Case2 Dcg

JCG GLOBAL AIR SERVICES Case Analysis

Submitted by: ST Vigneshwaran Nisha Sharma Kumar Shivam Bizza Srilatha Ishaat Zahidi Ramakant

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EXECUTIVE SUMMARY JCG global air services (AS) catered to the needs of transportation of JCG company headquarters, by operating four aircrafts. The company owned & operated 2 Cessna citation X aircraft (CE 750) & GV aircraft. The HQ, located at Moline housed around 2400 employees, who regularly travelled to company factories, marketing facilities & customer locations across the globe. Sam Bursk, the pilot/manager of this large corporate aviation department, has taken up the challenging job of preparing the fuel consumption plan for an upcoming four leg trip from Moline to Boston, New York City area; Dallas and back. The job is demanding as he has to consider many decision variables that include the amount of fuel to be purchased in accordance with the amount for ramp fee charged at each airport. Simple linear programming is used to optimize the fuel buying s.t the different constraints which are given below.

Gulfstream GV Decision Variables: Let Xi be the amount of fuel (pounds) be purchased at airport i. X1: Amount of fuel to be purchased at KMLI. X2: Amount of fuel to be purchased at KBOS. X3: Amount of fuel to be purchased at KTEB. X4: Amount of fuel to be purchased at KDAL. X5: Amount of fuel to be purchased at KMLI. Let Yi be binary decision variable for waving of the fees or not where, at a particular airport 1: Fee waved 0: Otherwise Y1: Binary Variable indicating if ramp fee has been waved or not at KBOS Y2: Binary Variable indicating if ramp fee has waved or not at KTEB

Y3: Binary Variable indicating if ramp fee waved or not at KDA Objective Function: Minimize total cost of travel by the aircraft. Z=(3.97X1+8.35X2+7.47X3+6.01X4+3.97X5)/6.7+(800Y1+450Y2+400Y3) Constraints s.t: Capacity of tank: KMLI: 7000+X1<=41300 KBOS: 7000+X1-6968+X2 <= 41300 KTEB: 7000+X1-6968+X2-2903+X3 <=41300 KDAL: 7000+X1-6968+X2-2903+X3-7694+X4<=41300 KMLI: 7000+X1-6968+X2-2903+X3-7694+X4-4500<=41300 Ramp Fee waiver: KBOS: 500*6.7*(1-Y1) <=X2 KTEB: 300*6.7*(1-Y2) <=X3 KDAL: 350*6.7*(1-Y3) <=X4 Minimum arrival fuel: KBOS: 48800 + X1-4800>= 2400 KTEB: 48800+X1-6968+X2-2903>= 2400 KDAL: 48800 +X1-6968+X2-2903+X3-7694>= 2400 KMLI: 48800+ X1-6968+X2-2903+X3-7694+X4-4500>=2400 Ramp weight: At KMLI: 48800+7000+X1+2*200<=90900 At KBOS: 48800+7000+X1-6968+X2+4*200<=90900 At KTEB: 48800+7000+X1-6968+X2-2903+X3+8*200<=90900

At KDAL: 48800+7000+X1-6968+X2-2903+X3-7694+X4+8*200<=90900 Landing weight: KBOS: 22200+7000+X1-4800+2*200<=75300 KTEB: 22200+7000+X1-4800+X2-2000+4*200<=75300 KDAL: 22200+7000+X1-4800+X2-2000+X3-5300+8*200<=75300 KMLI: 22200+7000+X1-4800+X2-2000+X3-5300+X4-3100+8*200<=75300 Integer and Non-negativity: X1, X2, X3, X4, X5, Y1, Y2, Y3 are non-negative and integers.

Excel snapshot

CE750 Decision Variables: Let Xi be the amount of fuel (pounds) be purchased at airport i. X1: Amount of fuel to be purchased at KMLI. X2: Amount of fuel to be purchased at KBOS. X3: Amount of fuel to be purchased at KTEB. X4: Amount of fuel to be purchased at KDAL. X5: Amount of fuel to be purchased at KMLI. Let Yi be binary decision variable for waving of the fees or not where, at particular airport

1: Fee waved 0: Otherwise Y1: Binary Variable indicating if ramp fee has been waved or not at KBOS Y2: Binary Variable indicating if ramp fee has waved or not at KTEB Y3: Binary Variable indicating if ramp fee waved or not at KDA

Objective Function: Minimize total cost of travel by the aircraft. Min, Z= (3.97X1+8.35X2+7.47X3+6.01X4+3.97X5)/6.7+(800Y1+450Y2+400Y3) Constraints s.t: Capacity of tank: KMLI: 7000+X1<=13000 KBOS: 7000+X1-4800+X2 <= 13000 KTEB: 7000+X1-4800+X2-2000+X3 <=13000 KDAL: 7000+X1-4800+X2-2000+X3-5300+X4<=13000 KMLI: 7000+X1-4800+X2-2000+X3-5300+X4-3100<=13000 Ramp Fee waiver: At KBOS: 500*6.7*(1-Y1) <=X2 At KTEB: 300*6.7*(1-Y2) <=X3 At KDAL: 350*6.7*(1-Y3) <=X4 Minimum arrival fuel: KBOS: 7000 + X1-4800>= 2400 KTEB: 7000+X1-4800+X2-2000>= 2400 KDAL: 7000 +X1-4800+X2-2000+X3-5300>= 2400

KMLI: 7000+ X1-4800+X2-2000+X3-5300+X4-3100>=2400 Ramp weight: KMLI: 22200+7000+X1+2*200<=36400 KBOS: 22200+7000+X1-4800+X2+4*200<=36400 KTEB: 22200+7000+X1-4800+X2-2800+X3+8*200<=36400 KDAL: 22200+7000+X1-4800+X2-2800+X3-5300+X4+8*200<=36400 Landing weight: KBOS: 22200+7000+X1-4800+2*200<=31800 KTEB: 22200+7000+X1-4800+X2-2000+4*200<=31800 KDAL: 22200+7000+X1-4800+X2-2000+X3-5300+8*200<=31800 KMLI: 22200+7000+X1-4800+X2-2000+X3-5300+X4-3100+8*200<=31800 Integer: X1, X2, X3, X4, X5,Y1, Y2, Y3 are integers. Excel snap

Q1. Consider the “no tankering” option, where just enough fuel is loaded at each stop to land at the next, with exactly 2400 lbs of fuel as required. What is the fuel plan for JCG’s upcoming four-leg trip with this option? Fuel plan with ‘no tankering’ option Where, X1: Amount of fuel to be purchased at KMLI. X2: Amount of fuel to be purchased at KBOS. X3: Amount of fuel to be purchased at KTEB. X4: Amount of fuel to be purchased at KDAL. X5: Amount of fuel to be purchased at KMLI after Journey Fuel to be bought Fuel price ($/pound) Ramp fee Minimum pounds to waive ramp fee Ramp fee waived (Y or N) Cost of buying fuel in dollars

X1 X2 200 2000 0.59 1.25 800 3350

118

X3 5300 1.11 450 2010

X4 3100 0.90 400 2345

No Yes 2500+800=3300 5883

Yes 2790

X5 4600 0.59

2714

X1 = 200 Fuel already in jet = 7000 pounds. Fuel burnt to fly from KMLI to KBOS = 4800 pounds Minimum fuel while landing = 2400 pounds Total fuel requirement from KMLI to KBOS = 7200 pounds Fuel to be bought at KMLI = 7200 – 7000 = 200 pounds Cost of buying fuel at KMLI = 200*0.59 = $118 X2 = 2000 Minimum Fuel while landing (already in jet) = 2400 pounds Fuel burnt to fly from KBOS to KTEB = 2000 pounds Total fuel requirement from KBOS to KTEB = 4400 pounds Fuel to be bought at KBOS = 4400 – 2400 = 2000 pounds Cost of buying fuel at KBOS = 2000*1.25 = $2500 Ramp fee = $800 (as minimum fuel to be bought to waive ramp fee is 3350 pounds) Total cost = 2500+800 = $3300 X3 = 5300

Total = 14805

Fuel already in jet = 2400 pounds. Fuel burnt to fly from KTEB to KDAL = 5300 pounds Total fuel requirement from KTEB to KDAL = 7700 pounds Fuel to be bought at KTEB = 7700 – 2400 = 5300 pounds Cost of buying fuel at KTEB = 5300*1.11 = $5883 (Ramp fee waived) X4 = 3100 Fuel already in jet = 2400 pounds. Fuel burnt to fly from KDAL to KMLI = 3100 pounds Total fuel requirement from KDAL to KMLI = 5500 pounds Fuel to be bought at KDAL = 5500 – 2400 = 3100 pounds Cost of buying fuel at KDAL = 3100*0.90 = $2790 (ramp fee waived) X5 = 4600 Fuel already in jet = 2400 pounds. Fuel to be available in jet for next trip = 7000 pounds Fuel to be bought at KMLI = 7000 – 2400 = 4600 pounds Cost of buying fuel at KMLI = 4600*0.59 = $2714 Total cost = sum of costs at X1, X2, X3, X4, X5 = 118+3300+5883+2790+2714 = $14805

Q2.

Excel snapshot: Constructing the Formula

Fuel to be bought Price of fuel ($/gallon)

X1 KMLI 6000 3.97

Ramp Fee Waiver Binary Variable Defining KBOS 1 KTEB 0 KDAL 0 Objective Function Minimize Constraints Relation between X and Y – 0 2010 2345 Fuel Tank constraints – 13000 8200 8210 5500 7000 Minimum Fuel Constraints – 8200 6200 2910 2400 Minimum Fuel on arrival at KMLI 7000 Maximum Ramp Weight 35600

11645.16418

0 2010 2590 13000 13000 13000 13000 13000 2400 2400 2400 2400

7000 36400

X2 KBOS 0 8.35

X3 KTEB 2010 7.47

800 450 400

Y1 Y2 Y3

X4 KDAL 2590 6.01

31200 32010 29300 Maximum Landing weight 30800 29200 26710 26200

36400 36400 36400 31800 31800 31800 31800