Answered Essay: Peapod an internet grocery distributor, delivers groceries to customers from its distribution center in Atlanta. Each delivery van can carry up to 200 cases of product.
Peapod an internet grocery distributor, delivers groceries to customers from its distribution center in Atlanta. Each delivery van can carry up to 200 cases of product. Peapod needs to coordinate the delivery of orders to 4 customers today. The distances and order quantities for each of these 4 customers are given in the table below. Use the procedure described in class to solve this vehicle routing problem. Provide details of the solution development. Identify each resulting route and its corresponding mileage. What is the total mileage of all routes in the solution?
Expert Answer
We don’t know what is the method discussed in the class. However, this problem can be solved by using the ‘time-saved’ algorithm.
First, we will find the time saved matrix. Here we will use distance in place of time.
| Time (distance) Saved matrix | |||
| Cust 2 | Cust 3 | Cust 4 | |
| Cust 1 | 10.6 | 21 | 18 |
| Cust 2 | 14.4 | 15.2 | |
| Cust 3 | 28 | ||
[Note: Calculation example – Time (distance) saved between Cust 1 and 2 = Time required from DC to Cust 1 + Time required from DC to Cust 2 – Time required from Cust 1 to Cust 2 = 12.0+7.8 – 9.2 = 10.6]
As per the above table, the sorted combination based on descending time-saved is:
| Rank of Pairing by time saved |
|
| Cust 3 – Cust 4 | 28 |
| Cust 1 – Cust 3 | 21 |
| Cust 1 – Cust 4 | 18 |
| Cust 2 – Cust 4 | 15.2 |
| Cust 2 – Cust 3 | 14.4 |
| Cust 1 – Cust 2 | 10.6 |
According to this table, we combine the destination and make routes. We also find the load after combination from the given demands and check whether they exceed 200 or not which is the capacity.
| Route | Load |
| DC – Cust 3 – Cust 4 – DC | 135 |
| DC – Cust 1 – Cust 3 – DC | 91 |
| DC – Cust 1 – Cust 4 – DC | 140 |
| DC – Cust 2 – Cust 4 – DC | 128 |
| DC – Cust 2 – Cust 3 – DC | 79 |
| DC – Cust 1 – Cust 2 – DC | 84 |
We find that none of these routes exceeds 200. So, they can be readily used as DC – Cust 3 – Cust 4 – DC and DC – Cust 1 – Cust 2 – DC. But these also not optimal choices because the further combination is possible. For example, if we combine the two routes DC – Cust 3 – Cust 4 – DC and DC – Cust 1 – Cust 3 – DC as DC – Cust 1 – Cust 3 – Cust 4 – DC, the load becomes 48+43+92 = 184, still less than 200.
So, the final solution is the following two routes:
DC – Cust 1 – Cust 3 – Cust 4 – DC (Distance = 12.0+7.6+3.6+15.0 = 38.2)
DC – Cust 2 – DC (Distance = 7.8+7.8 = 15.6)
Total distance is (12.0+7.6+3.6+15.0) + (7.8+7.8) = 53.8
