# Teaching Notes

 Page 2/16 Date conversion 20.05.2018 Size 0.96 Mb.

## Memo Writing Exercises

1. In most cases it is not feasible to perfectly balance a production line. First, there are technological constraints dealing with precedence and incompatibility issues. In other words, certain tasks have to be performed before others (precedence) and two tasks may not be performed at the same station due to their incompatibility (space and nature of operation considerations). Secondly, there are output constraints. Since most task times vary, output constraint determines whether an otherwise eligible task will fit at a work station because sum of the task times assigned to a station can not exceed the cycle time. As a result of both technological and output constraints, it is extremely difficult to achieve a perfectly balanced production line. The larger the number of tasks, the more difficult to achieve perfect balance.

2. Producing two products on the same assembly line allows the company to utilize the same workstations to produce the common parts and available to them. This results in reduced labour and capital costs. If one of the products is new, the company can shorten the period of time from design to actual production, and reduce the cost of manufacturing in the long run.

3. Fixed automation is utilized in a continuous flow/mass production environment. It enables the firm to manufacture a single or a few products at high volume and low cost. However, it is not flexible enough to produce a variety of parts and it is very costly to make changes to the process. Flexible automation is utilized in a batch environment, where a wide variety of products can be produced without significant changeover (setup) time/cost. Flexible machinery is not designed for high volume (mass) production.

## Solutions

 1. OT = 450 minutes a. Minimum cycle time = length of longest task, which is 2.4 minutes. Maximum cycle time =  task times = 18 minutes. b. Range of output: Maximum: @ 2.4 min.: 450 = 187.5 units 2.4 Minimum: @ 18 min.: 450 = 25 units 18 c. N = D x t = 187.5(18) = 7.5, Which rounds to 8 OT 450 d. Output = OT Solving for CT, CT = 450 = 3.6 minutes per cycle. CT 125 e. Output: (1) CT = 9 min.: OT = 450 = 50 units CT 9 (2) CT = 15 min.: 450 = 30 units 15

2.
 Desired output = 33.33 units per hour Operating time = 60 minutes per hour CT = Operating time = 60 minutes per hr. = 1.80 minutes per unit Desired output 33.33 units per hr.