Two Alternate Solutions

A Combined Inventory and Lateral Resupply Model for Repairable Items—Part II: Solution

5.2 Analysis

5.2.1 Two Alternate Solutions

In the solution presented in Table 5.5, we send six serviceable parts to depot 1, z₀₁ = 6. There can be at most nine failures at depot 1. Depot 1 can meet seven of them by using local resources (initial level + repair capacity = 1 + 6 = 7). Suppose depot 1 uses all of its available repair and inventory capacity to meet the failures. It still has a shortage of two serviceable parts. Since the maximum inventory capacity is four, we can replenish at most six serviceable parts before reaching capacity. Depot 1 decides to get four serviceable items out of the repair shop and two serviceable items out of the depot-0 inventory.

For depot 2, we can have at most 10 failures during that period. But we can sat- isfy 11 of them by using local resources. Depot 2 uses all of its repair capabilities and inventory to satisfy the failures. Since our maximum inventory level is four, we can hold at most four serviceable items from any depot in the system. Depot 2 decides to get four repaired units from depot 0 ( z₀₂ = 4).

For depot 3, we have nine failures; but it could meet all the failures by using its own local resources. Nothing is delivered to depot 3 because it has the lowest short- age cost among all depots. In fact it has a surplus of one. Here the binary variable y₃ is equal to one; suggesting it is ready to send the surplus to the other depots.

Table 5.3  The inventory cost data of FZ model with 10 nodes

Node no. Initial inv. level Minimum inv. level Maximum inv. level Inv. cost function

1 100 0 500 2600-4z1

2 50 0 100 437.5-5.75z2

3 150 0 400 2806.25-6.625z3

4 125 0 500 2500-4z4

5 40 0 150 1550-5z5

6 80 0 300 2100-5z6

7 120 0 250 1400-5z7

8 150 0 500 2400-4z8

9 75 0 375 1950-4z9

Table 5.4 The results of FZ model with 10 nodes y^{h i}ziTSP costInv. costTotal cost First iteration (Initial)

y0 1 = 1y

1 3 = 1y

2 2 = 1z = 0z = 50z = 2503914,80014,839123 ^{0 41 52 6}y = 1y = 1y = 1z = 0z = 0z = 200456 ^{0 71 82 9}y = 1y = 1y = 1z = 0z = 0z = 0789

Suggested routes for initial iteration

x0 07–x

0 71–x

0 14–x

0 40

x1 03–x

1 38–x

1 85–x

1 50

x2 09–x

2 92–x

2 26–x

2 60

Second iterationy

1 1 = 1y

1 2 = 1y

1 4 = 1z = 0z = 0z = 25030.514,80014,830.5123 ^{1 61 71 9}y = 1y = 1y = 1z = 0z = 0z = 100456 ^{2 32 52 8}y = 1y = 1y = 1Z = 100z = 0z = 0789

Suggested routes for second iteration

x1 09–x

1 94–x

1 41–x

1 17–x

1 72–x

1 26–x

1 60

x2 05–x

2 58–x

2 83–x

2 30

Third Iterationy

1 1 = 1y

1 5 = 1y

1 8 = 1z = 0z = 50z = 25030.514,80014,830.5123 ^{2 12 22 4}y = 1y = 1y = 1z = 0z = 0z = 100456 ^{2 62 72 9}y = 1y = 1y = 1Z = 100z = 0z = 0789

Suggested routes for third iteration

x1 05–x

1 58–x

1 83–x

1 30

x2 09–x

2 94–x

2 41–x

2 17–x

2 72–x

2 26–x

2 60

The assignment variables are consistent with the result. Since we are sending out of the main depot 0, y₀₁² = y₀₂¹ = y₀₃¹ = 1. Besides that, we deliver to depot 1 by vehicle 2. Therefore, assignment variable y₀₁² and associated vehicle-routing vari- ables such as x₀₁² and x₁₀² are all unity. y₀₂¹ is equal to one, suggesting vehicle 1 be servicing depot 2. The binary variables y₁, and y₂ equal zero. This means that depot 1 and depot 2 need serviceable items from other depots. In this case, depots 1 and 2 are supplied by depot 0. We send nothing from depot 3 to any other depots although y3 is equal to one, because we have already satisfied the demand requirements of depot 1 and 2. If depot 0 did not have enough to send, then depot 3 would be ready to satisfy the remaining demands. Note that depot 3 has only one surplus item to deliver to the other depots, as mentioned.

Vehicle 2 goes to depot 1 and comes back. The total travel time is the sum of the time from depot 0 to depot 1, 0.5 h for unloading, and the time from depot 1 back to depot 0—which is 6.5 h together. For vehicle 1, it is the travel time from depot 0 to depot 2, plus unloading time at depot 2, plus travel time from depot 2 to 3, and the time from depot 3 back to depot 0—summing up to 9 h and 8 min. Note that

Table 5.5  The result of the research model (w/o Benders)

First alternate solution Second alternate solution

Objective value z^* = 133.55 z^* = 133.55

Routes x¹₀₂ – x¹₂₃ – x¹₃₀ x¹₀₃ – x¹₃₂ – x¹₂₀ x²₀₁ – x²₁₀ x²₀₁ – x²₁₀

Depot 0 Initial level: 2 z₀₀ = 7 z₀₀ = 11

Max. level: 10 (x₀₀(u) = 7) (x₀₀(d) = 2, x₀₀(u) = 9) Repair Cap.: [0–15]

Depot 1 Initial level: 1 z₁₁ = 7 z₁₁ = 7

Max. level: 4 (x₁₁(d) = 1, x₁₁(u) = 6) (x₁₁(d) = 1, x₁₁(u) = 6) Repair Cap.: [0–6] z₀₁ = 6 z₀₁ = 6

Failure: [0–9] (x²₀₁(u) = 4, x²₀₁(d) = 2) (x²₀₁(u) = 6)

Depot 2 Initial level: 2 z₂₂ = 11 z₂₂ = 11

Max. level: 5 (x₂₂(d) = 2, x₂₂(u) = 9) (x₂₂(d) = 2, x₂₂(u) = 9) Repair Cap.: [0–9] z₀₂ = 4 z₃₂ = 4

Failure: [0–10] (x¹₀₂(u) = 4) (x¹₃₂(u) = 3, x¹₃₂(d) = 1) Depot 3 Initial level: 1 z₃₃ = 10

Max. level: 4 ( x₃₃( d) = 1, x₃₃( u) = 9) Repair Cap.: [0–9]

Failure: [0–9]

y²₀₁ = y¹₀₂ = y¹₀₃ = 1 y²₀₁ = y¹₃₂ = y¹₀₃ = 1 y₃ = y₀ = 1 y₃ = y₀ = 1

vehicle 2 delivers nothing to depot 3 because depot 3 is a candidate to deliver to the other depots.

There is an alternate solution to this problem. The objective function is at the same value of 133.55. But we satisfy the demands of depots 1 and 2 from depot 0 and depot 3. We are sending six repaired items out of depot 0 to depot 1 to satisfy the demand. Depot 3 delivers three repaired and one inventoried item to depot 2.

Depots 1 and 2 have the same number of failures as the previous solution, and they get the same number of serviceable items. But there is a big difference in how the demands are satisfied. Here, depot 3 can send six more units to the depot 2 if there is a need. The binary variables for delivery, y₀ and y₃, are consistent with the above results. y₀ = 1 means that we can deliver out of depot 0, similarly for y₃ = 1. The as- signment variables—y₀₁², y₃₂¹ and y₀₃¹—are also consistent. Since vehicle 2 goes out of depot 0, y₀₁² = 1. Similarly, vehicle 1 visits depots 3 and 2. The other assign- ment variables are consistent with the result as well.

One can interpret these two alternate results as follows. In the first run, we satisfy the demand out of depot 0. It means depot 0 is the supply source to get the optimal result. In the second run, the supply sources consist of depot 3 and depot 0. In this problem instance, we have two viable options to satisfy the demands. One can think of it as changing the supply sources. With the current newsvendor inventory-costs, we may deploy the available fleet of vehicles to deliver from depot 3 to depot 1 and/

or 2, treating depot 3 as a secondary supply depot.

The results of research problem with Benders’ decomposition procedure, two alternative solutions are obtained as it is shown in Table 5.6. The two alternative results consisted in terms of the quantity delivered to depot 1 by 3 units, 2 and 3 by 4 units.