example 3. authentic assessment 206aae operational ... · “authentic assessment should be...
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ASSESSMENT AND FEEDBACK CASE STUDY
by Kristina Marintseva
The online Portfolio: https://kristina691772122.wordpress.com/
7001CRB Developing and Enhancing Assessment and Feedback in Higher Education
Coventry University, 2020
Example 3. Authentic assessment
206AAE Operational Research for Aviation Management
“authentic assessment should be challenging with real-world problem-solving situations”
1.Model a range of situations encountered in airline, airport
and supplier operations for analysis by operational
research methods.
2.Solve a range of airline, airport and supplier operational
problems using appropriate software, analyse the output,
and interpret the results in context.
Learning Outcomes
On completion of 206 AAE module the student should be able to:
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Art of Modeling III:Transhipment Problem
Transportation Problem Formulation
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Steps Involved in Air Cargo LogisticsTransportation Problem Formulation
https://www.youtube.com/watch?v=hl_vsIjz3tsSea / Air TransshipmentServices via Singapore
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Transportation Problem Formulation
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50 largest air freight forwarders of 2018
The latest annual figures from Market analyst Armstrong & Associates show that the world’s top airfreight forwarders saw huge growth in 2017. The total tonnage across the leading 25 increasing by 11.1% to 16.1m metric tons.
1 DHL Supply Chain Germany 2,248,000
2 Kuehne + Nagel Inc. Switzerland 1,570,000
3 DB Schenker USA Germany 1,226,200
4 Panalpina Inc. Switzerland 995,900
5 Expeditors International of Washington United States 985,549
6 UPS Supply Chain Solutions United States 935,300
7 Nippon Express Co. Japan 705,478
8 Hellmann Worldwide Logistics Germany 654,104
9 DSV Air & Sea Ltd. Copenhagen, Denmark 635,655
10 Bollore Logistics France 569,000
11 Sinotrans Ltd. China 532,400
12 Kintetsu World Express Japan 495,947
13 Ceva Logistics The Netherlands 480,000
14 Agility Logistics Switzerland 415,000
15 Yusen Logistics Japan 368,198
Transportation Problem Formulation
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Air-Air Transhipments - Creating New ConnectionsAir-air solutions providers are typically freight forwarding companies that have the capability to handle customs documentation to accept an inbound shipment ‘under transit’ from an airline, and on forward a shipment to another airline.
When is it normally being used?A. Regional Transshipment
When are airfreight forwarder sends a shipment to a regional hub, which is then on forwarded to the region. Example: Frankfurt – Singapore Multidestination Consolidation – On-forwarding to Region Vietnam/Cambodia/Myanmar
B. Global Transshipment (Long Haul)When are airfreight forwarder sends a shipment to a global hub, which is then on forwarded to a long haul destination Example: Hong Kong – Amsterdam – Sao Paolo , Shanghai – Dubai – Nigeria
Transshipment hubs that are known to the industry are Dubai, Singapore, Miami, Amsterdam.
https://www.globalaircargoalliance.com/air-air.php
Transhipment Problem Transportation Problem Formulation
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The transportation model assumes direct shipments betweensources and destinations.This may not be the case in many situations where it may becheaper to tranship through intermediate nodes before reachingthe final destination.
Transhipment Problem Transportation Problem Formulation
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Formulation of the Transhipment Problem model Shortened form of problem recording Shortened form of problem recording where
Unbalanced TP
𝑴 = σ𝒊=𝟏𝒏 σ𝒓=𝟏
𝑹 σ𝒋=𝟏𝒎 𝒄𝒊𝒓𝒋𝒙𝒊𝒓𝒋 → 𝒎𝒊𝒏 (1)
St
σ𝒋=𝟏𝒎 σ𝒓=𝟏
𝑹 𝒙𝒊𝒓𝒋 ≤ 𝒂𝒊, 𝒊 = 𝟏, 𝒏 (availabilities) (2)
σ𝒊=𝟏𝒏 σ𝒓=𝟏
𝑹 𝒙𝒊𝒓𝒋 ≥ 𝒃𝒋, 𝒋 = 𝟏, 𝒏 (requirements)
(3)
σ𝒊=𝟏𝒏 σ𝒋=𝟏
𝒎 𝒙𝒊𝒓𝒋 ≤ 𝒅𝒓, 𝒓 = 𝟏, 𝑹 (4)
(transshipment point capacity)
𝒙𝒊𝒓𝒋 ≥ 𝟎, 𝒊 = 𝟏,𝒎; 𝒋 = 𝟏, 𝒏; 𝒓 = 𝟏,𝑹 (5)
Balanced TP
𝑴 =
𝒊=𝟏
𝒏
𝒓=𝟏
𝑹
𝒋=𝟏
𝒎
𝒄𝒊𝒓𝒋𝒙𝒊𝒓𝒋 → 𝒎𝒊𝒏
St
σ𝒋=𝟏𝒎 σ𝒓=𝟏
𝑹 𝒙𝒊𝒓𝒋 = 𝒂𝒊, 𝒊 = 𝟏, 𝒏 (availabilities)
σ𝒊=𝟏𝒏 σ𝒓=𝟏
𝑹 𝒙𝒊𝒓𝒋 = 𝒃𝒋, 𝒋 = 𝟏, 𝒏 (requirements)
𝒊=𝟏
𝒏
𝒋=𝟏
𝒎
𝒙𝒊𝒓𝒋 ≤ 𝒅𝒓, 𝒓 = 𝟏, 𝑹
𝒙𝒊𝒓𝒋 ≥ 𝟎, 𝒊 = 𝟏,𝒎; 𝒋 = 𝟏, 𝒏; 𝒓 = 𝟏, 𝑹
𝒙𝒊𝒓𝒋 = the amount
transported from supply point 𝒊 to requirement point j through transhipment point 𝒓 (optimizable parameter, unknown variable);
𝒄𝒊𝒓𝒋= unit costs of
transporting one unit of cargo from supply point 𝒊 to requirement point j through transhipment point 𝒓 ;
𝒂𝒊 = the supply levels inpoint 𝒊;
𝒃𝒋= the requirement levels in
point j;
𝒅𝒓 = capacity oftranshipment point 𝒓
Transportation Problem Formulation
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Transportation Problem Formulation
The Transportation Model plays an essential role as a decision-support tool, providing relevant and accurate information into planning and decision making.
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Practice of Modelling
LP – Problem: revision
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https://www.youtube.com/watch?v=xwTXA2IGQfs
https://www.youtube.com/watch?v=OINg49Dz8J0
https://www.youtube.com/watch?v=6p7_lU6GM8w
https://www.youtube.com/watch?v=JZu_gxi3sbs
Transportation of goods by air is justified only when thevalue of the cargo is high enough or when urgent delivery toits destination is required. The modern technology of freighttransportation provides for the use of containers andpallets. The consolidation of cargoes of various value,weight and dimensions is carried out with the aim ofincreasing the efficiency (including profitability) of airtransportation.Due to the large number and variety of cargoes, finding thebest solution to plan the loading of an airplane is not easyeven for an experienced specialist. The higher the cargovalue, the more expensive mistakes are made when loadingthe aircraft.This problem needs a specific technique in order to get thebest solution in an acceptable time for business.Using the examples of loading goods in an airplane andcontainer, consider the possible ways to find the bestoption.
Real life problem
Pallet - P1P. PAG
Internal VolumeBase: 88" x 125" Height: 64" (Passenger flight), 96" (Q6 Contour), 118" (Q7 Contour)Tare Weight - 114 kgMaximum Gross Weight - 4,626 kg (LD)6,033 kg (MD)Loadable Aircraft Type: 747, 747F, 777, Airbus
LP – Problem: revision
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Aircraft Payload, kg /Range, km Crew Deck Maximum
volume, m3
Cargo door size
Width, m Height, m
AN-124-100 120,000 4,500 6-13 Main deck 800 6.20 4.10
AN-225 200,000 4,000 6-13 Main deck 1,100 6.20 4.10
Boeing 747-8F 139,000 7,630 2 Main deck 692 3.30 3.05
Lower deck 150 2.64 1.68
Bulk 14 1.12 1.19
A300-600F 47,627 4,344 2 Main deck 280 3.58 2.56
Lower Deck 103 2.08 1.70
Bulk 10 0.95 0.95
Real life problemLP – Problem: revision
Transportation of goods by air: example of the cargo aircraft
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Antonov An-124 Condorhttps://www.popularmechanics.com/flight/g2467/11-of-the-largest-cargo-planes-in-the-sky/
Airbus A300-600ST (Super Transporter)
BOEING 737
BOEING 737
https://www.boeing.com/
Real life
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LP – Problem: revision
A cargo plane has three compartments for
storing cargo: main deck, lower deck and bulk.
These compartments have the following limits
on both weight and space: Deck Weight capacity (kg) Space capacity (m3)
Main deck 113,000 692
Lower deck 24,000 150
Bulk 2,000 14
139,000
Any proportion of these cargoes can be accepted. The objective is to determine how much (if any) of each cargo C1, C2, C3 and C4
should be accepted and how to distribute each among the compartments so that the total profit for the flight is maximised.
• Formulate the above problem as a linear program
• What assumptions are made in formulating this problem as a linear program?
Cargo Weight (kg) Volume (m3/kg) Profit (£/kg)
C1 21,000 0.0048 2.10
C2 70,000 0.0151 3.80
C3 56,000 0.0054 1.50
C4 120,000 0.0037 1.08
The following four cargoes are available for
shipment on the next flight:
Problem I
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P= σ𝒊=𝟏𝒎 σ𝒋=𝟏
𝒏 𝑷𝒊𝑪𝒊𝒋 → max
subject to
σ𝒊=𝟏𝒎 𝒂𝒊𝑪𝒊𝒋 ≤ 𝑨𝒋, 𝒋 = 𝟏, 𝒏; (volume (space) constraints)
σ𝒊=𝟏𝒎 𝑪𝒊𝒋 ≤ 𝑩𝒋, 𝒋 = 𝟏, 𝒏; (weight capacity constraints)
σ𝒋=𝟏𝒏 𝑪𝒊𝒋 ≤ 𝑫𝒊, 𝒊 = 𝟏,𝒎 ; (demand constraints)
𝑿𝒊𝒋 ≥ 𝟎, 𝒊 = 𝟏,𝒎 ; 𝒋 = 𝟏, 𝒏
Math Model
where
𝒊 = 𝟏,𝒎 is type of cargo;
𝒋 = 𝟏, 𝒏 is type of deck;
𝑪𝒊𝒋= the amount of cargo 𝒊
distributed to compartment j(optimizable parameter, unknown variable);
𝑷𝒊 = profit per unit of cargo 𝒊;
𝒂𝒊𝒋= volume per unit of cargo 𝒊 ;
𝑨𝒋 = deck j space capacity;
𝑩𝒋 = deck j weight capacity;
𝑫𝒊 = cargo 𝒊 available for shipment
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max
2100C11+2100C12+2100C13+3800C21+3800C22+3800C23+1500C31+1500C32+1500C33+1080C41+1080C42+1080C43
st
! cannot pack more of each of the four cargoes than we have available
C1) C11 + C12 + C13 <= 21
C2) C21 + C22 + C23 <= 70
C3) C31 + C32 + C33 <= 56
C4) C41 + C42 + C43 <= 120
!the weight capacity of each compartment must be respected
Main_W) C11 + C21 + C31 + C41 <= 113
Lower_W) C12 + C22 + C32 + C42 <= 24
Bulk_W) C13 + C23 + C33 + C43 <= 2
!the volume (space) capacity of each compartment must be respected
Main_S) 4.8C11 + 15.1C21 + 5.4C31 + 3.7C41 <= 692
Lower_S) 4.8C12 + 15.1C22 + 5.4C32 + 3.7C42 <= 150
Bulk_S) 4.8C13 +15.1C23 + 5.4C33 + 3.7C43 <= 14
END
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Problem Model for LINDO
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Cargo Main deck Lower deck Bulk
C1 6 15 0 21
C2 17 3 0 20
C3 43 6 2 51
C4 47 0 0 47
113 24 2
The solution of the problem
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types of goods , 𝑖 𝑤𝑖 , % 𝑚𝑖, кг 𝑣𝑖 , м3
1 (clothes) 10 300 0.5
2 (cosmetics) 20 250 0.7
3 (furniture) 30 200 1.5
4 (auto parts) 10 500 0.8
5 (medicines) 30 440 0.5
total 100 1,690 7.7
Let's say 5 types of goods arrived at the airport warehouse. The percentage distribution of value
(expected profit from transportation), weight and volume of cargo are given in the table. You need to
prepare the container for further air transportation by loading cargo maximizing the profit from
transportation. Each cargo can only be considered as a whole.
Problem II
•Internal Volume•152 cu. ft, 4.3mc•Tare Weight•73 kg / 100 kg•Maximum Gross Weight•1,588 kg
•Loadable Aircraft Type•747, 747F, 777, Airbus
Container - AKE
Notes:Gross weight is the total weight of goods, including the raw product, any packaging, and possibly the vessel transporting the goods. Net weight is the raw weight of the product only without any packaging. Gross weight = net weight + packaging/ container weight. Net weight = gross weight – tare (container) weight.
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Let’s denote:
𝒊= index of the goods (cargo), 𝒊 = 1,2, ..., 𝒏;
𝒘𝒊 = percentage distribution of cargo 𝒊 value;
𝒎𝒊 = the weight of cargo 𝒊;
𝒗𝒊 = cargo 𝒊 volume;
V = aircraft cargo compartment space (4.3 mc);
G = container load capacity (1,488 kg).
The problem of choosing the maximum possible value of the transported cargo is known in the
literature as the “Knapsack Problem”. Using the Boolean variables 𝒙𝒊 = 𝟎⋁𝟏, this problem can
be solved by the “branch and bound” method.
And define:
𝒙𝒊 = ൜1, if the goods i will be loaded into a container0, otherwise
;
𝒊 = 1, 𝑛;
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Mathematically, the problem is formulated as follows:
𝒎𝒂𝒙 𝒇𝟎 = σ𝒊=𝟏𝒏 𝒘𝒊 𝒙𝒊
at constrains
σ𝒊=𝟏𝒏 𝒎𝒊 𝒙𝒊 ≤ 𝑮
σ𝒊=𝟏𝒏 𝒗𝒊 𝒙𝒊 ≤ 𝑽
𝒙𝒊 = 𝟎⋁𝟏, 𝒊 = 𝟏, 𝒏 .
If you use the data in the Table, then this task can be written as follows:
𝒎𝒂𝒙 𝒇 = 𝟏𝟎𝒙𝟏 + 𝟐𝟎𝒙𝟐 + 𝟑𝟎𝒙𝟑 + 𝟏𝟎𝒙𝟒 + 𝟑𝟎𝒙𝟓
Subject to
𝟑𝟎𝟎𝒙𝟏 + 𝟐𝟓𝟎𝒙𝟐 + 𝟐𝟎𝟎𝒙𝟑 + 𝟓𝟎𝟎𝒙𝟒 + 𝟒𝟒𝟎𝒙𝟓 ≤ 1𝟒𝟖𝟎 (1588 kg-100 kg) (Net Weight)
𝟎. 𝟓𝒙𝟏 + 𝟎. 𝟕𝒙𝟐 + 𝟏. 𝟓𝒙𝟑 + 𝟎. 𝟖𝒙𝟒 + 𝟎. 𝟓𝒙𝟓 ≤ 4.3 (Internal Volume of container)
𝒙𝒊 = 𝟎⋁𝟏, 𝒊 = 𝟏, 𝟓
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LINDO report
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Interpretation of the LINDO report
Optimum of the objective function value is 90. In a given problem this is a profit percentage of the total possible profit for the transportation of goods. Following this is the optimal solution to the problem. That is, the strategy to set the decision variables in order to achieve the above optimal value (90). The value column contains the solution to the problem:Types of goods that should be loaded into the container for the given flight according optimal plan (obtained by solving the problem): X2 (cosmetics), X3(furniture), X4(auto parts), X5 (medicines).
Reduced cost for each variable. There are two valid, equivalent interpretations of a reduced cost.We can interpret a variable's reduced cost as the amount by which the objective coefficient of the variable would have to improve before it would become profitable to bring that variable into the solution at a nonzero value.
!!! Reduced costs are meaningful in linear and quadratic models, but, in general, should be ignored when solving integer programming (IP) models.
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Next, in the rows section of the report, you will find one line for each constraint giving the constraint's name (if any), its slack or surplus value, and its dual price. The SLACK OR SURPLUS column tells you how close you are to the right-hand side limit on each constraint. This quantity, on less-than-or-equal-to constraints, is generally referred to as slack. Similarly, on greater-than-or-equal-to constraints it is called a surplus. If a constraint is exactly satisfied as an equality, the SLACK OR SURPLUS value will be zero. If a constraint is violated, as in an infeasible solution, the SLACK OR SURPLUS value will be negative. Knowing this can help you find the violated constraints in an infeasible model.
SLACK: 90 kg and 0.8 mc – this is the remainder of the resource in terms of volume and payload capacity of the container, which we cannot use due to the indivisibility of the cargo.
The LINDO solution report also gives a DUAL PRICE figure for each constraint. You can interpret the dual price as the amount by which the objective would improve given a unit of increase in the right-hand side of the constraint. Dual prices are sometimes called shadow prices, because they tell you how much you should be willing to pay for additional units of a resource.
!!!Dual prices are meaningful in linear and quadratic models, but, in general, should be ignored when solving integer programming models.
Interpretation of the LINDO report