A tin-container manufacturing company uses the same machine to produce different

Question

A tin-container manufacturing company uses the same machine to produce different items such as popcorn tins and 30-gallon storage drums. The machine is set up to produce a quantity of one item and then is reconfigured to produce a quantity of another item. The plant produces a run and then ships the tins out at a constant rate so that the warehouse is empty for storing the next run. Assume that the number of tins stored on average during 1 year is half of the number of tins produced in each run. A plant manager must take into account the cost to reset the machine and the cost to store inventory. Although it might otherwise make sense to produce an entire year's inventory of popcorn tins at once, the cost of storing all the tins over a year's time would be prohibitive.

Suppose the company needs to produce 1.4 million popcorn tins over the course of a year. The cost to set up the machine for production is $1300, and the cost to store one tin for a year is approximately $1.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

(a) What size production run will minimize setup and storage costs? (Round your answer to the nearest integer.) _________ tins

1
(b) How many runs are needed during one year? (Round your answer to the nearest integer.) _______ runs
2
How often will the plant manager need to schedule a run of popcorn tins? (Round your answer to two decimal places.)

The plant manager need to schedule a run every ________ months

Please provide detailed answer. Thanks.

Explanation / Answer

Let there be x tins produced in a run. Then, we have carrying costs of 1/2x $1

As we manufacture 1.4 million popcorn tins in a year, and setup cost is $1300, we have

1.4*10^6/x setups, so the setup cost is 1.4*10^6/x * 1300

Thus, we minimize 1/2x* 1 + 1.4*10^6/x * 1300 =

1/2 x + 1.82*10^9/x

We calculate the first derivative to 0

1/2 - 1.82*10^9/x^2 = 0

1/2 = 1.82*10^9/x^2

x^2 = 2 * 1.82*10^9

x^2 = 3.64*10^9

x = 10^4 * sqrt(36.4) =

60332.4125159934

Rounding this is 60332

b) 1.4*10^6/60332.4125159934 = 23.2047740446129

This rounds to 23

2. This answer is in months.

12 months/year * 1 year/23.2047740446129 runs= 0.517131428571429 months/run

This rounds to .52

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