A plant produces 75 units per hour of an item with dimensions of 1\' Times 1.5\'
ID: 452964 • Letter: A
Question
A plant produces 75 units per hour of an item with dimensions of 1' Times 1.5' Times 1'. We want to store only one week supplies in containers measuring 5' Times 5 Times 5'. A minimum of 3" space is required between the products and the container walls (In all directions). The number of containers needed for a one day production output The containers may be stacked five high. If we allow 6" between stacks, determine total floor space required for this storage Allow 10% aisle apace. Figure 1 below, shows partial layout of a manufacturing plant In this plant, parts are produced in departments # 1, #2, and #3 The parts, after production, are transferred, by lift trucks, to the Temporary Storage. All material in the temporary storage are then transferred by large trucks to a warehouse near the plant. Suppose in this project you are only concerned about transfer of products from departments (1, 2, and 3) to the Temporary Storage. Use the information provided on the layout and table 1 below and determine how many lift trucks are needed Assume a lift truck speed of 200 ft/min and a congestion factor of 0.85Explanation / Answer
Q.#1 1) Assuming 8 hrs. in a day, total production = 75 x 8 = 600 units.
Volume of one unit = 1' x 1.5' x 1' = 1.5 cu. ft.
Total volume of 600 units = 600 x 1.5 = 900 cu. ft.
Volume of each container = 5' x 5' x 5' = 125 cu. ft.
Assuming a minimum of 3" space is required between product and container walls (in all directions), volume of container becomes 125 - 1/164, i.e., 124.994 cu. ft.
So, total number of containers required in a day = 900/124.994
= 7.2 = 8 (rounding off to next whole number) (Ans).
Q.# 1 2) Assuming 5 working days in a week, number of containers required = 8 x 5 = 40 containers.
Since containers may be stacked 5 high, number of stacks = 8.
Length of storage space required = 5' x 8 + 0.5' x 7 (assuming 6" between stacks) + 1.05' x 2 (assuming 10 % aisle space on 2 sides)
= 45.6'.
Width of storage space required = 5' + 1.05' x 2 (aisle space) = 7.1'.
Floor space required for storage = 45.6' x 7.1' = 323.76 sq. ft. (Ans)
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