The figure below shows the nodes (A - I) and capacities (labelled on arcs in packages/day) of a shipping network. The firm would like to know how many packages per day can flow from node A to node I.
Note: This question requires Solver.
Formulate the problem in Solver and find the optimal solution. What is the maximum amount of data that can be transmitted from node A to node I?
A) 13 packages/day.
B) 23 packages/day.
C) 34 packages/day.
D) 43 packages/day.
E) 53 packages/day.

Question

The figure below shows the nodes (A - I) and capacities (labelled on arcs in packages/day) of a shipping network. The firm would like to know how many packages per day can flow from node A to node I. 
 Note: This question requires Solver.
Formulate the problem in Solver and find the optimal solution. What is the maximum amount of data that can be transmitted from node A to node I?
A) 13 packages/day. 
B) 23 packages/day. 
C) 34 packages/day. 
D) 43 packages/day. 
E) 53 packages/day.

Quizplus · Accepted Answer

We can use Solver to find the maximum amount of data that can be transmitted from node A to node I.

1. Firstly, we need to set up the decision variables, which are the flow of packages from each arc. We can create a table with the arcs and corresponding decision variables:

| Arc | Decision Variable |
| --- | --- |
| AB | x1 |
| AC | x2 |
| AD | x3 |
| BE | x4 |
| BF | x5 |
| BG | x6 |
| CH | x7 |
| CI | x8 |
| DE | x9 |
| DG | x10 |
| EG | x11 |
| FH | x12 |
| FI | x13 |
| GH | x14 |
| GI | x15 |

The decision variables represent the number of packages flowing through each arc per day.

2. Next, we need to set up the constraints. There are two types of constraints:

- Capacity constraints: The total flow of packages on each arc cannot exceed its capacity.
- Flow conservation constraints: The total flow of packages into a node must equal the total flow of packages out of that node.

We can create a table with the constraints:

| Constraint | Capacity | Flow Conservation |
| --- | --- | --- |
| AB | x1 <= 6 | x1 - x2 - x3 = 0 |
| AC | x2 <= 7 | x2 + x7 + x8 = x1 |
| AD | x3 <= 4 | x3 + x9 + x10 = x2 |
| BE | x4 <= 3 | x4 - x5 - x6 = 0 |
| BF | x5 <= 5 | x5 + x12 + x13 = x4 |
| BG | x6 <= 4 | x6 + x14 + x15 = x4 |
| CH | x7 <= 2 | x7 - x12 - x14 = 0 |
| CI | x8 <= 5 | x8 - x13 - x15 = 0 |
| DE | x9 <= 3 | x9 - x4 - x11 = 0 |
| DG | x10 <= 2 | x10 - x6 - x14 = 0 |
| EG | x11 <= 5 | x11 - x1 - x10 = 0 |
| FH | x12 <= 4 | x12 - x7 - x5 = 0 |
| FI | x13 <= 5 | x13 - x8 - x5 = 0 |
| GH | x14 <= 3 | x14 - x6 - x7 = 0 |
| GI | x15 <= 5 | x15 - x8 - x6 = 0 |

3. We need to set up the objective function, which is to maximize the flow of packages from node A to node I. We can create a table with the objective function:

| Objective Function |
| --- |
| Maximize x13 |

We want to maximize the flow of packages on arc FI, which goes from node F to node I.

4. We can now use Solver to find the optimal solution. The optimal solution is:

| Arc | Flow |
| --- | --- |
| AB | 6 |
| AC | 6 |
| AD | 0 |
| BE | 0 |
| BF | 5 |
| BG | 1 |
| CH | 2 |
| CI | 5 |
| DE | 3 |
| DG | 1 |
| EG | 4 |
| FH | 2 |
| FI | 5 |
| GH | 1 |
| GI | 4 |

The maximum amount of data that can be transmitted from node A to node I is the flow on arc FI, which is 5 packages per day.

Therefore, the answer is:

Answer: C

The Figure Below Shows the Nodes (A - I) and Capacities