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Re: Determination warm up period, run length, number of replications

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2 weeks ago

I made an example for the

"drawn without replacement"

http://www.bangsow.eu/detail_en.php?id=808

in this way you can create randomly the exact number of parts of a sequence

Steffen Bangsowfreelance simulation specialist web: www.bangsow.eu mail: steffen@bangsow.net |

Re: Determination warm up period, run length, number of replications

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2 weeks ago

Re: Determination warm up period, run length, number of replications

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2 weeks ago

Dear @Nomden,

Thanks for your reply! Please find attached my Excel file in which I calculated the portions of my products while calculating the average order per day per product. As you can notice, the sum equals 8.46, which is the same as I used in the calculation of my previous post:

Spoiler

Spoiler

Let`s assume that for a certain product there are 120 orders per year. The system is operational for 5 days a week for 5.9 hours a day (+-1500 hours a year). This are (5/7)*365 = 261 days per year. Hence, 120/261 = 0.46 products will arrive per day from this product type. I repeat this calculation for the complete product scope of 76 products and sum up the total amount of products per day. Assume this is 8.86. Then, 5.9 (working hours per day) / 8.46 = 0.67 hours = 40 minutes. Hence, I will set the negative exponential interval with a mean of 40 minutes and products will be selected based on the portion calculated above (0.46 for the example product). Lastly, the "blocking" must be cleared (deactivated).

Could you please check my Excel file and let me know if you agree with my consideration? And do you mean with the absolute numbers that you just multiply values like 0.06 or 0.04 with 100 and include 6.0 and 4.0 to the ProductMixTable?

Many thanks in advance!!

Kind regards.

Re: Determination warm up period, run length, number of replications

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a week ago

Re: Determination warm up period, run length, number of replications

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a week ago

Dear @Nomden,

Thanks for your reply!! And what do you think of the following from my other post?

Spoiler

Let`s assume that for a certain product there are 120 orders per year. The system is operational for 5 days a week for 5.9 hours a day (+-1500 hours a year). This are (5/7)*365 = 261 days per year. Hence, 120/261 = 0.46 products will arrive per day from this product type. I repeat this calculation for the complete product scope of 76 products and sum up the total amount of products per day. Assume this is 8.86. Then, 5.9 (working hours per day) / 8.46 = 0.67 hours = 40 minutes. Hence, I will set the negative exponential interval with a mean of 40 minutes and products will be selected based on the portion calculated above (0.46 for the example product). Lastly, the "blocking" must be cleared (deactivated).

Many thanks in advance!!

Kind regards.

Re: Determination warm up period, run length, number of replications

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a week ago - last edited a week ago

Quick calculation for 2018:

261 working days / year * 5.9 hours/working day=1539.9 hours/year

1539.9 hours/year / 3088 orders/year=0.4987 hours/order

The Random product selection does the rest.

Re: Determination warm up period, run length, number of replications

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a week ago

Thanks for your reply! And then without blocking, right? The interval will then be 0.49 hours and the frequencies are just the total orders per year?

Re: Determination warm up period, run length, number of replications

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a week ago

To be sure create a seperate test model with a Source object and a Drain. The results should match!

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