Int J Adv Manuf Technol (1995) 10:269-276 © 1995 Springer-Verlag London Limited

Tolerance Allocation Using Neural Networks Parimal Kopardekar and Sam Anand Computer-Aided Manufacturing Laboratory, Industrial Engineering Division, Department of Mechanical, Industrial and Nuclear Engineering, ML 116, University of Cincinnati, Cincinnati, OH, USA

The purpose of tolerance allocation is to find a combination of tolerances to individual components such that the assembly, tolerance constraint is met with minimum production cost. There are several methods available to allocate or apportion the assembly tolerance to individual parts. Some of the most common methods use linear programming, Lagrange multipliers, exhaustive search and statistical distributions. However, all the methods have some limitations. Moreover, most of these methods cannot account for the frequently observed mean shift phenomena that occur owing to toot wear, chatter, bad coolant, etc. This paper presents a neural networks-based approach for the tolerance allocation problem considering machines' capabilities, and mean shifts. The network is trained using the backpropagation learning method and used to predict individual part tolerances'.

individual components in some rational manner. If the designer specifies excessively tight tolerances then the production cost increases. On the other hand, if the specified tolerances are too loose then the parts perform poorly. Also, there may be an increase in rejected parts as the total assembly tolerance may not be within the acceptable range. Obviously, both of these extremes will result in reduced productivity [1-5]. The objective of this paper is to present a new method for the tolerance allocation problem.

Keywords: Backpropagation; Neural networks; Tolerance

In this method, total assembly tolerance is allocated equally among the components to be assembled. This method is simplistic in nature as it does not consider the cost of manufacturing tolerances and machine capabilities. Consequently, the allocated tolerances are often unrealistic.

allocation

1.

2.

Tolerance Allocation Methods

2.1 Tolerance Allocation by Simple Division

Introduction

Interchangeability of parts refers to the random process of selection of parts for assembly. Interchangeability is essential to achieve higher productivity. The designer must specify tolerances to assemblies and to individual parts in order to achieve interchangeability of parts and account for machine or process variations. Tolerance can be defined as an acceptable deviation from nominal dimensions in order to facilitate interchangeability of parts. Tolerance allocation (also called tolerance distribution or tolerance synthesis) is a process where individual component tolerances are determined based on the total assembly tolerance. In the case of tolerance analysis, the component tolerances are known and assembly tolerance is to be determined. In the case of tolerance allocation, a more common process than tolerance analysis, the designer has to distribute or allocate tolerances to

Correspondence and offprint requests to: Dr Sam Anand, ComputerAided Manufacturing Laboratory, Industrial Engineering Division, Department of Mechanical Industrial and Nuclear Engineering, ML 116, University of Cincinnati, Cincinnati, OH 45221-0116, USA,

2.2 Tolerance Allocation by Proportional Scaling In this method, the designer assigns reasonable component tolerances by judgement, based on process or design guidelines. If the sum of component tolerances is not equal to the desired assembly tolerance then the component tolerances are scaled by a constant proportionality factor. The following two possible approaches can be used to apportion the total assembly tolerance:

1. Worst-limits analysis: In this method, the assembly tolerance is determined by summing the component tolerances linearly. Each component dimension is assumed to be at its maximum or minimum limit, resulting in the worstpossible assembly limits. 2. Statistical: In this method, component tolerances are added as the root sum squared. The low probability of the worst-case combination occurring is taken into account statistically, assuming normal distribution for component variations. The tolerances are assumed to correspond to six standard deviations.

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2.3 Tolerance Allocation Using Optimisation Techniques Optimisation techniques are based on minimising the cost of production of an assembly. For each component, a cost versus tolerance curve is defined. As the tolerance is reduced, the cost of production increases. The optimisation algorithm varies the tolerance for each component and searches systematically for the combination of tolerances that minimises the cost. A closed form solution for the least-cost component tolerances can be found by using the Lagrange multipliers method [3]. Dynamic and geometric programming approaches have also been used for optimal tolerance allocation, as discussed in [6]. However, the dynamic programming method cannot accommodate the true nonlinear nature of cost vs. tolerance relationship.

3. Limitations of Common Assembly Models The following points present some of the drawbacks of the existing tolerance allocation methods: 1. Too tight tolerances are assigned with worst-case-limit models, resulting in an increase in production costs. 2. Statistical models that allow loose tolerances predict higher assembly yields than those obtained in actual production. 3. Statistical models predict tolerances which are tighter than those predicted by worst-case models for: (a) assemblies with small number of components, and (b) assemblies in which one component tolerance is much greater than the remaining components. 4. Statistical models do not take into account skewness and bias in the component's distributions which are commonly observed. A shift in mean is often the result of bias due to variety of causes such as tool wear, improper coolant, etc. Tool wear may produce higher dimensions than expected whereas improper coolant may cause distortion owing to expansion of components because of increased temperature. 5. Statistical models assume that manufacturing variations follow normal distributions. However, this assumption is not valid as most of the processes may not follow normal distribution. 6. Closed form solutions for the least-cost tolerancing problem do not exist for component dimensions following nonnormal distributions (lognormal, beta, gamma) or a mix of normal and non-normal distributions [1, 7].

4. Objective The objective of this study is to use the neural networks approach for allocating individual component tolerances, with known total assembly tolerance, such that the total assembly tolerance is met with the least cost.

5.

Problem Description

A three-component assembly with three available machines is considered as an example. The tolerance to be achieved depends on the process and the part dimension. In general the higher the dimensions of the parts, the larger the tolerances are. This is due to the size and weight of the part. Also, secondary finishing processes such as honing and lapping, and grinding can achieve better tolerances than other rough machining operations. Further information about these processes can be obtained in any manufacturing processes text book, The following information about the part dimensions was assumed: Part Machine 1 2 3

Lathe 1 Lathe 2 Lathe 3

Dimension 1.5 in. 3.0 in. 4.75 in.

Half tolerance band for turning operation 0.0015 in. 0.002 in. 0.0025 in.

The upper and lower limits (twice the haft tolerance band) for component dimensions, which are called upper and lower natural limits, are taken from [7] and [8]. The advantage of this method is that it eliminates the need for assumptions about the underlying distribution of part dimensions which is unknown most of the time. The capabilities of the individual machines can be obtained from the manufacturer, or can be calculated from the records of the quality control department. The capabilities for various machines can be found in [7] and [8]. For each of the components, a cost ($) vs. tolerance (in.) relationship is defined in the form: cost = A + B/tolerance 2, where A and B are known constants. The following cost vs. tolerance equations were used: Component 1:

cost = 1.5 + 0.00003/t~

(1)

Component 2:

cost = 1.75 + 0.00004/t~

(2)

Component 3:

cost = 1.8 + 0.00005/4

(3)

The nature of these relationships is indicated in Fig. 1. The mean shift factors were assumed as 0.1, 0.2 and 0.3 for machines 1, 2 and 3, respectively. The mean shift factor indicates that the mean can shift in any direction with a given fraction of the half tolerance band (tolerance band is the difference between upper natural tolerance limit and lower natural tolerance limit). Such a mean shift factor can be obtained from the past data recorded by the quality control department in any organisation. With the consideration of mean shifts the new tolerance bands become 0.00165, 0.0024 and 0.00325 (in.), respectively. A n exhaustive procedure was used to identify the combination of tolerances leading to the least cost so that the assigned tolerances are greater than the natural tolerance limit of the machine. In other words, the natural tolerance limits indicate the maximum capability of the machine (or the limitation of the machine.) This is due to the fact that machine variability is difficult to control below the natural tolerance limits. The assignment of tolerances serves as an output for the neural network while the input is the natural tolerance limits of the three machines and total assembly tolerance.

Tolerance Allocation Using Neural Networks

27/

10 000

100C

-=- M a c h i n e

~R v

o

10C

÷

O

1

Machine 2

x Machine 3

1C

1 0.0001 0.0006 0.0011 0.0016 0,0021 0.0026 0.0031 0.0036 0.0041 0.0046

Tolerance (inches) Fig. 1. Cost vs. tolerance relationship,

6.

Significance of the Problem

Although tolerance allocation is a common problem in industry there seems to be a lack of methods and algorithms that take into account all the factors. Most of the algorithms developed are applicable in a restricted domain. If the underlying assumption about normality of part dimensions is not satisfied, the solutions are erroneous and often result in a high percentage of rejects. Another commonly observed phenomena in industry is the shift in the process mean. The statistical quality control methodologies try to identify the shift in the mean of the process. However, so far, tolerance allocation methods do not take into account the mean shift factor for dimensions that do not follow a normal distribution [1]. Lagrange multipliers or linear programming methods cannot be applied to solve tolerance allocation problems if different parts follow different statistical distributions and closed-form solutions cannot be generated. Therefore, an empirical neural network-based approach would be appropriate for solving the tolerance allocation problem in real-world situations.

tal knowledge [9]. The neural networks model consists of a number of interconnected processing elements which are called neurons. These neurons try to mimic the human information processing function. More precisely neural networks or artificial neural networks can be defined as: "artificial neural networks are massively parallel interconnected networks of simple (usually adaptive) elements and their hierarchical organizations which are intended to interact with the objects of the real world in the same way as biological nervous systems do" [10]. In natural or biological neural networks, the neurons (cells) process the information. The interconnections of neurons are callex axons. The synapse is where the neuron transmits its signal to the neighbouring neuron. Similarly, (artificial) neural networks are systems of simple but interconnected processing elements. The function of these elements is to act as receivers and filters to input signals, Originally, neural networks were developed for classification and for discrete mapping but now they are increasingly used for continuous mapping [11]. Further detailed information about neural networks can be obtained in [12-14].

7. Overview of Neural Networks as Applied to the Tolerancing Problem

8.

Artificial neural systems, or neural networks, are physical cellular systems which can acquire, store and utilise experimen-

A multilayer (three layered) fully connected network with a backpropagation learning method is used. The network

Network Architecture

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S. Anand and P. Kopardekar

is used for predicting the component tolerances. Several researchers have used the neural networks approach for prediction purposes [15,16]. The sigmoidal function, whose value lies between 0 and 1 only, is used at the hidden and output layers, respectively. For a three-component assembly, four input nodes are considered, one for each machine capability or natural tolerance and the fourth one for the total assembly tolerance. Three output nodes are considered for predicting the tolerance of each component (on one and only machine). A hidden layer with four nodes is considered. The inputs are normalised with respect to a fixed constant (102), which is equal to the maximum value of total assembly tolerance times 1000. The root mean square value (maximum) of acceptable error was assigned as 0.0005. The learning parameter was assigned a value of 0.4. Fig. 2 indicates the network architecture.

output nodes to hidden nodes and V is (J X t), a weight matrix connecting hidden nodes to input nodes. q+--l,

p+-l,

E+-0

(4)

Where q is the number of iterations, p is the number input patterns (we have 10 here), and E refers to the root mean square error. Step 2. Start training procedure. Each input pattern or vector is presented to the network (we have 4 nodes in the input layer). The output at output nodes is computed based on the randomly assigned weights and activation functions selected at hidden and output layers. In this example we have selected the sigmoidat activation function which is given as: f(net) = 1/(1 + e x p ( - h , n e t ) )

The value of this sigmoidal activation function lies between 0 and 1. In order to find the value at the hidden nodes, each input is multiplied by the weight of the carrying node and then these quantities are added at the respective hidden ]aodes, which is used to calculate the value of f(net). This value is used for calculating the activation function. The values obtained from these calculations are further forwarded after multiplying with the corresponding weights at the respective output nodes. Once again, at output nodes the activation function is used to find the final values at each node.

9. Training Method The following step by step procedure of backpropagation training algorithm is used [9]. Given are P training pairs {zl, dl, z2, d2, ..., zp, dp}, where zi is (I X 1), and di is (K X 1), and i = 1, 2 . . . . . P. Step 1. Select learning parameter (eta) and maximum allowable error (Emax). eta = 0.4 and Emax = 0.0005 Weights connecting the output nodes to hidden nodes and hidden nodes to input nodes are assigned random values. W is (K X J), a weight matrix connecting

machine capability 1

machine capability 2

machine capability 3

1,

tolerance

i

l, part 1

I~ tolerance part 2

Ib tolerance part 3 total assembly tolerance

v

Input layer

(5)

Hidden layer

Fig. 2. Network architecture.

Output layer

Tolerance Allocation Using Neural Networks Step 3. Calculation of error. As mentioned before, the values computed at the nodes may not be the desired values, therefore they have to be compared with the desired values. The difference between desired values and observed values gives the error. The error value is computed as follows: 1

e

Step 4. These error signal vectors at the output and hidden layers are computed at this stage. The error signal terms of the output layer in this step are given as: gok = (dk -- Ok) (1 -- Ok) Ok f o r k = 1,2, ...,k

(7)

The error signal terms at the hidden layer in this step are given as: K

~yj =yj(1 --yj)~

go~*

W,,4

(8)

1

for j = 1,2, ...,j Step 5. Based on the error signal vectors the output layer weights are adjusted as follows: W~j <-- W~I + ~lgokYj for k = 1, 2 ..... K j = l , 2 ..... j

(9)

Step 6. Based on the error signal terms the hidden layer weights are adjusted as follows:

Vii <-- TVTji-}- ~ y j Zi forj = 1, 2 ..... j i= 1,2,...,i

(10)

Step 7. If there are more patterns in the training set (we have ten here), then go to step 2 again, otherwise go to step 8. Step 8. The training cycle is completed. If Error (calculated) is less than specified Emax (in our case 0.0005) then terminate the training session. If the calculated error is more than specified then initiate the new training cycle by going to step 2, reinitialise E and p. Using the above-mentioned back-propagation learning algorithm the network was trained using 10 input patterns (assembly tolerances) or vectors, which were selected at random from a sample of 30 assembly tolerances. The output corresponding to these ten inputs was generated using an exhaustive search method to find the least-cost combination. As mentioned before, the input and output vectors served as a basis for training the network. As the desired outputs are also used during the training process, this type of training is also called supervised training. The training cycle was repeated until the error was reduced below the maximum root mean square value of 0.0005 (Emax). The initial weights of the node connections were randomly assigned. After the network was trained, it was used to predict the tolerances for each of the three components for all thirty data sets.

10.

273

Results and Discussion

Table 1 indicates the predicted tolerances obtained by the trained neural network and the actual tolerances obtained by the exhaustive search method. The objective of the exhaustive search method was to find a combination of tolerances such that the cost of production is minimum. Using the predicted tolerances, the predicted cost of production ($) was calculated using equations (1)-(3). The actual cost of production was obtained using the exhaustive search method. Fig. 3 depicts the comparison of the predicted and actual tolerances for all 30 assemblies. Interestingly, coordinate measuring machines (CMM) have a least count of 0.0001 in., which means that dimensions less than this value cannot be measured accurately [17]. The practical implication of this fact is that the predicted and actual tolerances have to be matched only to the fourth significant digit, with rounding up. If this criterion is applied then all of the predicted tolerances match with the actual assembly tolerances, indicating that this approach is quite successful in predicting tolerances. Fig. 4 clearly indicates this fact, as the predicted and actual assembly tolerance and cost curves are very close to each other. In this figure, the practical aspect of the least count (matching till fourth significant digit and rounding up) is taken into account so that the practical utility of this approach can be emphasised.

10.1 Advantages of the Proposed Method 1. This procedure does not need any assumption about distributions of the part dimensions unlike statistical techniques. 2. This approach can consider any mean shift in the process, although for simplicity we have considered a fixed mean shift for each machine. The method can also accommodate different mean shifts and also dynamic mean shifts which vary with time. If this approach is combined with the information about the mean shifts from the quality control department then dynamic tolerancing can be possible depending upon the machine, tool, lubricant and other considerations. 3. This approach can be extended to assemblies of large numbers of parts. In fact, as the number of parts to be assembled increases, this approach will be much faster than the exhaustive method, owing to the parallel processing ability of neural networks. 4. In the above example, three machines and three parts were considered. This approach can be easily extended to m-machines and n-parts tolerance allocation problem.

10.2 Limitations of the Proposed Method 1. This approach requires some data with known outputs. Therefore the necessity of the exhaustive search cannot be completely eliminated. 2. This approach was not faster than the exhaustive search method for the study considered. However, the utility of this approach will be more evident when the number of parts to be assembled (e.g. 5 or more) increases. This is

274

S. Anand and P. Kopardekar

.C

+ Part 1 (actual) = o

-4- Part 2 (actual) - ~ Part 3 (actual)

0

4l- Part 1 (predicted) C}.

-~ Part 2 (predicted) 41" Part 3 (predicted)

.-9_ .>

0 ~-;"

0.0081

0.0077

0,0073

0.0085

0,0089

0,0093

0,0097

0,0t01

Total assembly tolerance (inches)

Fig. 3. Comparison of actual and predicted tolerance.

30

0.0105 1

25 ~" 0.0095

0.009

2, ~ &0085

8

iiiii!ii ii

"~- Actual "6

o.oo8

0,0075

0.007 5

7

9

-4-Actual total cost -)K- Predicted total cost

8

3

Predicted

11

13

15

17

19

21

23

25

27

29

Combination

Fig. 4. Comparison of actual and predicted assembly tolerances.

Tolerance Allocation Using Neural Networks Table 1, Comparison of predicted and actual tolerances and resulting cost of production Assembly Actual tolerances and number resulting cost of production

Predicted tolerances and resulting cost of production

1

Part 1:0.001649 Part 2:0.0024 Part 3:0.003251 Assembly: 0,0073 Cost: 27.7579 Part t: 0.001748 Part 2:0.0024 Part 3:0.003252 Assembly: 0.0074 Cost: 26.5407 Part 1:0.001845 Part 2:0.0024 Part 3:0.003255 Assembly: 0.0075 Cost: 25.5267 Part 1:0.001942 Part 2:0.0024 Part 3:0.003258 Assembly: 0.0076 Cost: 24.65% Part 1:0.002041 Part 2:0.0024 Part 3:0.003259 Assembly: 0.0077 Cost: 23.90377 Part 1:0.002139 Part 2:0.0024 Part 3:0.003261 Assembly: 0.0078 Cost: 23.2532

Part 1:0.0016440 Part 2:0.00239762 Part 3:0.0032633 Assembly: 0.007304 Cost: 27.8069 Part 1:0.0017551 Part 2:0.0023886 Part 3:0.00325245 Assembly: 0.007397 Cost: 26.5250 Part 1:0.0018569 Part 2:0.0023908 Part 3:0.00324761 Assembly: 0.007494 Cost: 25.49766 Part 1:0.0019496 Part 2:0.0024022 Part 3:0.0032425 Assembly: 0.007593 Cost: 24.63649 Part 1:0.0020340 Part 2:0.0024212 Part 3:0.00323939 Assembly: 0.007695 Cost: 23.8843 Part 1:0.0021107 Part 2:0.0024465 Part 3:0.00323822 Assembly: 0.0077947 Cost: 23.23959

Part 1:0.002237 Part 2:0.0024 Part 3:0.003263 Assembly: 0.0079 Cost: 22.6855 Part 1:0.02235 Part 2:0.0025 Part 3:0.003265 Assembly: 0.008 Cost: 22.1460 Part 1:0.002333 Part 2:0.0025 Part 3:0.003267 Assembly: 0.0081 Cost: 21.64637 Part 1:0.002327 Part 2:0.0026 Part 3:0.003273 Assembly: 0.0082 Cost: 21.1748 Part 1:0.002424 Part 2:0.0026 Part 3:0.003276 Assembly: 0.0083 Cost: 20.7317 Part 1:0.002416 Part 2:0.0027 Part 3:0.003284 Assembly: 0.0084 Cost: 20.3127

Part 1:0.0021806 Part 2:0.00247691 Part 3:0.0032390 Assembly: 0.007895 Cost: 22.64847 Part 1:0.0022444 Part 2:0.0025113 Part 3:0.0032417 Assembly: 0.007997 Cost: 22.10819 Part 1:0.002303127 Part 2:0.00254912 Part 3:0.00324633 Assembly: 0.008098 Cost: 21.60706 Part 1:0.002357299 Part 2:0.00258940 Part 3:0.0032527 Assembly: 0.008199 Cost: 21.142 Part 1:0.00240769 Part 2:0.0026316 Part 3:0.00326092 Assembly: 0.0082989 Cost: 20.7087 Part 1:0.00245492 Part 2:0.0026752 Part 3:0.0032707 Assembly: 0.008399 Cost: 20.29764

2

3

4

5

6

7

8

9

10

11

12

275

Table 1. Continued

Assembly' Actual tolerances and number lesulting cost of production

Predicted tolerances and resulting cost of production

13

Part 1:0.002508 Part 2:0.0027 Part 3:0.003292 Assembly: 0.0085 Cost: 19.9201

Part 1:0.0024995 Part 2:0.00271989 Part 3:0.00328215 Assembly: 0.00850095 Cost: 19.90266

14

Part 1:0.0025 Part 2:0.0028 Part 3:0.0033 Assembly: 0.0086 Cost: 19.5434

Part 1:0.00254197 Part 2:0.00276516 Part 3:0.003295031 Assembly: 0.0086010 Cost: 19.5336

15

Part 1:0.002591 Part 2:0.0028 Part 3:0.003309 Assembly: 0.0087 Cost: 19.18722

Part 1:0.00258266 Part 2:0.00281077 Part 3:0.00330928 Assembly: 0.0087012 Cost: 19.1813

16

Part 1:0.002588 Part 2:0.002899 Part 3:0.003313 Assembly: 0.0088 Cost: 18.84405

Part 1:0.002621938 Part 2:0.00285646 Part 3:0.003324819 Assembly: 0.0088018 Cost: 18.8440

17

Part 1:0.00268 Part 2:0.002899 Part 3:0.003321 Assembly: 0.0089 Cost: 18,51989

Part 1:0.002600 Part 2:0.00290205 Part 3:0.00334153 Assembly: 0.00890353 Cost: 18.5176

18

Part 1:0.002678 Part 2:0.002999 Part 3:0.003323 Assembly: 0.009 Cost: 18.20856 Part 1:0.002769 Part 2:0.002999 Part 3:0.003332 Assembly: 0.0091 Cost: 19.9137

Part 1:0.00269732 Part 2:0.00294736 Part 3:0.0033593 Assembly: 0.009003 Cost: 18.20979 Part 1:0.00273386 Part 2:0.0029922 Part 3:0.0037815 Assembly: 0.00910385 Cost: 17.9t399

20

Part 1:0.002768 Part 2:0.003099 Part 3:0.003333 Assembly: 0.0092 Cost: 17.6314

Part 1:0.0027698 Part 2:0.0030366 Part 3:0.00339787 Assembly: 0.0092035 Cost: 17.63113

21

Part 1:0.00286 Part 2:0.003099 Part 3:0.003341 Assembly: 0.0093 Cost: 17.36205 Part 1:0.00286 Part 2:0.003099 Part 3:0.003441 Assembly: 0.0094 Cost: 17.1054

Part 1:0.00280545 Part 2:0.00308049 Part 3:0.00341841 Assembly: 0.0093043 Cost: 17.35583 Part 1:0.00284073 Part 2:0.00312366 Part 3:0.00343970 Assembly: 0.0094033 Cost: 17.09517

Part 1:0.002859 Part 2:0.003199 Part 3:0.003442 Assembly: 0.0095 Cost: 16.84927 Part 1:0.00295 Part 2:0.003199 Part 3:0.003451 Assembly: 0.0096 Cost: 16.6043

Part 1:0.00287577 Part 2:0.00316614 Part 3:0.00346164 Assembly: 0.009502 Cost: 16.84233 Part 1:0.00291064 Part 2:0.0032079 Part 3:0.00348418 Assembly: 0.0096020 Cost: 16.59852

19

22

23

24

276

S. Anand and P. Kopardekar

Table 1. Continued

References

Assembly Actual tolerances number and resulting cost of production

Predicted tolerances and resulting cost of production

25

Part 1:0.00295 Part 2:0.003298 Part 3:0.003452 Assembly: 0.0097 Cost: I6.3707

Part 1:0.0029453 Part 2:0.0032489 Part 3:0.0035072 Assembly: 0.0097011 Cost: 16.36336

26

Part 1:0.00295 Part 2:0.003298 Part 3:0.003552 Assembly: 0.0098 Cost: 16.13782

Part 1:0.0029800 Part 2:0.00328920 Part 3:0.00353073 Assembly: 0.00979973 Cost: 16.13682

27

Part 1:0.003042 Part 2:0.003298 Part 3:0.00356 Assembly: 0.0099 Cost: 15.9146

Part 1:0.00301459 Part 2:0.00332872 Part 3:0.0035546 Assembly: 0.0098973 Cost: 15.91963

28

Part 1:0.003042 Part 2:0.003398 Part 3:0.00356 Assembly: 0.01 Cost: 15.70142

Part i: 0.3049090 Part 2:0.336748 Part 3:0.00357884 Assembly: 0.0099953 Cost: 15.70818

29

Part 1:0.003042 Part 2:0.003398 Part 3:0.00366 Assembly: 0.0101 Cost: 15.48877

Part 1:0.00308352 Part 2:0.00340549 Part 3:0.00360334 Assembly: 0.010091 Cost: 15.50639

Part 1:0.003134 Part 2:0.003398 Part 3:0.003668 Assembly: 0.0102 Cost: 15.28497

Part 1:0.0031178 Part 2:0.00344277 Part 3:0.00362805 Assembly: 0.010189 Cost: t5.30821

30

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Proceedings of the Second International Neural Networks Conference, pp. 699-705, 1991.

due to the fact that the exhaustive search procedure will be time consuming and computationally intensive as the number of parts in the assembly increase.

16. J. M. Twomey, A. E. Smith and M. S. Redfern, "A neural network model of the dynamic coefficient of friction", Proceedings of 2nd Industrial Engineering Research Conference, pp. 187-191, May 1993. 17. D. D. Bedworth, M. R. Henderson and P. M. Wolfe, Computer Integrated Design and Manufacturing, McGraw Hill, New York, 332 pp., 1991.