Due to the continuous use of machine tools, all parts will be damaged due to wear and tear, resulting in process obstacles. If production continues at this time, it is zero.

A large number of unqualified products will appear in the parts, resulting in losses. In order to reduce losses, parts should be inspected regularly.

Check and repair the machine tool if unqualified products are found. And the failure caused by tool damage is the cause of process failure.

95%, so you can consider changing tools according to a certain strategy. The above operation requires a certain maintenance fee, but it can be

Under the action of this pair of contradictions, there must be the best inspection interval and tool changing strategy.

Minimize the sum of maintenance cost and loss of nonconforming products.

According to the known tool failure record of 100, it is known through inspection that the number of parts completed before the two tool failures meets the requirements.

Normal distribution. Because other faults only account for 5% of the process obstacles, they have little influence on the best inspection gap and tool change strategy.

. In order to simplify the calculation, it can be assumed that when other faults occur, the number of completed parts meets the uniform distribution and other faults occur.

It has nothing to do with the occurrence of tool failure.

The random process of two adjacent tool updates is defined as an update cycle, and its value is the machine tool generation in the two update processes.

Number of parts produced. Is the total cost of the update cycle. Then our goal is to find the best and minimize it. This can be

In order to get a better solution through computer search. Monte Carlo method is used to simulate the actual process and the scheme is tested.

Test and make some adjustments.

Three data analysis

The title gives 100 tool failure records (the number of completed parts), and the more direct idea is to analyze this 100 data.

Conduct mathematical statistics. The statistical results show that the average value is 600 and the standard deviation is 196.63, and the significance level is given.

, tool failure can obey the normal distribution after inspection, recorded as:

However, the probability density of normal distribution is valuable, and the number of parts completed when the tool fails is positive.

The explanation is as follows:

According to the principle of normal distribution, the known tool failure probability is 99.7%, which falls within other ranges.

The probability is very small, 0.3%, which can be considered as a small probability event, which does not happen in practice and can be explained.

National contradiction.

From the average value of 600, it can be known that a tool accident will occur every 600 parts produced on average; The topic also talks about the process reasons.

Among the obstacles, 95% are tool damage failures, and 5% are other failures. Combined with the law of large numbers, other faults can be inferred.

The mathematical expectation of the number of parts produced is 600.

Record the probability density of the number of parts produced when other faults occur:

Four-parameter description

T: inspection interval; D: The average cost (including the tool change fee) of adjustment to make it return to normal in case of failure is 3,000 yuan.

Yuan;

K: the cost of replacing a new tool when no fault is found, 65,438+0,000 yuan H: the loss caused by shutdown due to wrong judgment of working procedure, 65,438+0, 500 yuan.

K0: the specified tool change interval, which is constant; F: Loss of 200 yuan due to the production of a waste product;

N: The number of inspections contained in k0. In the computer solution, we think that K0 includes an integer number of checks.

(t, K0 measures the length of time by the number of parts produced)

Five basic assumptions

1. Tool failures and other failures are independent of each other;

2. If employees check out nonconforming products, they should stop for inspection.

Establishment of six models

Through the above analysis, we know that this problem belongs to optimization, and it is necessary to determine the optimal inspection gap and tool replacement.

Interval to minimize the expected cost. Objective function:

Seven model solutions

Question 1: It is assumed that all the parts produced when the process fails are unqualified, and all the parts produced under normal conditions are qualified.

Design the most effective inspection gap and tool replacement strategy for this process.

Arrange employees to check and change tools according to the following steps:

1. Check with parts as gaps. If the first inspection finds that the parts are unqualified, go to 3; Otherwise, continue.

Continue until the second inspection is completed, and go to 2;

2. Replace the cutter and go to1;

Adjust it and make it return to normal. If the fault is caused by a tool fault, go to 1. If the fault is caused by other faults,

Then complete the remaining second check and go to 2. (If nonconforming products are found, handle them according to the method in 3). We need the best products now.

And minimize it.

Obviously, when the number of parts produced by the machine tool is greater than the failure of the tool, the cycle will be updated, and when the tool fails, the opportunity will come.

When the number of parts produced by the bed is less than, it is updated circularly, and the conclusion is drawn:

Note: The number of parts produced before failure is a probability density function.

Change the clearance of the tool, that is;

Check the gap.

Is the probability that the tool will fail during the i-th inspection interval.

Secondly, approximate determination, that is, the average cost in a tool changing cycle, tool failure and other failures will cause maintenance costs.

Use, so we should consider it comprehensively.

1. Considering the cost caused by tool failure:

Consistent with the idea of seeking the same thing, we can write the expression of cost expectation caused by tool failure.

Description: indicates the average cost of tool failure when the number of tool changes is. Indicates that the tool failure occurred in the first

The probability in the inspection gap approximately represents the loss caused by unqualified products.

2. Consider the cost of other failures.

Because other failures only account for 5% of process failures, we can consider simplifying the processing, assuming that in any update cycle.

, other faults occur at most once.

Assuming that other faults occur in the first inspection gap, the probability is. Produce parts for machine tools before other faults occur.

Probability density function of numbers.

Because the occurrence of other faults will not affect the update cycle, no matter which inspection interval other faults occur, check

The cost is and the maintenance cost is. Therefore, the mathematical expectation of expenses caused by other faults is:

;

Considering comprehensively, the approximate expression of can be given.

A function that aims at the sum of two variables can be found by computer search.

Minimum value, K0=450, T= 18.

Question 2

If the parts produced during the normal operation of this process are not all qualified products, 2% are unqualified products; And zero output when the process fails.

40% of the parts are qualified and 60% are unqualified. The loss caused by normal working procedures and faulty shutdown is 15.

00 yuan/piece. The best inspection interval and tool replacement interval are designed for this process.

To solve this problem, we need to determine the objective function.

The first expression:

1, considering the loss caused by tool failure:

Similar to the solution to the problem 1, the tool change cycle x >；; K0, its probability is that this situation is found now.

Under loss.

Because the probability of other faults on (0, k0) is very small, other faults can be considered when considering tool faults.

The fault does not appear at (0~k0). Therefore, it can be regarded as (0,k0) if the above-mentioned tool failure does not occur at (0,k0).

There is no process failure. The expenses include: inspection fee, tool change fee, and 2% failure when the machine tool is trouble-free.

1500 yuan's loss caused by loss of qualified products and misjudgment of machine tool failure, namely.

Tool change cycle x < K0 hours, in order to find the loss expectation in this case, it can still be similar to the solution of the problem 1, and put (0

, k0) subsection, which is the probability of tool failure in the first section. The following is the introduction of tool failure in the first section.

The losses are discussed as follows:

(1) maintenance fee after failure d=3000 yuan,

(2) Although the machine in the previous section (approximately the fault occurred in the middle of the first section) works normally, there is a test fee.

% loss of nonconforming products and 1500 yuan machine failure misjudgment loss, namely

flow chart

(3) After the machine tool fails, the unqualified products are produced with a probability of 0.6 and the qualified products are produced with a probability of 0.4, so the I section.

Finally, at the I-th checkpoint, the probability of machine tool failure is 0.6, and the loss comes from the loss of half-unqualified products.

, and inspection fee: However, it is also possible that the machine failure cannot be detected at the i-th checkpoint until the i+ 1 th checkpoint.

Can find out the fault situation, the probability is 0.4*0.6, loss; Similarly, there is a problem that a failure is not detected until the i+2nd checkpoint.

Situation, possibility, loss. Although there may be I+3, I+4 ... checkpoints to find out the fault, but by

Because of the probability of these situations, they are no longer considered.

To sum up, the loss expectation in the first paragraph can be written as:

Therefore, based on the above considerations, we can get the loss expectation caused by tool failure:

2. Consider the losses caused by other faults:

In 1, we neglected to consider the losses caused by other faults, because other faults are in the tool changing cycle.

The probability of occurrence is small, which is reasonable. But in order to make the model more complete, we also introduce the loss of other faults approximately.

Lose expectations. Similar to the analysis of the first question, take the maximum value K0 of the tool change cycle and approximately calculate the cost expectation at this time:

It is necessary to point out the gains and losses. Compared with S 1, the value of S2 is relatively small, which has no significant influence on the results.

1. Determine the mathematical expected value of the tool change cycle;

The mathematical expectation of tool change cycle is also determined by tool failure (repairing other failures without changing tools), so

The form is the same as that in Problem One.

K0 and t when the objective function reaches the minimum value can be calculated by computer.

k0=324，T=39 .

Question 3:

On the premise of the second question, the loss can be reduced or improved by correctly adjusting the inspection interval and tool change interval.

Look at ways to get higher benefits.

The controllable part of the loss is the downtime loss caused by misjudgment. Analyze the causes of misjudgment, and calculate the waste as follows

95% is produced under normal working procedure, and 5% is produced under fault working procedure. However, under normal working procedures, two kinds of waste products will be produced continuously.

The probability is 0.0004, and in the case of abnormal process, the probability of two consecutive failures is 0.36. When there are two consecutive

Waste can be considered as an abnormal process. Therefore, there is an improvement plan.

1. When the part is genuine, the inspection is over.

2. When the part is rejected in the inspection, check the next one; When they are rejected, stop the machine for inspection; When it is true, it will not stop.

This process is considered normal.

Although this scheme increases the inspection cost, it greatly reduces the cost of downtime due to misjudgment.

Eight model analysis

In order to test the influence of different changes of H, T, K and F on the loss sf, we give them different values and calculate them.

The corresponding sf values are shown in the following table: (sf is the sum of losses in the production of 60,000 parts)

The influence of 1 and h on sf

Rmb13001400150016001700

Sf (ten thousand yuan) 59.870 60.216 60.550 60.706 61.05438+03

K0 (unit) 330 324 324 3 12 3 12

T (piece) 33 36 36 39 39

N (times) 10 9 8 8

M (ten thousand yuan) 1.3 1

2. The influence of the change of T on sf

T (element) 8 9101112

Shunfeng (ten thousand yuan) 60.200 60.375 60.550 60.553.199896966666

K0 (parts) 324 324 324 3 12 3 12

T (piece) 36 36 36 39 39

N (times) 9 9 9 8 8

Ten thousand yuan 4. 10

3. The influence of the change of K on sf

Yuan 800 900100011001200.

Sf (ten thousand yuan) 56.633 58.743 60.550 62.438+052 638+0

K0 (unit) 287 304 324 324 340

T (piece) 4 1 38 36 36 34

N (times) 7 8 9 9 10

Ten thousand yuan 16.94

4. The influence of the change of F on sf

Rmb100125150175 200 225 250 275 300

Shunfeng (ten thousand yuan) 44.116 48.570 52.656 56.468 60.5550 64.2568.5538+0.9438+0.75.5008556866

K0 (unit) 343 344 328 320 324 315 306 310 297

T (piece) 49 43 4140 36 35 34 3133

N (times) 7 8 8 8 9 9 9 10 9

Ten thousand yuan 23.86

In order to judge the influence of parameters on the loss cost function, the relative change evaluation index is introduced.

set up

That is, the influence of changing a unit relative quantity (such as 1%) on the loss cost.

When the relative quantity of the parameter unit changes, the greater the loss cost, the higher the sensitivity of the parameter, and the order is:

That is, sensitivity to damage.

Part loss expense > tool updating expense > shutdown loss > inspection expense.

Suggestion:

In order to seek maximum economic benefits and reduce production losses.

1, and the optimal inspection cycle and tool change cycle are adopted.

2, as far as possible to reduce the cost of parts loss, tool change can greatly reduce the cost of loss.

Nine model test

We use computer simulation to test the model results. The idea of the simulation program is briefly described as follows:

Firstly, according to the probability distribution of tool failure and other failures, a series of sample points are generated, and then within a certain range, no

Try to get the inspection cycle and tool change interval in different places, and simulate the actual production process to get a series of loss costs, the most of which.

The minimum value and its corresponding inspection period and tool change interval are taken as the optimal solution.

Please refer to Appendix 4 and Appendix 5 for specific procedures.

The problem 2 was simulated many times, and the following series of results were obtained:

Losses sf (ten thousand yuan) 59.55438+06 59.696 60.129 58.146 60.058+0.0438 59+0 58.80.50080.50080506066

8 59.720

Tool change cycle K0 336 368 360 294 234 336 365 438+02 294 240 400

Inspection period T 48 46 40 42 39 48 39 42 42 40

Analysis: The reason why the simulation results fluctuate in the range of [234-400] is that a group of tool loss zeros are randomly selected for each execution.

The number of pieces, because the variance is 196.62, fluctuates greatly, but it is still nearby, and the cost function is stable at 580,000-

6 1000, and the sum of the solutions obtained by the model accurately falls in the above areas. The model has good stability.

refer to

Zhu Wenyu, Mechanical Reliability Design, Shanghai Jiaotong University Press.

1992

Xu Zhong et al. Probability Theory and Mathematical Statistics Sichuan Science and Technology Publishing House

988

Matlab source program (omitted)