Supply chain management
Case Study (SCM):
Fowler Distributing Company
Project Title
Case (SCM): Fowler Distributing Company Logistics and Distribution Planning Supply Chain Delivery Plan
Table of Contents
Abstract: 5
Problem statement: 6
Executive Summary: 7
Introduction 8
Literature review: 9
Methodology: 10
Analysis: 13
Finding, and results: 21
Conclustion: 23
References: 24
Appendices : 25
List of Figures
Figure 1 10
Y
Figure 2: 1 X, Y coordinates 12
Figure (3: 1) 5 routes design for SCM CASE 1 (2) 13
Figure (4:1): Optimum design for SCM CASE 1 (2) 14
Figure (5:1) 5 route design for SCM CASE 1 6(2) 15
Figure (6:1) 5 route design cancel 50 cases and less. 16
Figure (7:1) Without a time window. 18
Figure (8:1) New warehouse location (3) 19
Figure (9:1) No overtime…………………………………………………………………….20
List of tables
Table (1:1) Represent the how to calculate distance, and time from one point to another 12
Table (2:1) 5 routes design for SCM CASE 1(2) 13
Table (3:1) Optimum design for SCM CASE 1(2) 1
Table (4:1) Trucks at 600-cases of capacity 1. 15
Y
Table (5:1) 5 routes design cancels 50 cases and less. 16
Table (6: 1): Without time window. 17
Table (7:1) New warehouse (3). 18
Table (8:1) Without time window…………………………………………………………….19
Table (9:1) Major results. 22
Abstract:
Distribution system is a branch field from supply chain management which has generated much interest in recent years. They are inherent monopolies and therefore they have generally been regulated to protect customers and to ensure cost-effective operation. In the modern world, the management science is playing a major role in every company and organization and the right management decisions can make a significant difference between the competitors. In addition, supply chain management is the knowledge that applied to design or optimize the flowing of goods or services for the most benefit out of the process by decreasing the time of the process, increase the efficiency of the process, etc. Many organizations are suffering from the wrong optimization, or design for their supply chain, and every organization check and apply changes if needed to its supply chain design. Roy is a distributor for a juice and water who face some obstacles regarding delivering of his products to pre-sell accounts customers on time without delaying their orders This paper is going to show how he has managed the supply chain distribution for his company and what are the mistakes that occurred in the distribution plan, what is its impact on the distribution process in regard to spent time and the total cost. It will also show how to optimize the distribution design for the company using the sweep method and changing multiple factors in the distribution elements to reach to the best optimization possible and giving the recommendations regarding that distribution plan.
Keywords:
Supply chain management, cost-effective operation, sweep method
Problem Statement:
Fowler Distributing Company provides an analysis of the current distribution network. The purpose of the assignment is to investigate the current routes designed to formulate proposals or optimize new designs that will contribute to reductions in the amount of cost of the current design. The optimization solution has given a methodology of how to optimize the number of tracks, trucks capacity, and satisfy the customer time window. In the First scenario, the current routes are of Fowler’s trucks are used and optimized to meet customers’ requirements. In the second scenario, using the current trucks or larger capacity trucks to minimize total cost. In the third scenario, the impact of the customer variables, stop sequence for the trucks, truck capacities, and time windows are analyzed to determine their impact on total costs.
Executive Summary:
This paper is going to discuss how Roy Fowler designed his distribution plan to meet customer’s time windows. The plan was using five trucks that have 500 cases capacity and eight hours of working time before the overtime starts for the drivers. The time window for each customer is different than the others but all customers participate in a period between 8 am and 5 pm. The cost of the distribution depends on multiple factors the driver’s working cost, the truck operating cost, and the truck initial price. The drivers are paid 13 dollars per hour and it increases to be 20 dollars per hour for overtime. The truck has an operating cost of 0.9 dollars per kilometer and its initial cost is 20000 dollars for each with a seven years useful life. To optimize the routes, the first route must be analyzed to see the weaknesses, then other routes can be optimized easily and to make them more efficient than the current Roy Fowler design, which was costs 771.86 dollars per day. The optimizations can be modified in three variables that affect the route’s design, these variables are the time window, the number of tracks, and their capacities. The first optimization is changing the stop sequences to reduce the overall time spending during the distribution working time, the daily cost is 752.81 dollars, and the annual cost is 201,062 dollars. The Second optimized design is “no time windows restrictions” for each customer by using 5 trucks of 500 cases capacity, and its daily cost is 714.13 dollars while its annual cost is 191,393 dollars. The third optimized design is “600 tracks capacity” which have an operating cost of 0.95 dollars per mile with the initial cost is 35000 dollars for each truck. The number of routes and trucks is four and the final cost is 726.06 dollars per day and 199,517 dollars per year. The fourth optimized design is” NO time windows restrictions”, the design is to make sure all cases are delivered within the working time for the drivers without having any overtime by using 5 trucks of 500 cases capacity and the result is 750.65 dollars per day and 200,522 dollars per year. The fifth optimized design is changing the warehouse location to be in the center which results in changing the distance, and total time work which results in minimizing the total cost. The cost of a new warehouse is 679.74 dollars per day, and the annual cost is 182,796 dollars again by using 5 trucks of 500 cases capacity. Furthermore, the changing of the warehouse location will be one-time payment which is cost 15000 Dollars. The sixth optimized design is to” Cancel 50 cases and less” route design, the annual cost is 206408$, according to the analysis of the outcome result, which result in outside transport service is not a good choice comparing to the current design. Finally, the “new warehouse” is the best choice to decrease the cost.
Introduction
Fowler is a local beer and wine supplier started by Vietnam War veteran Roy Fowler. He owns 5 trucks and is responsible for operating expenses such as maintenance, and fuel costs related to his trucks. The five trucks are operated by union drivers who earn $13 per hour which includes a 30 percent fringe benefits package. Drivers might be paid overtime if they work more than eight hours. However, Mr. Fowler does not want to pay overtime. Each of the five delivery vehicles has a capacity of 500 cases.
Mr. Fowler provides distribution services from the local warehouse to 21 customers who they are requesting demands are increased. Each of these 21 “pre-sell” accounts requires distribution services 250 days a year and are located anywhere from 1 to 35 miles from the warehouse. There are different time windows for every one of the 21 “pre-sell accounts” which must be complied with. Besides, each of the “pre-sell” accounts have different demands that must be met by Mr. Fowler. Distribution trucks must leave the warehouse between 6:30 and 8:00 am.
With the current five trucks and route configuration, he is capable to meet time window, and also capacity of the truck’s restrictions. Nevertheless, he would like to find more economical ways to meet his customers’ demands to decrease costs.
Literature review:
Delivery products are one of the most important things for any company, therefore failure, as well as the success of any company depends on the services that are provided to its customers. Thus, logistic planning is one of the key factors that can achieve valuable services to make sure customer satisfaction[1], by finding the optimum solution for the vehicles routes problems, which is “belongs to the class of NP-hard combinatorial problems”[2] The main basics reasons are minimizing the number of trucks, and driven miles to minimize the total cost. Anyway, in this case, the fundamental formulation of this problem is how to manage the vehicle capacity, optimize the route’s design, and the period in which every customer has to bed served within a time window [2].
For solving this case problem, there were diversity of procedures that has been proposed.” Older methods developed for the vehicle routes problem are described in 16 Vehicle Routing Problem the survey (Cordeau et al., 2002) and (Laporte, 1992)”[2]. Most of the latest techniques tested on Solomon’s benchmarks are consist of in (Bräysy & Gendreau, 2005a; Bräysy & Gendreau, 2005b) [2]. The ways that used the two-stage methodology for solving vehicle routes problems are observed to be extremely successful (Bräysy & Dullaert, 2003) [2]. For the Period Of the first stage, the constructive heuristic procedure is used to create a possible initial solution. In the second stage, an iterative improvement heuristic is applied to the initial solution. The method for avoiding the local best-case scenario is often applied in the second stage [2].
However, there are also much research which is already done by some scientist such as, Anderson, which also propose better ways to optimize the performance of supply chain [3]. So they suggest many principles, however, we will mention some of them which are related to our case study, and also typically as the previous descriptions, so they segmenting the customer according to their requested and providing them all the recruitment services which have been asked [3].
Customizing the logistics network to solve the vehicle routing problem with split distributions was presented by Dror and Trudeau (1989, 1990) [2]. Who presented the mathematical formulation of the problem and analyzed the economy that can be made when it is permitted that a customer is satisfied by more than one vehicle, economy both related to the number of vehicles and total distance traveled [2] in (1992, 1995) Frizzell and Giffin have developed structural heuristics enhancement for the vehicles routing distribution through grid network distances. They also considered time windows constraints [2].
Methodology:
To get this case study done we follow the sweep method by locating all stops including the depot on a map or grid. Furthermore, extend a straight line from the depot (warehouse) in any direction, then sweep it clockwise or counterclockwise, and when it touches the first stop a straight line from the warehouse.
At the end of each route, the trucks have to go back to the warehouse to refill by forming the teardrop method for the best formation and minimum distance as possible.
Therefore, the main objectives of the case are analyzing and optimizing different vehicle routing relative to the warehouse, to minimize costs, moreover meeting various constraints such as customer time windows, as well as truck capacity. Figure [2] shows the number of cases demanded, the required time (in minutes) to serve the account, and the time window in which the account can be served, if restricted. By using five trucks which have a capacity of 500, and then optimized the routes based on the customer time window. Also, other choices as using larger capacity trucks for the previous reasons.
The best optimization for the number of tracks and routes can be done by using the data in figure [2] and applying it in the following mathematical equations:
This is the way which we follow to get our results and finding the answers to the case’s questions.
- The mathematical equation below expresses how to get maximum number of the cases that can be separate in each stop sequences.
Figure 1
- For getting the distance from point to another point we used Pythagoras’s theorem to find the results, where “a” is the difference between X coordinates for the two points, “b” is the difference between Y coordinates for the two points, and “c” is the distance between these two points.
- The second important parameter is the interval period during traveling from one point to another point.so to find the time we used the following equation:
Where the average speed is 25 miles/hour.
- The third critical parameter is finding the operational cost, where the operational cost represents the cost of drivers, and the cost of the tracks, where the cost of the driver is equal to time distance from one point to another, also the delivery multiply by 13$ per hour “, where the cost of the truck is 20000$
distance multiply by 0.9$/mile this cost only for the truck’s capacity 500 cases, however, the cost of the trucks which have a capacity of 600 cases is the distance multiply by 0.95$/mile
Where the salvage value of the truck is 10% of its initial cost
- Salvage value for truck 500 cases=20000*0.1=2000.
- Salvage value for truck 600 cases=35000*0.1=3500.
- The cost of 600 capacity’s truck for one year is = (20000-2000)/7=2572 $.
- The cost of 600 capacity’s truck for one year is = (35000-3500)/7=4500 $.
Furthermore, we considering the overtime cost for the driver, which is cost 20$ per hour
for instance, the cost for the current design for one year is equal to:
Where,
Total transportation cost
the load expressed as weight, number of trips or units being shipped from the proposed site and location
: coordinate location of the proposed site.
: cost shipping one unit for one mile between the facility and market or supply source
: quantity to shipped between facility and market or supply warehouse
: coordinate of the existing facility.
Figure 2: 1 X, Y coordinates
| A | Distance | Time | From -to |
| 1 | =SQRT(((I5-K5)^2)+((J5-L5)^2)) | =(M5/25)*60 | w to 7 |
| 1 | 8.732125 | 20.9571 |
Table (1:1) Represent the how to calculate distance, and time from one point to another
Analysis:
To start our optimum design analysis, we have to analyze the current case study routes by determining the total distance and time. To assess our optimum design. where the total current design cost is 205,827 dollars per year and to meet all the requirement’s demands of fowler’s twenty-one pre-sell customers, they use all the five tracks for the current design. Besides, we need to analyze the overtime cost for the current design which is costs them 54.3 $ per day, therefore according to the current analytical design’s data we create our design to optimize the current designs.
- The table below which illustrate the number of routes which are as shown five routes, also the start time for each route, furthermore, the stop sequences for each route
| Stop sequence | Start time | Rout |
| 12,15,1,14,5 | 7:45 A.M. | 1 |
| 2,3,4 | 7:33 A.M. | 2 |
| 6,16,17,8,19 | 7:22 A.M. | 3 |
| 11,20,18,21,9 | 8:00 A.M. | 4 |
| 7,13,10 | 7:39 A.M. | 5 |
Table (2:1) 5 routes design for SCM CASE 1(2)
On the right figure is the current design routes which display the path of how the driver will go from one point to another through these stop sequences frame.
As it clearly has shown the five routes is covered by five tracks, therefore the total distance for these routes is 339.23 mile, and the total time 2124.143 minutes, including the overtime which is 54.3 minutes.
Figure (3: 1) 5 Route design for SCM CASE 1 (2)
Figure (4:1): OPTIMAM design FOR SCM CASE 1 (2)Figure (3: 1) 5 route design FOR SCM CASE 1 (2)
The operational cost of the current design as the following, first we need to find out the total distance which already shown how it is calculated in the methodological part, also total time for route’s design, besides to determining the overtime work. The operational cost for a day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year)
$.
- According to the analytical data information from the current design analysis, we optimized the current design, so the following table and figure illustrate how many routes, the stop sequence for each route, and the start time.
| Stop sequence | Start time | Rout |
| 7,13,10 | 7:40 A.M. | 1 |
| 9,11 | 12:45 P.M. | 2 |
| 20,21,18,16,19 | 7:09 A.M. | 3 |
| 2,17,8,6 | 8:00 A.M. | 4 |
| 12,1,15,14,5 | 7:45 A.M. | 5 |
| 4,3 | 8:00 A.M. | 6 |
Table (3:1) Optimum design for SCM CASE 1(2)
As it is shown in the right figure (4) we adjust six routes which is covered by five tracks, also the total distance for these routes is 331.57 miles, and the total time 2075.77 minutes, including the overtime which is 39.86 minutes.
Therefore, the operational cost of the optimum design can be found by calculating the total distance, total
Figure (4:1): Optimum design FOR SCM CASE 1 (2)
time for route’s design, and the overtime work. The operational cost for a day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year)$.
- Within this case study, we had another option of using larger tracks which have a capacity of six hundred cases to reduce the amount of total cost as will be explained below.
| Stop sequence | Start time | rout |
| 7,13,10,9,11 | 7:39 A.M. | 1 |
| 20,21,18,19,16 | 7:09A.M. | 2 |
| 17,2,8,6,5 | 6:49A.M. | 3 |
| 12,1,15,14,3 | 8:00A.M | 4 |
| 4 | 15:38 P.M. | 5 |
Table (4:1) Trucks at 600-cases capacity 1.
The second option designing is the 600 hundred track’s capacity, the number of the routes are five routes, and number of tracks is 4 tracks, as it shown on the table 4 above, and in figure 5.
Figure (5:1) 5 route design for SCM CASE 1 6(2) Also, the total distance for these routes is 300.5721 mile, and the total time 2031.373 minutes, including the overtime which is
132 minutes
The operational cost for the 600 hundred track’s capacity designing founded by
determining the total distance, and total time for the route’s design, to calculate the overtime work. The operational cost for a day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year).
- The key factors for this case study are minimizing the total cost as well as satisfy the customer’s demands. Using an outside transport service to deliver all accounts with the demand of 50 cases or less, the price has to be $35.00 per account according to the case assignment.
| Stop sequence | Start time | Rout |
| 7,13,10 | 7:40 A.M. | 1 |
| 9,20 | 12:41 P.M. | 2 |
| 21,17,16,19,8 | 8:00 A.M. | 3 |
| 6,2,15 | 8:00 A.M. | 4 |
| 12,1,3,4 | 8:00 A.M | 5 |
Table (5:1) 5 routes design cancels 50 cases and less
By doing all the mathematical operations to determine the total distance for these routes which is 264.49258 mile, and the total time is 1824.782 minute, also including the overtime which is 94.874, besides the number of tracks which are four in this
Figure (6:1) 5 route design cancel 50 cases and less
particular design on right figure. So here are the outcomes of 5 routes design cancel 50 cases and less. The operation cost for outside transport service per day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year).
Depend on our analysis of the outcome result, we believe that, it is not a good choice to use outside transport service because the cost of using outside transport service is more expensive than the current design, as well as our optimum design. Therefore, it useless to use it for the main purpose that we need to achieve.
- To create the best design route, we follow the sweep method by locating all stops including the depot on a map or grid. Besides, the Heuristics method which load points closest together on the same track, build routes starting with points farther from the depot first, and the routes should not cross, and create a form of teardrop pattern routes .
| Stop sequence | Start time | Rout |
| 7,13,10,11,5 | 7:40 A.M. | 1 |
| 9,20,21,18,19 | 7:30A.M. | 2 |
| 8,17,16,6 | 8:00 A.M. | 3 |
| 12 | 14:30 A.M. | 4 |
| 14,15,2 | 8:00 A.M. | 5 |
| 1,3,4 | 8:00 A.M. | 6 |
Table (6: 1): Without time window
In figure 7 we neglect the time window restrictive, create a form of teardrop pattern routes, we followed the previous method.
However, we have design
Figure (7:1) Without a time window the routes which display the path of
than other driver will go from one point to another which is shown on the figure 7.
As it visibly shown we have six routes which is covered by five tracks, so the total distance for these routes is 303.02725 mile, and the total time 2037.265 minutes
The operational cost of the teardrop form design as the following, as the same finding out the total, in addition total time for route’s design. The operational cost for a day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year)$
| Stop sequence | Start time | Rout |
| 7,13,10 | 7:27 A.M. | 1 |
| 9,11 | 13:28 P.M. | 2 |
| 20,21,18,19,16 | 8:00 A.M. | 3 |
| 17,2,6,8 | 8:00 A.M. | 4 |
| 12,1,15,14,3 | 7:19 A.M. | 5 |
| 5,4 | 8:00 A.M. | 6 |
Table (7:1) New ware house (3)
This figure represents a new central location for the warehouse at coordinates X
Figure (8:1) New ware house (3) = 20, Y = 25. However, the rent cost for a new central warehouse is the same as the current location, we followed the same previous method to designed the distributions Flower’s new warehouse which is finding total distance for these routes which is 276.70484 mile, and the total time is 1984.092 minute, also including the overtime which is 7 minutes, besides, the number of tracks which are 5 in the new central location warehouse. therefore, the cost results of the new location as the following,
The operational cost for a day is equal to tracks cost plus drivers cost. =276.70484*0.9$+(1977.092/60*13$+7/60*20$) =679.7455$
, where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year) =250*679.7455$+5*2572$=182,796$
| Stop sequence | Start time | Rout |
| 7,13,10 | 7:40 A.M. | 1 |
| 9,11 | 12:41P.M. | 2 |
| 20,21,18,19,8 | 8:00 A.M. | 3 |
| 2,16,17,5 | 8:00 A.M. | 4 |
| 12,1,15,14, | 7:45 A.M | 5 |
| 4,3,6 | 8:00 A.M | 6 |
Table (8:1): No over time
The No over time designing as the following the number of the routes are six routes and number of tracks is five trucks, as it shown on the figure above, and in the table 8.
Furthermore, the total distance for these routes is 328.7449 mile, and the total time is 2098.988 minutes.
The operational cost for a day is equal to tracks cost plus drivers cost. , where the annual cost is equal to (250* the operation cost of a day) + (track’s cost for one year).
Figure (9:1): No over time
Finding, and results:
By reviewed the current route design configuration and used it as a reference to evaluate our optimum design modifications. Table 8 clarifies, and compare all routes which are given by the case study. as clearly shown in table 8 the annual cost of the current design is. 205,827$, the current design used five tracks to meet all the required needs for the customers. The daily overtime cost is 1086$. However, the annual cost of the optimized route design, using five trucks is 201063$, and the daily overtime cost is 800$. The enhancement of the optimum design here is the amount of money saved annually which is 4763.6$ comparing to the current design. By reducing the amount of distance, time, and overtime work.
On the other hand, the annual “No time windows route design” cost is 191,393$ using five tracks which is a good solution, because there is no overlapping in any single route and all routes have the classic teardrop shape.
The annual cost for the “six hundred Trucks capacity route design” is 203274 dollars, although this route has 600 tracks capacity, it supposes to cost lower. However, it is still expensive to the optimum design because of two main reasons which are initial cost of the trucks, the initial cost for one truck is 35000$, so the total initial cost for this design is increased by 40000$, the second reason is the overtime work is increases as well, and the amount cost of the overtime is 20$ per hour. Furthermore, it is cheaper than the current design. However, in our point of view, it is not a good solution according to the analysis data costing. The six-row in the table above is shown the “No overtime route design”, the annual cost for the No overtime route using five tracks is 200,522 dollars, depending on the rest of all designing routes, it is somehow a good solution to target the case’s objectives. Last but not least, the findings of the annual cost for designing “new warehouse” location using five tracks are 182,796 dollars. Depend on the analysis of the results, we believe that, the “New warehouse design” is the best choice because the annual cost is 182,796 dollars. Which is the cheapest design comparing to all the rest designing routes as shown in table 8. The annual amount of money saved is 23031 dollars comparing to the current design. Therefore, it is the best choice design to select for the main purpose that we need to achieve.
Final, the “Cancel 50 case and less” route design, the annual cost is 206408 dollars, according to the analysis results, we believe that, it is not a good choice to use outside transport service because the cost of using outside transport service is more expensive than the current design, as well as our optimum design. Therefore, it useless to use it for the main purpose of the case’s study.
Finding, and results:
| Annual cost | Cost for a day | Number of routes | trucks | Over time cost | Over time | time | distance | Run type |
| 205,827 | 771.8662 | 5 | 5 | 1086 | 54.3 | 2124.143 | 339.22641 | Current design |
| 201063.4 | 752.814 | 6 | 5 | 800 | 40 | 2075.769 | 331.57056 | Optimized current |
| 191,393 | 714.132 | 5 | 0 | 0 | 2037.265 | 303.02725 | No time windows | |
| 5 | 4 | 2640 | 132 | 2031.373 | 300.5721 | Trucks at 600-casa capacity | ||
| 200,522 | 750.6511 | 5 | 0 | 0 | 2098.988 | 328.74491 | No over time | |
| 182,796 | 679.7455 | 5 | 140 | 7 | 1984.092 | 276.70484 | New warehouse location | |
| 644.48 | 4 | 1880 | 94.874 | 1824.782 | 264.49258 | Cancel 50 case and less | ||
| 784.48 |
Table (9:1) Major results
Conclusion:
The case study shown how changing the factors of supply chain distribution can make a major impact on the process and the outcomes of it. As the paper shows there is multiple optimizations that surpasses the current design of the distribution plan and it varies from one route to another. The best optimization the distribution plan can benefit from is the new warehouse location and it shows a significant difference from the current design so it is the optimization that must be followed due to the high decrease in the cost. The paper shows that the worst optimization is the “Cancel 50 cases or less” it is even costing higher than the current design then this optimization is not a choice to have a solution for enhancing the distribution plan. The rest of the optimizations in this paper can enhance the current design in different ways and if the optimizations ways have been done in this paper merged it may show a better overall optimization for the distribution plan. In addition, there are some quality issues that may affect the current design, for example, the quality of the roads from the warehouse to the customers locations, accuracy of the location coordinates and the availability of the customer in his location according to his time window. On the other hands, some risks may affect the designed distribution plan like traffic, accidents in the road, technical breakdowns in trucks and weather fluctuations. Finally, Performance measures and metrics are essential for effectively managing logistics operations, particularly in a competitive global economy. So,ppropriate KPIs must be set to measure productivity and performance as ordered capture, inventory management, warehousing, costs of goods sold and transportation.
References:
[1] C. Ratke, H. Hoffmann, and H. Nehring, “Routing of Vehicles Using CSP: Case Sudy,” in 2015 Ninth International Conference on Complex, Intelligent, and Software Intensive Systems, Santa Catarina, Brazil, Jul. 2015, pp. 250–253, doi: 10.1109/CISIS.2015.34.
[2] A. H. El Hassani, A Hybrid Ant Colony System Approach for the Capacitated Vehicle Routing Problem and the Capacitated Vehicle Routing Problem with Time Windows. 2008.
[3] S. Chidambaram, L. Whitman, and S. H. Cheraghi, “A Supply Chain Transformation Methodology,” p. 6.
Appendices:
| Rout1 truck 1 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | distance | y2 | x2 | y2 | x1 |
| 70 | 08:00 | 07:39 | PWH-P7 | 140 | 20.95 | 8.73 | 38.5 | 23 | 35 | 15 |
| 50 | 09:32 | 09:10 | P7-P13 | 80 | 21.89 | 9.124 | 40 | 32 | 38.5 | 23 |
| 90 | 10:42 | 10:22 | P13-P10 | 180 | 19.68 | 8.2 | 33.5 | 27 | 40 | 32 |
| 12:41 | 12:12 | P10-PWH | 29.02 | 12.09 | 35 | 15 | 33.5 | 27 |
Total driving time for rote1 =91 mint
Total time for rote1=91+210=301mint
Total distance for rote1 = 38.144 mile
Total delivery time=210 mint
| Rout2 truck 1 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | distance | y2 | x2 | y2 | x1 |
| 60 | 01:52 | 12:41 | PWH-P9 | 110 | 30.734 | 12.80625 | 25 | 23 | 35 | 15 |
| 20 | 03:08 | 02:52 | P9-P11 | 30 | 16.099 | 6.708 | 28 | 29 | 25 | 23 |
| 70 | 03:44 | 03:28 | P11-P20 | 150 | 16.099 | 6.708 | 22 | 32 | 28 | 29 |
| 05:45 | 04:54 | P20-PWH | 51.362 | 21.4 | 35 | 15 | 22 | 32 |
Total driving time for rote2 =114 mint T
Total delivery time=150mint
Total time for rote2=114+150+30=294mint
Total time for rote2 = 294 mint
Total distance for rote2 = 47.62 mil
| Rout3 truck2 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | distance | y2 | x2 | y2 | x1 |
| 40 | 08:00 | 06:54 | PWH-P21 | 80 | 65.287 | 27.20 | 13 | 31 | 35 | 15 |
| 30 | 08:57 | 08:40 | P21-P18 | 50 | 16.97 | 7.071 | 8 | 36 | 13 | 31 |
| 60 | 09:48 | 09:27 | P18-P17 | 140 | 21.6 | 9 | 8 | 27 | 8 | 36 |
| 60 | 10:58 | 10:48 | P17-P16 | 100 | 9.6 | 4 | 8 | 23 | 8 | 27 |
| 70 | 12:25 | 11:58 | P16-P19 | 90 | 27.364 | 11.4 | 1 | 32 | 8 | 23 |
| 03:36 | 02:05 | P19-PWH | 91.231 | 38.013 | 35 | 15 | 1 | 32 | ||
| 460 |
Total driving time for rote1 =232 mint
Total delivery time=260mint
Total time for rote3=232+260+30=522mint
Total distance for rote3 = 96.68 mil
| Rout4 truck 3 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | distance | y2 | x2 | y2 | x1 |
| 40 | 08:00 | 07:12 | PWH-P8 | 60 | 48.373 | 20.155 | 16.5 | 23 | 35 | 15 |
| 90 | 09:16 | 08:40 | P8-P2 | 200 | 36.01999 | 15.00 | 9 | 10 | 16.5 | 23 |
| 50 | 11:10 | 10:46 | P2-P6 | 50 | 24.621 | 10.259 | 16.5 | 17 | 9 | 10 |
| 90 | 12:30 | 12:00 | P6-P15 | 160 | 29.686 | 12.369 | 13.5 | 5 | 16.5 | 17 |
| 03:27 | 02:30 | P15-PWH | 56.90 | 23.711 | 35 | 15 | 13.5 | 5 |
Total driving time for rote4 =195mint
Total delivery time=270mint
Total time for rote3=195+270+30=495mint
Total distance for rote4 = 81.5 mil
| Rout5 truck4 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | distance | y2 | x2 | y2 | x1 |
| 50 | 08:00 | 07:44 | PWH-P12 | 90 | 15.3674 | 6.403 | 40 | 11 | 35 | 15 |
| 60 | 09:19 | 08:50 | P12-P1 | 120 | 28.8499 | 12.020 | 28.5 | 7.5 | 40 | 11 |
| 60 | 10:34 | 10:19 | P1-P3 | 120 | 15.2735 | 6.3639 | 24 | 12 | 28.5 | 7.5 |
| 80 | 11:49 | 11:34 | P3-P4 | 150 | 14.5986 | 6.0827 | 30 | 13 | 24 | 12 |
| 01:52 | 01:09 | P4-PWH | 12.9243 | 5.385 | 35 | 15 | 30 | 13 |
Total driving time for rote5 =87mint
Total delivery time=250mint
Total time for rote5=87+250+30=367mint
Total distance for rote5 = 36.25 mil
| Rout6 truck4 | ||||||||||
| location | point 2 | point1 | ||||||||
| DELIVERY | END TIME | STAR TIME | from-to | CASES | Time | Distance | y2 | x2 | y2 | x1 |
| 30 | 02:37 | 01:52 | PWH-P14 | 50 | 44.59 | 18.5 | 18 | 7.5 | 35 | 15 |
| 40 | 03:48 | 03:07 | P14-P5 | 50 | 40.6 | 16.9 | 34 | 13 | 18 | 7.5 |
| 04:33 | 04:28 | P5-PWH | 5.36 | 2.23 | 35 | 15 | 34 | 13 |
Total driving time for rote6 =90mint
Total delivery time=70mint
Total time for rote5=70+90=160mint
Total distance for rote6= 37.63 mil