Distribution patterns
2.1 Modelling with tours
Consider a vehicle replenishing a set of customers S. In terms of travel distance, the most e ective way to supply these customers is to travel along the shortest tour that visits the depot and all of the customers in S, or the `TSP-tour'
through S plus the depot, T SP (S + ). Cyclically repeating this tour gives a solution for the in nite time horizon. The time between two consecutive iterations of the tour is called the `cycle time' T .
The vehicle needs some time for loading at the depot, denoted t, for driving around in the tour to all customers, denoted TT SP (S+) and for unloading at all of the customers, denotedP
j2Stj. The tour cannot be restarted before it is nished, so the total time needed to complete a tour gives a lower bound on the cycle time.
Sometimes customers restrict their delivery frequency fj. This results in an-other lower bound on the cycle time. Suppose a tour can be nished within one hour, but customer j in that tour wants to be replenished at most once every two hours (fj = 1=2). Then the cycle time for the whole tour has to be at least two hours (= 1=fj).
Thus, the minimal cycle time for replenishing the set of customers S in a single tour, denoted by Tmin, is given by the following formula.
Tmin= max
The cycle time of a tour is not only bound from below. Due to capacity restrictions, there is also an upper bound.
The rst capacity restriction is related to the limited vehicle capacity. The maximum quantity a vehicle can distribute over the customers in a tour is exactly the vehicle capacity . To avoid stock-outs, the vehicle has to return to its customers before they have consumed this full truckload. Knowing that the vehicle load is divided over the customers in a tour according to a fair-share mechanism, the vehicle capacity divided by the cumulative demand rate of all customers in the tour, P
j2Sdj, gives the maximal time between two consecutive deliveries without customer stock-outs, and thus an upper bound on the cycle time of the tour.
Other capacity restrictions are related to the limited storage capacities of the customers. If the storage capacity jof a customer is smaller than the maximal delivery quantity resulting from the vehicle capacity, this further reduces the maximal cycle time. E.g. consider two customers with demand rates of 20 and 10 units per hour, served in a single tour by a vehicle with a capacity of 120 units. The maximal cycle time of this tour is 4 hours, with a delivery of 80 units to the rst customer and a delivery of 40 units to the second customer.
Now suppose the rst customer can only hold up to 60 units of stock. Then the maximal cycle time reduces to 3 hours, bringing 60 units to the rst customer, and 30 units to the second customer. In this case, the available vehicle capacity cannot be fully used and it makes sense to add a third customer to the tour, if possible.
2.1 Modelling with tours 13
The following formula gives the maximal cycle time for replenishing the set of customers S in a single tour, denoted by Tmax.
Tmax= min P
Obviously, a tour is only feasible if its minimal cycle time is not greater than its maximal cycle time: Tmin Tmax.
The cost rate of a tour, denoted C, consists of the ve components that we discussed in Section 1.2.
1. The rst component is the xed operating cost of the vehicle, euro per period.
2. The second cost rate component is the variable transportation cost. Per iteration, the vehicle travels along the TSP-tour through the depot and the set of customers once, at a cost of CT SP (S+) euro. This results in a variable transportation cost of CT SP (S+)=T euro per period.
3. The third cost rate component is the xed vehicle loading and dispatching cost. The vehicle is loaded and dispatched once at a cost of 'euro per iteration of the tour, giving '=T euro per period.
4. The fourth cost rate component is the xed unloading cost at the cus-tomers. Per cycle, there is one delivery at each of the customers, so this cost component amounts to P
j2S'j euro per iteration of the tour, or P
j2S'j=T euro per period.
5. The fth and last cost rate component is the stock holding cost at the customers. The quantity qj delivered to customer j 2 S covers demand for a whole cycle (i.e. until the next delivery), so qj = djT units. The average stock level at customer j during a cycle is then qj=2 units. With a storage cost of j euro per unit per period, the stock holding cost at customer j is: jqj=2 or jdj T=2 euro per period. The total stock holding cost at all customers in S is then T2 P
j2Sjdj euro per period.
The total cost rate of the tour replenishing the set of customers S is given in the following formula.
This tour cost rate varies with the tour cycle time. Each tour thus has a theoretical optimal cycle time, for which the cost rate is minimal. This occurs when @C@T = 0, where holding costs at the customers (component 5), which
increase when cycle time increases, are balanced with the sum of transportation costs and xed tour costs (components 2, 3 and 4), which decrease when cycle time increases. This is a generalization of the well-known `Economic Order Quantity' model, and therefore this optimal cycle time is called the EOQ cycle time and denoted Teoq.
Teoq=
Unfortunately, this EOQ cycle time is not always feasible. In many cases, Teoq will be outside the interval [Tmin; Tmax] of feasible cycle times. When this is the case, the best feasible cycle time T is the feasible cycle time closest to Teoq, which is exactly the maximal or minimal cycle time.
T=
Throughout this chapter, a 4-customer example is discussed to illustrate the opportunities and complexities of using the di erent routing concepts for cyclic inventory routing. Distances and demand rates of the example are shown in Figure 2.1. A vehicle with a capacity of 120 units is available for product replenishment from the depot. For simplicity, it is assumed here that (i) the loading and unloading times of the vehicle are negligible (t = tj = 0), (ii) customers have no storage capacity restrictions (j = ), and (iii) customers do not impose delivery frequencies (fj = +1).
The cost parameters are as follows:
1. The xed vehicle cost: = 20 euro per hour.
2. For the variable transportation cost, an average speed of 50 km per hour and a cost of 1 euro per kilometer are assumed.
3. No vehicle loading and dispatching cost is accounted: '= 0.
4. No vehicle unloading cost is accounted: 'j= 0; j = 1::4.
5. All four customers have the same storage cost of 0:15 euro per unit per hour.
2.1 Modelling with tours 15
Figure 2.1: Illustrative example with 4 customers
The shortest tour through the depot and all four customers goes from the depot to customer 2, on to customers 1, 3 and 4 next and then back to the depot. The minimal cycle time of this tour is its travel time, i.e. 18 hours. The maximal cycle time due to the limited vehicle capacity is 120=(4 + 3 + 2 + 1) = 12 hours.
Because the minimal cycle time exceeds the maximal cycle time, this simple tour solution is infeasible. Therefore, when using the single tour concept for routing vehicles, a second vehicle would be necessary for replenishing the four customers in this example. Alternatively, a bigger vehicle can be considered.
Note that the suggested tour becomes feasible with a vehicle capacity of 180 units. However, when generalizing the routing concept in the following sections, we will show that no second vehicle or larger vehicle is needed at all.
2.1.1 Tours under driving time restrictions
So far, we have assumed that a vehicle can make a tour at any time of day.
This assumption may be valid for some industrial applications, such as the distribution of crude oil derivates or lique ed gases, raw material replenishment, etc. However, in most real-life situations, deliveries can only occur during the day. In Chapter 5, a real-life application of our cyclic inventory routing tool is presented. In that case, the products are consumed day and night every day, while replenishments are allowed only during the day, on weekdays, so a vehicle can only drive 8 hours a day, and 5 days a week.
When driving time restrictions of 8 hours per day apply, we assume that the cycle time of a tour has to be an integer number of days and thus that a tour cannot be made more than once per day. Similarly, when driving is only permitted on weekdays, we assume that the tour cycle time is an integer number
of weeks and thus that a tour is made at most once per week. Cycle times therefore have to be `translated' from hours to a number of days.
The minimal cycle time under driving time restrictions is denoted by TMIN. It is 1 day if completing the tour takes less than 8 hours. A minimal cycle time of more than one day is not permitted, because we assume that the vehicle drivers have to be back in the depot at the end of their 8-hour working day.
TMIN =
1 if Tmin 8;
+1 otherwise: (2.6)
If driving time restrictions are imposed, we assume that customer demand rates are expressed in units per day (or week) instead of units per hour. The maximal cycle time under driving time restrictions, denoted TMAX, is then an integer number of days (or weeks) given by a straightforward extension of the original formula for the maximal cycle time Tmax. Obviously, a maximal cycle time of zero days (or weeks) indicates that a tour is infeasible.
TMAX = min
$P
j2Sdj
%
; min
j2S
j dj
!
(2.7)
The optimal cycle time TEOQ is given by rounding Teoq to the closest integer number of days (or weeks).
Illustrative example
In the illustrative example, the driving time restrictions that we impose are the 8-hours driving per day constraints. Further, it is assumed that the customers in the example are retail outlets such that demand only occurs during these 8 hours per day.
The single tour solution suggested above takes 18 hours. This violates the 8-hour driving constraint, such that this tour is always infeasible, even if the vehicle capacity would be 180 units or more.