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This paper addresses production-inventory problem for the manufacturer by explicitly taking into account multistage and varying demand. A nonlinear hybrid integer constrained optimization is modeled to minimize the total cost including setup cost and holding cost in the planning horizon. A genetic algorithm is developed for the problem. A series of computational experiments with different sizes is used to demonstrate the efficiency and universality of the genetic algorithm in terms of the running time and solution quality. At last the combination of crossover probability and mutation probability is tested for all problems and a law is found for large size.

Production-inventory control plays a vital role in the management of manufacturing enterprises. During production process we often hope that the volume of a given product is just enough to satisfy customers’ demand without overextending the production line and manufacturing too many. The redundant inventory will cut down our net profit. On the other hand, lacking supply will also make a heavy loss when a big order is placed. So there must be an optimal production quantity.

Many researchers are interested in this problem and it has been investigated from various perspectives. Goyal and Giri [

Considering uncertainty, Doğru et al. [

Ouyang et al. [

Chao et al. [

Many researches of perishable inventory problem are investigated. Broekmeulen and Donselaar [

Some researchers are interested indiscussing backorder problem in inventory. Yao and Chiang [

The method for assembly line balancing problem is very similar to the inventory problem. Sotskov et al. [

The traditional deterministic inventory models are based on the assumption that the demand rate is constant and order cycle is inflexible. As we all know the total cost is the integral of the production-inventory function because the production time is the same and the maximum inventory in every stage is also equal. But actually there is more than one customer. Demand cycles of these customers are different and the demand rate of every customer is also varying. This is multistage and varying demand production-inventory problem without stockout. In the planning horizon, which is the sum of all stages, what we face to decide is also when and how many quantity to produce in order to minimize the total cost subject to meeting every stage demand. Under the circumstance the traditional unconstrained optimization method is not suitable for the issue although it is still linear. Because according to every demand and its cycle there are different maximum inventory in every stage under different production schedules which means production frequency and production time. We have to define the maximum inventory in every stage for the sake of obtaining the objective function. And this definition relation is only determined by the constraint. So constrained optimization method will be applied to solve the problem. Motivated by the above case we model a hybrid integer programming to characterize the problem and solve it by using genetic algorithm.

The remainder of the paper is organized as follows. Section

In multistage production-inventory system, every stage extent is versatile, demand rate is also different and production rate is a constant and is more than demand rate in every stage. The production-inventory status is drawn in Figure

Production-inventory status.

Twice production

Another two times production

In fact there are four production schemes under the circumstance of twice production. Except the above two schemes mentioned, one is that the first production is in Stage 1 alone and the second production starts from Stage 2 to Stage 5. The other is that the first production begins from Stage 1 to Stage 3 and the second production starts from Stage 4 to Stage 5.

One extreme case is five times production in planning horizon, that is, production occurs in every stage and the quantity as inventory in every stage only satisfies the current stage demand. The other extreme case is only once production in planning horizon. The best production times and quantity are specified by the total cost which bears setup cost of production and carrying cost held in inventory. And we will find the best production scheme under all production times.

First we define some denotations in our model to express the above conception.

Parameters are defined as follows.

Variables are defined as follows:

Because the stockout is not permitted, the total quantity should be equal to the sum of demands in the planning horizon. And the quantity in every production in a certain stage interval should satisfy the demand in the same stage interval:

Production times in the planning horizon are calculated by the following formula:

Production time should be less than demand time:

The logic variables should satisfy the following:

The inventory level

The relationship between the logic variables

The maximum inventory level

when

The sum cost holds setup cost of production and carrying cost held in inventory. So the objective is to minimize the sum cost.

When producing

In stage

The case from stage

So given production times

For every possible production times

We solve the solution by employing Genetic Algorithm (GA). GA is a stochastic search method that works on a population of the solutions simultaneously and searches large and complicated fields based on the mechanics of natural genetics and evolutionary principles. In addition, it is particularly suitable for optimization problems with an objective function subject to numerous constraints. GA has demonstrated considerable success in these optimization problems and received more and more attentions during the past decades.

We encode each chromosome as an

The decoding procedure is as follows. When a chromosome is determined, we could decode this chromosome to get

Crossover generates offspring by operating parent chromosomes. Let parameter

Example of single parent crossover operating.

Mutation modifies a chromosome to form an offspring. Let parameter

Example of mutation operating.

Case 1:

Case 2:

Evaluation function is to evaluate the quality of a chromosome. We define evaluation value of a chromosome as the objective function of the corresponding solution.

The selection strategy means how to choose the chromosomes in the current population will create offspring for the next generation. We take the roulette wheel as selection mechanism, in which each chromosome is assigned a slice of a circular roulette wheel and the size of the slice is proportional to the chromosome’s evaluation value.

In this section we perform a series of computational experiments to evaluate the GA proposed in Section

In order to test the efficiency and universality of the GA, we generate four groups of data randomly in which stages are 10, 20, 50, and 100, named problems 1, 2, 3, and 4, respectively. Tables

Demand data in Problem 1.

Stage number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|

Demand extent | 14 | 14 | 12 | 3 | 13 | 17 | 12 | 15 | 7 | 12 |

Demand rate | 102 | 73 | 264 | 152 | 119 | 61 | 29 | 292 | 215 | 211 |

Demand data in Problem 2.

Stage number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|

Demand extent | 14 | 3 | 11 | 17 | 6 | 15 | 9 | 13 | 16 | 13 |

Demand rate | 169 | 60 | 54 | 240 | 246 | 68 | 202 | 18 | 123 | 107 |

| ||||||||||

Stage number | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |

| ||||||||||

Demand extent | 15 | 4 | 4 | 3 | 8 | 6 | 8 | 5 | 14 | 8 |

Demand rate | 24 | 14 | 280 | 48 | 265 | 218 | 216 | 27 | 49 | 22 |

Demand data in Problem 3.

Stage number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|

Demand extent | 4 | 14 | 4 | 5 | 4 | 12 | 10 | 7 | 7 | 9 |

Demand rate | 229 | 230 | 84 | 263 | 121 | 167 | 254 | 70 | 121 | 188 |

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Stage number | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |

| ||||||||||

Demand extent | 13 | 9 | 20 | 20 | 23 | 3 | 7 | 12 | 14 | 9 |

Demand rate | 26 | 158 | 233 | 227 | 153 | 146 | 27 | 149 | 280 | 165 |

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Stage number | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |

| ||||||||||

Demand extent | 4 | 3 | 6 | 7 | 12 | 9 | 9 | 4 | 6 | 10 |

Demand rate | 267 | 194 | 295 | 12 | 224 | 290 | 274 | 77 | 246 | 281 |

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Stage number | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |

| ||||||||||

Demand extent | 7 | 12 | 9 | 10 | 13 | 9 | 3 | 7 | 6 | 7 |

Demand rate | 167 | 282 | 17 | 184 | 105 | 214 | 37 | 194 | 161 | 161 |

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Stage number | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |

| ||||||||||

Demand extent | 6 | 14 | 11 | 3 | 6 | 3 | 6 | 14 | 8 | 13 |

Demand rate | 244 | 145 | 144 | 160 | 117 | 173 | 18 | 236 | 268 | 246 |

Demand data in Problem 4.

Stage number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|

Demand extent | 4 | 8 | 7 | 14 | 12 | 3 | 12 | 12 | 6 | 13 |

Demand rate | 70 | 202 | 53 | 22 | 78 | 231 | 203 | 22 | 219 | 239 |

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Stage number | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |

| ||||||||||

Demand extent | 11 | 10 | 10 | 9 | 9 | 6 | 14 | 10 | 7 | 11 |

Demand rate | 149 | 23 | 239 | 72 | 231 | 168 | 111 | 234 | 79 | 87 |

| ||||||||||

Stage number | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |

| ||||||||||

Demand extent | 14 | 12 | 11 | 5 | 9 | 7 | 12 | 3 | 14 | 6 |

Demand rate | 203 | 210 | 106 | 145 | 32 | 24 | 136 | 115 | 245 | 93 |

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Stage number | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |

| ||||||||||

Demand extent | 11 | 12 | 6 | 4 | 4 | 4 | 13 | 9 | 7 | 10 |

Demand rate | 97 | 209 | 289 | 238 | 174 | 114 | 174 | 194 | 124 | 94 |

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Stage number | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |

| ||||||||||

Demand extent | 6 | 3 | 8 | 14 | 9 | 13 | 13 | 7 | 13 | 7 |

Demand rate | 238 | 149 | 80 | 95 | 93 | 161 | 229 | 58 | 198 | 188 |

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Stage number | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |

| ||||||||||

Demand extent | 4 | 8 | 6 | 10 | 6 | 4 | 7 | 4 | 9 | 8 |

Demand rate | 242 | 250 | 293 | 159 | 180 | 198 | 185 | 83 | 285 | 262 |

| ||||||||||

Stage number | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 |

| ||||||||||

Demand extent | 8 | 7 | 5 | 7 | 10 | 14 | 8 | 10 | 13 | 4 |

Demand rate | 273 | 183 | 228 | 50 | 93 | 211 | 121 | 70 | 78 | 236 |

| ||||||||||

Stage number | 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 |

| ||||||||||

Demand extent | 6 | 7 | 9 | 10 | 3 | 4 | 12 | 5 | 5 | 13 |

Demand rate | 133 | 218 | 167 | 30 | 127 | 48 | 195 | 222 | 213 | 238 |

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Stage number | 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 | 90 |

| ||||||||||

Demand extent | 4 | 10 | 6 | 5 | 4 | 10 | 12 | 11 | 9 | 14 |

Demand rate | 22 | 298 | 170 | 28 | 283 | 127 | 72 | 210 | 274 | 59 |

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Stage No. | 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 |

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Demand extent | 10 | 8 | 10 | 13 | 3 | 8 | 3 | 7 | 8 | 6 |

Demand rate | 75 | 59 | 222 | 202 | 239 | 105 | 113 | 16 | 100 | 299 |

We execute the GA in 30 independent runs on each problem. The evaluation value is recorded for every trial in every problem and shown in Figure

Computational results statistics.

Problem number | 1 | 2 | 3 | 4 |
---|---|---|---|---|

The number of generations | 100 | 500 | 1000 | 2000 |

The best found evaluation value | 1409.4977 | 1366.0474 | 1665.7842 | 1715.5158 |

The average evaluation value | 1411.9887 | 1417.7661 | 1695.2008 | 1746.4505 |

Standard deviation | 1.6308 | 32.4708 | 21.1338 | 13.45 |

The average running time (sec.) | 0.45 | 1.8 | 4.22 | 8.48 |

Computational results.

Because the parameters in the proposed GA, crossover probability and mutation probability, play a crucial role of the result, we also test the best combination of the two parameters for all problems. Both crossover probability and mutation probability vary from 0.1 to 0.9 when the step is 0.1. So there are 81 combinations in every problem. The average evaluation value is obtained from 30 independent trials under all combinations. Figure

The best combination of (

Problem number | 1 | 2 | 3 | 4 |
---|---|---|---|---|

The best combination | (0.4, 0.5), (0.9, 0.8) | (0.4, 0.2) | (0.6, 0.1) | (0.6, 0.1) |

The average evaluation value | 1410.8025 | 1398.782 | 1596.6894 | 1676.5948 |

Comparison results with combination of

For Problem 1

For Problem 2

For Problem 3

For Problem 4

For Problem 1 the best combination parameters are (0.4, 0.5) and (0.9, 0.8). This indicates that Problem 1 is not sensitive to the two parameters. Owing to the small size of Problem 1, if the number of generations increases we could find better solution than the former under any combination parameters. For example, when the number of generations reaches 1000, the average evaluation value is 1419.8273 from the 30 trials under the worst combination (0.1, 0.2) whose average evaluation value is 1482.9856 and number of generations is 100 in Figure

We find a law in Figures

Table

The partial best solutions and evaluation values.

Problem number | The stages in a continuous production cycle | Evaluation value |
---|---|---|

1 | (3, 4), (6, 7), (8, 9) | 1409.4977 |

2 | (1, 2), (5, 6), (7, 8), (14, 15, 16), (17, 18, 19, 20) | 1332.6984 |

3 | (1, 2, 3), (4, 5), (7, 8), (16, 17), (19, 20), (21, 22, 23, 24), (26, 27, 28), (30, 31), (32, 33), (36, 37), (44, 45), (46, 47), (49, 50) | 1487.9055 |

4 | (2, 3), (6, 7, 8), (11, 12), (13, 14), (18, 19), (24, 25), (29, 30), |
1584.0483 |

The most cases are that one production cycle includes two stages in Table

By increasing the number of stages and demands rate, the total cost must increase. But the average cost may not always be so. Although the stage number in Problem 2 is twice of it in Problem 1 and the demand rate in Problem 2 is also more than that in Problem 1, the average cost of Problem 2 is less than that of Problem 1. This suggests that the best production mode (production times and volume of production) could reduce the average cost. So the total cost must cut down.

In this paper it is found that the efficient and effective production-inventory strategy could reduce unnecessary funding, inventory cost, and production cycle. Furthermore, the production-inventory problem with multistage and varying demands, in which every stage extent is different, is investigated. Moreover, we generate a hybrid integer model to optimize the production-inventory process and a GA to find a better solution. Computational results illustrate that the proposed GA yields a high-quality solution relatively fast. According to the tests of combination parameters we find the law between the average evaluation value and mutation probability under the large size.

This work is supported by the National Natural Science Foundation of China (Grant nos. 60870008 and 61164003) and the Lanzhou Jiaotong University Young Scientific Research Fund Project (2011020).