Supply chain model

Problem statement

The company has two factories, one situated at Liverpool and one at Brighton. In addition, it has four depots with storage facilities at Newcastle, Birmingham, London and Exeter. The company sells its product to six customers C1, C2… C6.

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Task

The task was to create a distribution network schematic diagram showing all the potential connection/ flow then use linear programming to minimize the cost of transportation and finally come up with another distribution network of the optimal solution.

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Tools used

This is an optimization problem, so to that end, I used solver in excel to conduct the linear programming, Microsoft word was used to detail the explanation of the linear model. At the end of this write up is the excel sheet which was used to create and optimize the model alongside the answer report, sensitivity report and limits report.

Solution

So to understand the solution I would be explaining the solutions with the terminologies used.

So a linear programming problem is represents one example subset of optimization problem. An optimization problem refer to the selection of the best elements from some of the available alternatives.

So in a linear program, there are terminologies such as objective functions, decision variables, parameters, and constraint and feasibility regions.

There are 4 components of a linear programming model and these are:

• The objective function

• The fixed variable

• The decision variables

• The constraints

The objective function: This is what we want to achieve. The name function may be a bit misleading because it is normally not like a function in a programming sense but this is the objective of our model. For example it may be the number we are trying to minimize or maximize. The objective function is to minimize the cost associated with transporting the unit ton of materials from a place of production through the transshipment routes then to the various customers. So, there could be many variations of product flow with respect to the cost and to that end, I decided to make the objective function, the least possible amount for transportation. This became the sum product of the decision variable which was the flow and the cost of transportation.

The fixed variables: In problems such as this, there are numbers that can’t be changed. This may include individual cost per tonnes, number of demand by the customers as well as maximum supply.

The decision variables: These are the cells we need to change or the variables we need to change to come up with the best solution to our problem. So in this particular problem, the variables were the number of materials to move to ensure minimum spending. Our model had two to three routes. The routes from the factory to the transshipment routes, the routes from the transshipment routes to the customers and finally the routes from the factory to the customers. With respect to the formulation of the decision variables, the table was created with all the routes and the total demand and total supply were stipulated on the table alongside the cost of transportation by the possible routes of transportation. This table was copied and the labels for cost changed to flows. The cost of transportation was used to select the number of flows which then allowed for optimization of the problem. Some routes were not possible so to that end an enormous amount was put so that solver won’t use those routes

Constraints: The constraints are the conditions which must be met and limits the decision variables from going up to infinity. The constraint of the problems were that:

The total number of the factories’ capacities would be greater than or equal to the demand of the transshipment routes or possible customers supplied.

The total number of inflow of the customer level and the transshipment level would be equal to their throughput. This was done because the goal was to see the maximum throughput of the routes.

The total number of throughputs from the transshipment routes would be greater than the number of throughputs at the customer level.

The number of supply from the transshipment routes should be equal to the number of inflow. This was done because there is no demand at the transshipment route and the sum of outflow and inflow should be equal to zero. It was also done so as to minimize amount of product moving from the factories to the warehouses.

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Conclusion

So this is a transshipment problem with warehouse and final customers as already highlighted in the problem statement. So in conclusion, with respect to the model created, it showed that the supply was more than the demand.  Newcastle for example had no outflow and subsequently no inflow in the model because the cost of transportation through Newcastle was high when compared to places like Exeter.

Recommendation

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So with respect to the model, more attention should be paid to the outflow than the inflow to reduce movement of products through routes that had higher cost of transportation. More attention should also be placed on the Exeter route as its cost of transportation is low but its throughput rate is the lowest.