Linear Programming Case Study Solution
Linear Programming
CE750 Aircraft’s Fuel Plan
In the CE750 Aircraft’s Fuel Plan, the decision variables that are taken are mentioned in below table.
| X1 | The fuel needs to be purchased in number of pounds at KMLI |
| X2 | The fuel needs to be purchased in number of pounds at KBOS |
| X3 | The fuel needs to be purchased in number of pounds at KTEB |
| X4 | The fuel needs to be purchased in number of pounds at KDAL |
| Y1 | A binary variable made to define that the ramp fee has been waved or charged at KBOS |
| Y2 | A binary variable made to define that the ramp fee has been waved or charged at KTEB |
| Y3 | A binary variable made to define that the ramp fee has been waved or charged at KDAL |
Here, the binary numbers 0 and 1 are used, 1 is representing that the fee is waved while the 0 representing that the fee is charged.
Here, the objective function is used to minimize the total cost of travel which is
Minimum, Z= (3.97*X1+8.35*X2+7.47*X3+6.01*X4+3.97*X5) / 6.7 + (800*Y1+450*Y2+400*Y3)
In addition to this, there are multiple constraints are used which are the
| Constraints | |||
| Relation between X and Y | |||
| KBOS | (1-Y1) * 500 * 6.7 | ≤ | 0 |
| KTEB | (1-Y2) * 300 * 6.7 | ≤ | 2,010 |
| KDAL | (1-Y3) * 350 * 6.7 | ≤ | 2,590 |
| Fuel Tank constraints | |||
| KMLI | X1+ 7,000 | ≤ | 13,000 |
| KBOS | X1 + X2+ 7,000 - 4,800 | ≤ | 13,000 |
| KTEB | X1 + X2+ 7,000 - 4,800+ X3- 2,000 | ≤ | 13,000 |
| KDAL | X1+ X2+7,000 - 4,800 + X3 + X4- 5,300- 2,000 | ≤ | 13,000 |
| KMLI | X1+ X2+7,000 - 4,800 + X3 + X4- 5,300 - 2,000 + X5 - 3100 | ≤ | 13,000 |
| Minimum Fuel Constraints | |||
| KMLI | 7,000 – 4,800+ X1 | ≥ | 2,400 |
| KBOS | 7,000 + X1 + X2– 4,800 – 2,000 | ≥ | 2,400 |
| KTEB | 7,000 + X1+ X2 + X3– 2,000 – 5,300 – 4,800 | ≥ | 2,400 |
| KDAL | 7,000 + X1 + X2 + X3 + X4– 3,100 – 4,800- 2,000 – 5,300 | ≥ | 2,400 |
| Maximum Ramp Weight | |||
| KMLI | 22,200 + 7,000 + 2 * 200+ X1 | ≤ | 36,400 |
| KBOS | 22,200 + 7,000 + 4 * 200 – 4,800 + X1+ X2 | ≤ | 36,400 |
| KTEB | 22,200 + 7,000 + X1 + 8 * 200 – 2,000 + X3 – 4,800 + X2 | ≤ | 36,400 |
| KDAL | 22,200 + 7,000 + X1 + 8 * 200 – 2,000 + X3 – 4,800 + X2– 5,300 + X4 | ≤ | 36,400 |
| Maximum Landing weight | |||
| KMLI | 22,200 + 7,000 + 2 * 200 – 4,800+ X1 | ≤ | 31,800 |
| KBOS | 22,200 + 7,000 + X1 + 4 * 200 – 2,000 – 4,800 + X2 | ≤ | 31,800 |
| KTEB | 22,200 + 7,000 + X1 + 8 * 200 – 2,000 – 4,800 + X2 + X3– 5,300 | ≤ | 31,800 |
| KDAL | 22,200 + 7,000 + X1 + 8 * 200 – 2,000 – 4,800 + X2 + X3– 5,300 + X4– 3,100 | ≤ | 31,800 |
In addition to this, the constraint has been used that all the decision variables must be an integer.
The linear programming model is looking like
| X1 | X2 | X3 | X4 | X4 | |
| KMLI | KBOS | KTEB | KDAL | KMLI | |
| Fuel need to bought | 6000 | 0 | 2010 | 2590 | 4600 |
| Price of fuel ($/gallon) | 3.97 | 8.35 | 7.47 | 6.01 | 3.97 |
| Ramp Fee Waiver Binary Variable Defining | |||||
| KBOS | 1 | 800 | Y1 | ||
| KTEB | 0 | 450 | Y2 | ||
| KDAL | 0 | 400 | Y3 | ||
| Objective Function | |||||
| Minimize | 11,645 | ||||
This model gives us the minimum fuel cost of $ 11,645 in which the KMLI will need to bought 6,000 pounds fuel, KBOS will need to bought 0 pounds fuel, KTEB will need to bought 2,010 pounds fuel, KDAL will need to bought 2,590 pounds fuel and KMIL will need to bought 4,600 pounds fuel.
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