by the EPA. We need to find a way for the first two factories to meet the original demand. We can do this by using a least squares approximation. Let A be a 3 x 2 matrix representing the outputs of the first two factories and b the demand vector. A least squares approximation can be found for the input vector x with the following formula:

(AT*A) -1AT*b = x

This method gives an approximation for the inconsistent solution, but because it is an approximation, the estimate can be inexact. The table in the next slide demonstrates that in this case it is impossible for the two factories to produce exactly enough gasoline without severely overproducing diesel oil and motor oil.

Page 13: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Plant Three Shut DownFactory 1 Factory 2 Demand/Total Barrels

Motor Oil 20 4 5000

Diesel Oil 10 14 8500

Gasoline 5 5 10000

Barrels of Oil (exact) 145.8075 646.7391 792.5466

Barrels of Oil (approximate) 146 647 793

Motor Oil Diesel Oil Gasoline0

2000

4000

6000

8000

10000

12000

DemandApproximation

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Factory 1 Factory 2 Factory 3 Factory 4 Demand/Total Barrels

Motor Oil 20 4 4 4 5000

Diesel Oil 10 14 5 5 8500

Gasoline 5 5 12 12 10000

Barrels of Oil (exact) 48.75 331.25 675 - t t 1055

Barrels of Oil (approximation) 49 332 675 – t t 1056

Adding a 4th PlantThe situation with the EPA has caused enough concern that the CEO of our company is considering buying another plant identical to the third. We can create another system of equations for this situation and solve it for our original demand:

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Factory 1 Factory 2 Factory 3 Paraffin

Paraffin Produced 147 1660 1350 3157

Disposing of Paraffin WasteEach factory produces a certain amount of paraffin as waste per barrel of oil used. Specifically, factory #1 produces 3 gallons per barrel of oil, while factory #2 produces 5 gallons, and factory #3 produces 2 gallons.

Supposing our petroleum company finds a candle company that will purchase the waste paraffin, we can use the amount of oil used for the original demand to determine how much paraffin can be supplied to the candle company.

Page 16: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Due to aging equipment and high labor costs, the company is considering selling factory #1. They want to know how this would affect production capability.

Reverting to two factories would prevent the company from meeting the initial demand. This is because the initial demand is not a linear combination of the output of the two factories.

The company could again attempt to approximate production, but as shown earlier, approximation can result in extreme overproduction or underproduction.

Eliminating the First Plant

Page 17: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

With two factories producing three products, the demands that can be satisfied are severely limited.

When a system of equations has more equations than variables, that system is said to be overdetermined. An overdetermined system usually has no solutions. Eliminating the first plant would produce such a system.

In this situation, if the company added another plant that was identical to plant #2 or #3, the system would still be generally inconsistent since the output of the new plant would be a scalar multiple of an existing plant. To provide versatility, the output of any new factories should be linearly independent of existing ones.

Overdetermined Systems

Page 18: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Conversely, when a system of equations has more variables than equations, that system is said to be underdetermined. The earlier example with a 4th factory was an underdetermined system.

Underdetermined systems generally have an infinite number of solutions.

As far as our petroleum company is concerned, having more factories than products would allow for increased flexibility when meeting demand, especially if the output of the additional factories were linearly independent.

Underdetermined Systems

Page 19: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

Finally, when a system of equations has the same amount of variables as equations, it generally has a unique solution. The first few situations in this presentation were all examples of unique solutions.

Unique Solutions

Page 20: Jack Porter Jarrod Phillips Scott Simon Weon Kim.

In conclusion, it is best for the company’s production to be an underdetermined system. This allows for increased flexibility, since a problem in one factory doesn’t prohibit the company from meeting demand.

A unique solution to the company’s production is also good, but in this case a factory shutdown will put production into an overdetermined situation, in which case it is impossible for the company to meet demand without overproducing any products.

Conclusion