Видео

You're welcome!

@@The_Business_Doctor 😊

Thanks!

Hi Armando, That's a good question, but the centre of gravity model is not intended to be used for more than one centre of gravity (which would be like having two suns in the solar system!). If you're looking to add two distribution centres to a network, presumably each is intended to support a specific geographic region so I think you would want to look at the entire network and decide which ones to group together (such as northern region and southern region) and then use the model to find the centre of gravity for those regions separately. Each DC could conceivably act as a backup for the other region incase something happens to one (say it blows up or something), but it won't be in an optical location. Mark

Go for it!

Thanks. Yes there are many ways to solve a problem like this. I did it this way for a particular reason to demonstrate that particular Excel function.

You’re welcome! Good luck on your test!

You're welcome 😊

Based on the textbooks I have used, cycle time is "the maximum time allowed at each workstation to complete its set of tasks on a unit." (Stevenson & Hojati, 2011; Heizer, Render, Munson, & Griffin, 2020). This problem gives us tasks which have yet to be organized into workstations and the cycle time represents the maximum amount of time that a unit can spend in a workstation in a given cycle. The total sum of all the tasks (process if you like) to complete a product is the task duration time and we divide the duration time by the cycle time to determine the optimal number of workstations. So in this case, if the task duration time (in seconds for an entire hour) is 3600 seconds to produce a unit through all the tasks and the cycle time is a maximum of 20 seconds per unit, that means we need to try to organize the tasks into 3 workstations to allow the unit to be produced. It could be that different textbooks and sources may interpret the cycle time slightly differently. Here was have a WORKSTATION cycle time of 20 seconds per unit and a total (ideal) PRODUCTION cycle time of 60 seconds (which is what I think you're trying to get at). However, sometimes the theoretical number of workstations cannot be achieved because of the precedence relationships and times of the tasks, so you might need 4 workstations. In this case the actual PRODUCTION cycle time would be greater than the WORKSTATION cycle time. I guess it depends on the source you're referring to.

Hi there. Funny you mention that. I don't know of any of I was actually thinking about putting together a course for lawyers! I know trust accounting would be a key element but would have to consult with some lawyers that I know to determine what appropriate content should be. Also, by "cheap" what do you mean? What do you think a reasonable price point is for a program that would cover what you're looking for? You are welcome to email me direct at mark@strategema.ca

@@The_Business_Doctor To be honest, I haven't seen an online accounting course that isn't associated with a University. And all of those programs are very, very expensive. So, a quarter of the price point would be very attractive.

You’re welcome!

Thanks!

I don't know what you mean by the single word comment.

Hi there. Absolutely you can -- it involves working backwards using roots. For example, if you have two components in series, each with a reliably of 95%, the reliability of the system is 0.95 x 0.95 = 0.9025 So if you know the system reliability, and there are 2 components, you would take the square root of 0.9025 and get 0.95! If there are 3 components in series and the reliability of the system is 0.8574, then you would take the third root of 0.8574, or if you have a calculator that is not capable of anything higher than a square root, you would take 0.8574^(1/3) (i.e. to the exponent of 1/3) and you'll get 0.95. Proof: 0.95 x 0.95 x 0.95 = 0.8574. For 4 components each with a 95% reliability: System reliability = 0..95^4 = 0.8145 (using the yx button on a calculator) If you know the system, then each component is 0.8145^(1/4)( also using the yx button) For parallel and more complicated systems, this can be more difficult and requires a lot of careful working backwards. Hope this helps. Mark

You’re welcome! Operations management is international!

Sorry, but I don't have the time to solve extra problems and these requests are often from students asking for help with homework or assignment problems, which I don't provide.

Great news!

Thanks very much!

Thanks!

Hi Adrienne. The company would have to add 8 machines in year 1 to make up for the capacity shortage, then 1 more in years 2 through 5. Hope this helps. Mark

Glad you found it useful

Thanks a lot!

You’re welcome!

Awesome! Hope your exam went well!

Good question. From the MAPE you can sort of really assess if the forecasting is “good” or bad per se. The it’s the average absolute percentage error which means that the forecast is “off” by the MAPE and that could mean over or under relative to the actual demand. It’s the size of the MAPE that’s important. For example if the MAPE is 5% then we can say the forecasts differ from actual demand by an average of 5% which is not very much in many cases since you would expect some differences. If, however, the MAPE is 15% that might not be acceptable because the higher the MAPE the less reliable the forecast is. Hope this helps.

@@The_Business_Doctor understand your point. One more thing to ask- how we can increase our projection accuracy, is there another technique or methods?

Awesome! Hope your exam went well!

Hi there. The crash time values are provided, so there are no calculations required. You can use any unit you like: days, weeks, months, etc.. Hope this helps. Mark

Yes, you're right, that video isn't clear that the 0.577 comes from the A2 factor in a table. However, the formula does specify A2 and, presumably, students taking an OM course where a problem like this appears would be familiar with the formula and what A2 relates to. Still, I should have included a snip of the tables. As for the sample size, I believe you're confusing the number of samples (12) with the sample SIZE (5) and that's clarified at the beginning of the problem.

Unfortunately it is available only when you purchase the textbook.

I presume you're referring to the averge production in Plan C. The solution table shows Production (Avg Demand), so that would be the avergae of all the demands given in the problem data. Add them all up and divide by 12 and you end up with 1,775.

Good question! This is because when the production starts, they have too many staff (to produce 1500 units). It's after the first period where there is a single round of layoffs that results in stable employment to meet the production of 1,350 for every period. Hope this helps. Mark

I’m glad you liked it!

I’m glad you found it helpful.

You’re welcome!

Thanks!

I don’t know what that means

This is the ISBN for the current version of the textbook: 9780137319084, 0137319088 Operations Management: Sustainability and Supply Chain Management (Canadian Edition) 4th Edition by Jay Heizer; Barry Render; Chuck Munson; Paul Griffin

Hi Sara, Sorry for the late reply to this. In part A we want the single-factor productivity in tires per HOUR. If we wanted the single-factor productivity per LABOUR DOLLAR, we would then take 1000 tires per day and divide by (400 x $12.50) and end up with 0.20 tires per labour dollar. That's a different single-factor metric. In part B we DO multiply that to get the total labour cost of $5,000 which is added to the other costs for multi-factor productivity on per-dollar basis. The key is to know what you want to calculate: productivity per hour, per dollar, etc. Hope this helps. Mark

You’re welcome!