9.5. Objective Function and Revenue Calculation

Finally, we have this minimum requirement that between A and B, we must apply at least 2, between B and C we must apply at least 2 and between A and C we must apply at least 2 and these are the sums of the indirect and direct, we do not distinguish between them. And of course, every capacity must be non negative, so this gives us all are constraints.

In this part, we outline the minimum bandwidth requirements for each connection between the users A, B, and C. We need to ensure that at least 2 Mbps is allocated to each pair (A to B, B to C, and A to C). This requirement is crucial as it ensures that all users receive adequate service. Additionally, it's essential to note that none of the allocated bandwidth can be negative; hence, all flow values must be nonnegative.

Think of this scenario like a food delivery service that guarantees a minimum amount of food (e.g., two meals) to every customer (A, B, and C). Just like the service ensures that each customer gets at least the minimum they ordered, the bandwidth allocation ensures that each connection meets the bare minimum requirement.

What are the objective functions? The objective function is the revenue that we realize. So, the A to B connection is x A B plus y A B, this is the total volume that gives us 300. Similarly, x B C and y B C gives us 200 and x A C plus y A C gives us 400. So, we multiply 300 into x A B plus y A B 200 into x B C plus y B C and 400 into x A C plus y A C and add it up, this is our total revenue and you want to maximize this revenue.

Here, we define the objective function for our linear programming problem, which is to maximize revenue from bandwidth allocation. Each pair of connections generates different revenues depending on how much bandwidth is allocated. For example, 'xAB + yAB' represents the total bandwidth from A to B and yields 300 rupees for each megabit transmitted. Similarly, connections to B from C and A from C contribute 200 and 400 rupees respectively. By finding the optimal values for 'x' and 'y' variables (the amounts of bandwidth used), we aim to achieve the highest possible total revenue.

Imagine managing a snack stand at a sports event. You charge different prices for different snacks: 300 rupees for sandwiches, 200 for chips, and 400 for drinks. Your goal is to sell the right amount of each snack to maximize your total earnings while ensuring that you meet customers' minimum snack requirements (like in our bandwidth example)!

So, for these particular numbers, these are the answer that we get, that we have nothing flowing directly from A to B, we have 7 going from A to B directly and so on. So, if you look for the example at this link. So, this link lies on the direct route from A to B, so that is 0, it lies on the indirect route from A to B, that is 7, it lies in the indirect route from B to C, that is 1.5 and it lies on the indirect route from B to C, that is other 1.5. If you see that this is 10, therefore this link as a total capacity of 10 and all 10 units are utilized, given the combination that it gives.

This part describes the specific results obtained from our bandwidth allocation model. For instance, while no bandwidth is being directly routed from A to B, 7 Mbps is indirectly going from A to B. This flow is part of the total capacity allowed on the link, which is capped at 10 Mbps. Each path's flow is tracked, and by summing them up, we conclude that the total utilization of the link is effectively maximized within its limits.

Consider a busy highway with multiple lanes. Even though one lane might be empty (no direct traffic from A to B), cars are still using alternate routes (the indirect routes from A to B and B to C). By managing the overall traffic flow and coordinating which routes are being used, city planners can ensure that the total number of cars (bandwidth) does not exceed what the highway can handle (its capacity).

There is however, another problem which is the way we are actually set up the linear program. So, we have the set up the linear program is to take each possible way of outing the traffic. So, we have A to B, we have a link, we have A to B like this, we have another paths and for each path, we have a variable here, x A B, y A b and so on. So, the every path is represented by the quantity flowing through that path.

The problem highlighted here suggests that the method used to design the linear program could be inefficient. For every possible path in the network, a variable needs to be created. With more nodes or users in the network, the number of paths increases exponentially, leading to a more complex and unwieldy model. This complexity can complicate the analysis and computation of optimal solutions.

Think of a courier service where every possible route to deliver packages is charted out. If you only have a couple of deliveries each day, that map isn’t a problem. However, if you suddenly have hundreds of deliveries and routes, the task of coordinating every possible path becomes overwhelming and inefficient, much like the exponential increase in variables with each additional user or connection in our network.

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