The document discusses optimizing delivery routes for a transportation company through applying the Kruskal's algorithm. It begins by establishing distances between delivery points. It then sorts the point-to-point distances in descending order and selects pairs to create an optimal route that minimizes total distance. Implementing this algorithm for a sample of 10 customers reduced the delivery route from 20.75 km to 17.21 km. While effective for small routes, scaling the algorithm to a full delivery schedule could provide greater efficiency gains for the company.

1. SUPPLY CHAIN STRATEGY
AND OPERATIONS
Individual Assignment
Alexandre Lopez
Alexandre.lopez2@student.anglia.ac.uk
MSc International Business
ID: 1415246

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MSc International Business Individual Assignment
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1) Simulation
In this first part, it will be question of transport. Transport represents one of the most important
part in the supply chain since it is a big cost for the manufacturer and allow to costumer to have
the product in good conditions and times. Nowadays, with the rise of globalisation and
capitalism, transportation has become more complex and technical. The new methods of
transportation allowed to facilitate and accelerate the delivery of products in a world
continuously in expansion.
First of all, it is important to define the notion of transport: “A category of stocks relating to the
transportation of goods or customers. The transportation sector is made up of airlines, railroads
and trucking companies” (Investopedia).
Transport is one of the three main activities in the supply chain management: purchase,
manufacture and transport (Thomas et al, 1996).
Transport affects the results of logistics activities, modifies the structure of the supply chain
and influences production and sale.
Transport can be seen today as a key factor. Indeed, transportation enables the link between
“the extraction of natural resources; the fabrication of industrial, commercial and consumer
products; and the final distribution of goods to wholesalers, retailers and end users” (SAP
company, 2006). Then, transportation is able to build a bridge between produces and customers
as showed in the figure 1.
Figure 1: Transportation in the adaptive supply chain.
Nowadays, it is not only question to deliver the product; the chain who benefits from the
transportation (suppliers to customers) wants a service transparent, efficient and of good
quality. Therefore, all the processes of transportation are an integrant part of the quality
management (Yung-Yu Tseng, Wen Long Yue, Michael A P Taylor, 2005).
Now, what are the elements who can determine the best grid of transportation?
Regarding transport, companies want to make a choice and choose the best offer in function of
different points. Then, the best grid of transportation among all of them rests on several criteria

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as the basis of the overall cost, service quality and deadlines. More precisely, firms need to
emphasise on several points who are the availability of the logistics means at each point of
transport (storage space, and handling equipment users). They also need to be focused on the
risk of the transportation (hazardous products…). Then, they have as well to determine the best
way of transportation (maritime, air or land).
These three types of transportation represents advantages and inconvenient for each. It is
important to detail their particularities. Firstly, the maritime transportation represents a high
carrying capacity at a cheap price. However, delays are longer while the weather strongly
influences the schedule. Moreover, most of the time, the customer cares more about the service
quality than the delivery price, especially for the consumer goods. It is essential to find a
compromise between quality service and delivery time regarding transportation. The best way
of transportation is going to be this who realise this compromise.
Concerning the air freight logistics, it represents the best way of transportation in terms of
“speed, lower risk of damage, security, flexibility, accessibility and good frequency for regular
destinations” (Yung-Yu Tseng, Wen Long Yue, Michael A P Taylor, 2005). Nonetheless, it
remains the most expansive.
More, “when the value per unit weight of shipments is relatively high and the speed of delivery
is an important factor” (Reynolds-Feighan, 2001).
Now, land is the most complex way of transportation because it implies a lot of actors and ask
to adapt in function of the land and the diversity of delivery addresses (last mile delivery).
Nonetheless, it is an advantage because the transporter can access at whatever place, because
of the very high accessibility. It allows also to extend the delivery services for air and maritime
transport from airports and seaports.
Then, once networks of distribution are determined, it admits to organise the transport operation
that it means prepare the route of deliverers.
The next step is about the fleet management that it means administrative and technical
management concerning the transport vehicles.
Now, it is interesting to be focused on links simulation in order to make the connection with
the reality and theories explained above.
First of all, the objective is to choose the best carrier in terms of price and delivery service.
All the carriers proposed as way of delivery two possibilities: air shipment and surface
shipment.
In our team Synergy, our choice was made on the carrier I and M within the region 2 and 3.
We knew that “an average of 70% of surface transported volume arrives at regional DCs in time
to meet current-month orders” (links manual). We also knew that the air shipment was the best
way of transportation (100%).
So, we wanted to find a good balance between air and surface shipments. In the two regions,
we had a large amount of air shipment because in order to have a good reliability of delivery
and consequently a good customer satisfaction.
In the region 2, we choose carrier I Air, because it was the most efficient in terms delivery and
cost. Indeed, we had 100% of delivery, and the price was the same than surface shipment. Below

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this price, our products were not sure to be delivered on time. Our policy was all the customers
had their products on time.
For the region 3, the air transportation was more expensive than the surface shipment. So, we
decided to split in order to have at least 67% of deliveries on time, so 3% under the current-
month order.
The ratio was around:
- 46% air shipment
- 54% surface shipment
Then, only the half of the goods (54%) were likely to have 3% of unfilled orders, hence a good
fill rate during the game.
Finally, we tried to optimise the delivery to have the best price and the most filled orders in
order to have a good customer satisfaction. Then, a good transport system performing brings
benefits to service quality and company competitiveness.
2) Scenario evaluation
In this part, I’m going to talk about a situation I met in an organisation during an internship
carried out. First of all, it is important to present and talk about the company in order to situate
at best the context in which I was. Then, it will be enable to introduce the supply chain issue
and see which solution is best to counter this problem.
Paqu’express is a company where I effected my internship for my undergraduate. The
commercial activity of the company is goods road transport. Then, this company counts 8
deliveries men and the manager. The company is in charge to deliver products in the city of
Marseille in several quarters. Paqu’express is a subcontractor who is responsible for delivering
the merchandise of big firms in France like Chronopost, specialised in the transport of goods in
France and around the world.
Figure 2: organisation chart of Paqu’express
Personnel management
- Accountant
- Commercial
Vehicle fleet
Reverse logistics
Replacements
Schedule
Manager 8 deliveries men

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In this market, the concurrence is fierce with price pull down. Then, the price is not fixed by
the company but by the market itself. The demand is important that it allows a transparent and
fair market even if the concurrence is strong.
In this way, the company cannot play on the price. The company needs to be focused on another
point: be focused on the supply chain and optimised at the maximum to increase the profit.
Given that it is a goods road transport, it is important to be focused on the optimisation of the
transportation of products.
During my internship, while the delivery round, the deliveries did not follow a delivery circuit
well defined. Then, this leads to a waste of time and money.
The company are not going to optimise its transport and consequently its supply chain.
The expenses can be high in terms of petrol, tires wear and the most important regarding the
waste of time: if the time of delivery was optimised, deliveries men could have a delivery round
most dense with the same amount of hours worked.
Therefore, I was in charged to implement a method who could optimise the delivery round of
each delivery man. This method is called the Kruskal’s algorithm or “method of distance”.
This method is used in the transport management system in order to suggest routing solutions.
Knowledge of delivery points, the average frequency and average tonnage delivered to each
recipient are the essential elements for the development of delivery rounds.
Often it is the intuition and experience that allow drivers to organize their delivery rounds. But
this practice does not necessarily serve to achieve the objectives of minimal cost and customer
satisfaction.
Optimization software exist and now allow to properly address this issue. They mostly use
Kruskal's algorithm, capable of manual application for a small number of customers to deliver.
Its objective is to minimize the distances by the transport vehicles or the corresponding
durations. The Kruskal's algorithm is a heuristic method that does not give the perfect solution.
It is based on the simple notion of distance defined as follows:
O is a deposit and two clients A and B. We want to find the shortest way to deliver A and B
from O.
There are two possibilities:
-Deliver A, return filing and B and deliver back to O.
-Include A and B in the same tour.
Distances are written in this way, d (x, y) mean distance from x to y.
1st solution: 2 * d (O, A) + 2 * d (O, B)
2nd solution: d (O, A) + d (A, B) + d (B, O)
The gain or the spacing of the pair (A, B) relative to the centre O is the difference between these
two quantities, namely:
e (A, B) = d (O, A) + d (O, B) - d (A, B)

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3) Plan and risks
In this third and last part, we are going to implement the method introduced in the second part.
Firstly, it is important to follow few steps in order the method is effective.
1. Establish a distance of the points to be delivered, which will indicate distances in km between
points taken two by two (actual distance course given the vehicles).
2. Calculate the distances of all pairs of points from the deposit.
3. Then Sort in descending order.
4. Select each pair of the list; abandon those that are incompatible with those previously
selected.
5. Prepare by bringing end to end, in the order of their value apart, successive links. Stay of
proceedings when all customers were served.
6. Close the chain by bringing the ends from point O (warehouse).
First of all, the first week, I noticed all delivery points and took the top ten. So I got based on a
sample of ten clients.
After applying this method, I noticed saves time on the tour of delivery. Indeed, the tour type
driver was 20.75 km. But with this method, I managed to get by doing the same tour the previous
lower mileage: 17.21 km.
Note that in this method, the tonnage to be taken into account. However, each delivery run, the
payload is never exceeded, because in our case, the average load of 6 kg parcel and found a
hundred parcels in the delivery van. (= 1000 kg payload constraint so not exceeded weight).
In this case, we gain approximately 3 km to time and money.

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1) Distance between points
2) Calculating the distances between each point
AB: OA+OB–AB= 5.1+4.7-0.85=8.95
AC: OA+OC-AC= 5.1+4.7-0.35=9.45
AD: OA+OD-AD=5.1+5.3-0.5=9.9
AE: OA+OE-AE=5.1+4.5-0.65=8.95
AF: OA+OF-AF=5.1+5.8-1.6=9.3
AG: OA+OG-AG=5.1+6.1-2=9.2
AH: OA+OH-AH=5.1+5.3-2.1=8.3
AI: OA+OI-AI=5.1+5.1-1.8=8.4
AJ: OA+OJ-AJ=5.1+5.2-1.8=8.5
BC: OB+OC-BC=4.7+4.7-1=8.4
BD: OB+OD-BD=4.7+5.3-1.5=8.5
BE: OB+OE-BE=4.7+4.5-0.9=8.3
BF: OB+OF-BF=4.7+5.8-2.6=7.9
BG: OB+OG-BG=4.7+6.1-2.9=7.9
BH: OB+OH-BH=4.7+5.3-2.1=7.9
BI: OB+OI-BI=4.7+5.1-2=7.8
BJ: OB+OJ-BJ=4.7+5.2-2=7.9
CD: OC+OD-CD=4.7+5.3-0.6=9.4
CE: OC+OE-CE=4.7+4.5-0.4=8.8
CF: OC+OF-CF=4.7+5.8-1.7=8.8
CG: OC+OG-CG=4.7+6.1-2.1=8.7
CH: OC+OH-CH=4.7+5.3-1.9=8.1
CI: OC+OI-CI=4.7+5.1-1.8=8

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CJ: OC+OJ-CJ=4.7+5.2-1.8=8.1
DE: OD+OE-DE=5.3+4.5-0.9=8.9
DF: OD+OF-DF=5.3+5.8-1.3=9.8
DG: OD+OG-DG=5.3+6.1-1.7=9.7
DH: OD+OH-DH=5.3+5.3-1.8=8.8
DI: OD+OI-DI=5.3+5.1-1.6=8.8
DJ: OD+OJ-DJ=5.3+5.2-1.7=8.8
EF: OE+OF-EF=4.5+5.8-2=8.3
EG: OE+OG-EG=4.5+6.1-2.3=8.3
EH: OE+OH-EH=4.5+5.3-1.6=8.2
EI: OE+OI-EI=4.5+5.1-1.4=8.2
EJ: OE+OJ-EJ=4.5+5.2-1.5=8.2
FG: OF+OG-FG=5.8+6.1-0.35=11.55
FH: OF+OH-FH=5.8+5.3-0.5=10.6
FI: OF+OI-FI=5.8+5.1-0.8=10.1
FJ: OF+OJ-FJ=5.8+5.2-0.85=10.15
GH: OG+OH-GH=6.1+5.3-0.16=11.24
GI: OG+OI-GI=6.1+5.1-0.5=10.7
GJ: OG+OJ-GJ=6.1+5.2-0.55=10.75
HI: OH+OI-HI=5.3+5.1-0.55=9.85
HJ: OH+OJ-HJ=5.3+5.2-0.6=9.9
IJ: OI+OJ-IJ=5.1+5.2-1=9.3
3) Distance in descending order
FG 11,55
GH 11,24
GJ 10,75
GI 10,7
FH 10,6
FJ 10,15
FI 10,1
AD 9,9
HJ 9,9
HI 9,85
DF 9,8

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DG 9,7
AC 9,45
CD 9,4
AF 9,3
IJ 9,3
AG 9,2
AB 8,95
AE 8,95
DE 8,9
CE 8,8
CF 8,8
DH 8,8
DI 8,8
DJ 8,8
CG 8,7
AJ 8,5
BD 8,5
AI 8,4
BC 8,4
AH 8,3
BE 8,3
EF 8,3
EG 8,3
EH 8,2
EI 8,2
EJ 8,2
CH 8,1
CJ 8,1
CI 8
BF 7,9
BG 7,9
BH 7,9
BJ 7,9
BI 7,8

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4) Development of the optimal route
Standard course taken by the delivery driver before the study:
OADCBGFEHJIO= 5,1+0,5+0,6+1+2,9+0,35+2+1,6+0,6+1+5,1= 20,75 km

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Course recommended after study:
OFGHJIADCBEO= 5,8+0,35+0,16+0,6+1+1,8+0,5+0,6+1+0,9+4,5= 17,21 km
Detailed calculation:
Useful information:
Diesel prices at the time of the study: € 1.29/Litters
Consumption of a pickup truck: 6.5 Litters / 100 km
Before study:
Distance raised: 20.71 km
Fuel consumption (in litters) = 6.5 * 20.71 / 100 = 1.35 litters
Fuel consumption (in €) = 1.29 * 1.35 = 1.74 €
Converter
1€ = £0.78

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After study:
Distance raised: 17.21 km
Fuel consumption (in litters) = 6.5 * 17.21 / 100 = 1.12 litters
Fuel consumption (in €) = 1.29 * 1.12 = 1.44 €
If we extend it to an entire tour (60 clients), eight delivery drivers, and over a year, the gains
will be enormous.
In conclusion this method is very successful. We know that in some companies, this algorithm
has been computerized to optimize delivery routes (transport management system: TMS).
However in the company where I did my internship, this software does not exist. So I suggested
to the manager to invest in the future in software that would optimize the delivery of touring all
delivery drivers of the business. The risk is certainly a high cost to afford this software.
It is advised to calculate the price of acquisition of this software and determined the savings
achieved if the company use this software.
Because this method works for ten customers, it would then be able to develop computationally
for a hundred customers (establish a large database). To do this, it should establish a
specification, containing all the customers deliver in database form. This computer program
would then do the work I do manually (all steps of the method), and each day he would
determine the ideal course. At my level, I could not afford to take more than ten clients, because
the reasoning became too tedious.

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