Logistics planning technique and applications

Extract

[...] To build a route we have to start each route from the DC and sum them to be in accordance with the transported quantity. We choose a random first Retailer to deliver then we try to deliver as much as possible in the time clock wise. Route Route Mileage Quantities Number Mileage Once done we try by starting again and form the beginning from the first next retailer on the right. Route Route Mileage Quantities Number Mileage As quantity is limiting the possibilities we just have two different routes and the second one is the best suited. [...]

[...] London: Geraldine Lyons. 540.) That is why we have to adapt our constraints. We want to satisfy the maximum of retailers in despite of the few unit available so we have to deliver all the unit available: D = A and we want that these retailers get the unit available so D B Here there is a feasible area defined by 740] ∩ 1065] meaning that the solutions ensemble is equal to 740]. So in order to solve this problem later on we have to define the constraints like these: X11+X12+X13+X14+X15 = 180 X21+X22+X23+X24+X25 = 250 Constraints due to availability X31+X32+X33+X34+X35 = 310 X11X21X31 170 X12X22X32 220 X13X23X33 300 Constraints due to demand X14X24X34 230 X15X25X35 145 Now we just have to define the objective function to complete this model. [...]

[...] Let X the name of the variable We have in this case 15 variables. We have now defined the constraints. We can find those out with the "Table & "Table of the Course work. Distribution Capacity Retailer Demand (Units Usually for this type of problems the constraints would be defined by: Sum of the delivery for a Retailer Number of units available Sum of the delivery for a Retailer Demand for this particular retailer But here, the total demand is much more than the number of unit available, so if we let the previously given constraints we will not found any solution because there will be no feasible area. [...]

[...] The target cell is now B20, we are looking for the minimum and the changing cells are B10 & C10. Then when we'll press solve, EXCEL is giving us the perfect location for the new Distribution Centre. X-Coordinate Y-Coordinate Location Retailers distance from DC Retailers Distance So the best Location for the new DC is on the coordinates 13). We can figure it out on a graphic: Now that the DC is well placed we should evaluate the required capacity. [...]

[...] both have advantage and convenience. The saving algorithm need a vehicle for each route created, that mean that here we'll need five vehicle. Moreover, the saving algorithm don't take hold of the quantity transported but the reading told us that we dispose of an unique vehicle with a maximum capacity of 400000 units. So in our mind the best suited algorithm is the sweep one. C.2: Determination of a routing plan As we choose the Sweep Algorithm, we'll now use it to solve the routing problem. [...]