Assess the Impacts of Discount Policies on the Reliability of a Stochastic Air Transport Network
Abstract
:1. Introduction
2. SATN Model and Reliability Evaluation
2.1. Assumptions
2.2. Nomenclature
X ≤ Y | (x1, x2,..., xz) ≥ (y1, y2,..., yz) if xk ≥ yk for all k |
X < Y | (x1, x2,..., xz) > (y1, y2,..., yz) if X ≥ Y and exists at least one xh > yh |
X ≹ Y | neither X ≥ Y nor X < Y |
3. Reliability Assessment
3.1. Minimal Paths and Feasible Flow Vectors
3.2. Traveling Costs under Discount Policies
3.3. Generate All Lower Boundary Points
3.4. Solution Procedure
- Step 1: Generate all feasible minimal paths (MPs) as follows.
Set j = 1, Ω = Ø, Ə = {o}, and, μ = length(Ə)//*j is an index of a feasible MP (Pj), Ω stores Pj’s flights, and Ə stores airports that Pj visits.
Function Search (Ω, Ə, o, d, μ, θ)(1.1) IF μ ≠ 1, let wh be the μ–1th element of Ω
Select one available flight wk such that ≡ and ∉ Ə(1.2) IF ≥ + θ (1.3) Ω ← Ω ∪ {wk}, Ə ← Ə ∪ {}, and μ = μ + 1 (1.4) IF = d then Pj ← Ω, output Pj, j = j + 1 (1.5) ELSE, Call search (Ω, Ə, o, d, μ, θ) (1.6) ELSE, select one available flight wk that departs from o (1.7) Go to step (1.2) (1.8) Ω ← Ω\{wk}, Ə ← Ə\{}, and μ = μ − 1 (1.9) Next selection (1.10) END
- Step 2: Through the following equation, obtain set F of all flow vectors F fulfilling the travel demand E.
- Step 3: Through Equation (15), convert each flow vector F ∈ F into capacity vector X and store it in Δ.
- Step 4: Calculate the traveling cost of capacity vector X ∈ Δ on MPs.
- (4.1) Under policy 1:
- (4.2) Under policy 2:
- Step 5: Through the following comparison process, obtain a set of all exact lower boundary points from for o = 1, 2.
Initially, Θ stores θ indexes of Xs in (5.1) FOR k = 1 to θ ∈ Θ (5.2) FOR h = k + 1 to θ ∈Θ (5.3) IF Xk ≤ Xh then Xh is not a lower boundary point, Θ = Θ\{h}
ELSE IF Xk > Xh then Xk is not a lower boundary point, Θ = Θ\{k}, and BREAK(5.6) = {Xk| k ∈Θ}
4. A Numerical Example and Reliability Analysis
4.1. Evaluate the Reliability of Numerical Example
- Step 1: Generate all feasible minimal paths (MPs). Totally, four MPs: P1 = {w1, w5, w8}, P2 = {w1, w7}, P3 = {w3, w6, w7}, P4 = {w3, w8} are accepted.
- Step 2: Through Equations (20) and (21), obtain set F of all flow vectors F fulfilling the travel demand E
- Step 3: Through Equation (22), convert each flow vector F ∈ F into capacity vector X and store it in Δ, referring to column 2 of Table 3.
- Step 4: Calculate the traveling cost of each capacity vector X ∈ Δ on MPs. Then, reject from Δ any capacity vectors through Constraints (24) and (26) to get corresponding sets of all lower boundary point candidates under policies 1 and 2.
- (4.1) Under policy 1:
- (4.2) Under policy 2:
- Step 5: Through the following comparison process, obtain the sets of all exact lower boundary points = = {X1, X2,…, X21, X26, X30, X33, X35, X36, X40, X43, X45, X46, X49, X51, X52, X54}. Column 5 of Table 3 remarks why a certain X does not belong to or/and .
- Step 6: Utilize the RSDP algorithm into Equation (27) to compute the reliabilities under two policies.
4.2. Analyze the Impact of Discount Policies on the Reliability
5. Conclusions
Funding
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Conflicts of Interest
References
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Capacity | (i) | 1 | 2 | 3 | 4 | 5 |
Quantity Discount Rate | (gi) | 0.05 | 0.075 | 0.1 | 0.15 | 0.2 |
Flight | Fare | Departure Time | Arrival Time | Departure—Arrival Airport | Probability Pr (xk) | |||||
---|---|---|---|---|---|---|---|---|---|---|
wk | ck | tk | (ak–) | xk = 5 | xk = 4 | xk = 3 | xk = 2 | xk = 1 | xk = 0 | |
1 | 900 | 7:00 | 8:00 | DAD-HAN | 0.81 | 0.1 | 0.05 | 0.02 | 0.01 | 0.01 |
2 | 900 | 9:45 | 10:45 | DAD-HAN | 0.80 | 0.1 | 0.05 | 0.02 | 0.01 | 0.02 |
3 | 950 | 7:15 | 8:30 | DAD-SGN | 0.81 | 0.05 | 0.06 | 0.05 | 0.01 | 0.02 |
4 | 950 | 10:15 | 11:30 | DAD-SGN | 0.80 | 0.08 | 0.05 | 0.05 | 0.01 | 0.01 |
5 | 1200 | 8:45 | 10:15 | HAN-SGN | 0.935 | 0.035 | 0.01 | 0.01 | 0.01 | |
6 | 1200 | 9:00 | 10:30 | SGN-HAN | 0.94 | 0.04 | 0.005 | 0.005 | 0.01 | |
7 | 3250 | 11:00 | 14:30 | HAN-TPE | 0.81 | 0.1 | 0.05 | 0.02 | 0.01 | 0.01 |
8 | 3250 | 11:30 | 15:00 | SGN-TPE | 0.90 | 0.055 | 0.02 | 0.01 | 0.005 | 0.01 |
Step 2 | Step 3 | Step 4.1 | Step 4.2 | Step 5 | |||
---|---|---|---|---|---|---|---|
(F) | (Δ) | () | |||||
F1 = | (0, 0, 0, 5) | X1 = | (0, 0, 5, 0, 0, 0, 0, 5) | √ | √ | √ | √ |
F2 = | (0, 0, 1, 4) | X2 = | (0, 0, 5, 0, 0, 1, 1, 4) | √ | √ | √ | √ |
… | |||||||
F6 = | (0, 1, 0, 4) | X6 = | (1, 0, 4, 0, 0, 0, 1, 4) | √ | √ | √ | √ |
F7 = | (0, 1, 1, 3) | X7 = | (1, 0, 4, 0, 0, 1, 2, 3) | √ | √ | √ | √ |
… | |||||||
F11 = | (0, 2, 1, 2) | X11 = | (2, 0, 3, 0, 0, 0, 2, 3) | √ | √ | √ | √ |
F12 = | (1, 0, 1, 3) | X12 = | (2, 0, 3, 0, 0, 1, 3, 2) | √ | √ | √ | √ |
… | |||||||
F22 = | (1, 0, 1, 3) | X22 = | (1, 0, 4, 0, 1, 1, 1, 4) | = 5036 | √ | X22 > X6 | |
F23 = | (1, 0, 2, 2) | X23 = | (1, 0, 4, 0, 1, 2, 2, 3) | √ | √ | X23 > X7 | X23 > X7 |
… | |||||||
F26 = | (1, 1, 0, 3) | X26 = | (2, 0, 3, 0, 1, 0, 1, 4) | √ | √ | √ | √ |
F27 = | (1, 1, 1, 2) | X27 = | (2, 0, 3, 0, 1, 1, 2, 3) | = 5001 | √ | > B | X27 > X11 |
F28 = | (1, 1, 2, 1) | X28 = | (2, 0, 3, 0, 1, 2, 3, 2) | √ | √ | X28 > X12 | X28 > X12 |
… | |||||||
F37 = | (2, 0, 1, 2) | X37 = | (2, 0, 3, 0, 2, 1, 1, 4) | = 5083 | √ | > B | X37 > X26 |
F38 = | (2, 0, 2, 1) | X38 = | (2, 0, 3, 0, 2, 2, 2, 3) | √ | √ | X38 > X11 | X38 > X11 |
F39 = | (2, 0, 3, 0) | X39 = | (2, 0, 3, 0, 2, 3, 3, 2) | √ | √ | X39 > X12 | X39 > X12 |
… | |||||||
F54 = | (4, 1, 0, 0) | X54 = | (5, 0, 0, 0, 4, 0, 1, 4) | √ | √ | √ | √ |
Travel Demand and Limited Budget EB = (E, B) | Policy 1 Quantity Discount | Policy 2 Contractual Discount | ||||
---|---|---|---|---|---|---|
g = g1 | g = g2 | g = g3 | g = g4 | g = g5 | ||
EB = (3, 5000) | 0.99786556 | 0.99591643 | 0.99787289 | 0.99787289 | 0.99787289 | 0.99787289 |
EB = (5, 5000) | 0.98755955 | 0.97779929 | 0.98755955 | 0.98755955 | 0.98755955 | 0.98755955 |
EB = (7, 5000) | 0.90026408 | 0.8862315 | 0.90026408 | 0.90026408 | 0.90026408 | 0.90026408 |
EB = (9, 5000) | 0.67142394 | 0.66424293 | 0.67142394 | 0.67142394 | 0.67142394 | 0.67142394 |
EB = (3, 4500) | 0.99591643 | 0.99591643 | 0.99591643 | 0.99591643 | 0.99591643 | 0.99787289 |
EB = (5, 4500) | 0.98335734 | 0.97779929 | 0.97779929 | 0.97779929 | 0.97779929 | 0.98755955 |
EB = (7, 4500) | 0.89911623 | 0.8862315 | 0.8862315 | 0.8862315 | 0.8862315 | 0.90026408 |
EB = (9, 4500) | 0.67142394 | 0.66424293 | 0.66424293 | 0.66424293 | 0.66424293 | 0.67142394 |
EB = (3, 4000) | 0.99591643 | 0.99591643 | 0.99591643 | 0.99591643 | 0.99591643 | 0.99591643 |
EB = (5, 4000) | 0.97779929 | 0.97779929 | 0.97779929 | 0.97779929 | 0.97779929 | 0.97779929 |
EB = (7, 4000) | 0.8862315 | 0.8862315 | 0.8862315 | 0.8862315 | 0.8862315 | 0.8862315 |
EB = (9, 4000) | 0.66424293 | 0.66424293 | 0.66424293 | 0.66424293 | 0.66424293 | 0.66424293 |
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Nguyen, T.-P. Assess the Impacts of Discount Policies on the Reliability of a Stochastic Air Transport Network. Mathematics 2021, 9, 965. https://doi.org/10.3390/math9090965
Nguyen T-P. Assess the Impacts of Discount Policies on the Reliability of a Stochastic Air Transport Network. Mathematics. 2021; 9(9):965. https://doi.org/10.3390/math9090965
Chicago/Turabian StyleNguyen, Thi-Phuong. 2021. "Assess the Impacts of Discount Policies on the Reliability of a Stochastic Air Transport Network" Mathematics 9, no. 9: 965. https://doi.org/10.3390/math9090965