Question

A plant produces 75 units per hour of an item with dimensions of 1'x 1.5' xl'. We want to store the one week supplies in containers measuring 5'x 5 x 5'. A minimum of 3" space is required between the products and the container walls (In all directions). 1). The number of containers needed for a one day production output 2). 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.

Answer 1

No. of units produced/week = 75 units * 8 hours per day * 5 days of working days = 3000 units

Number of products per container = (60/12) * (60/18) * (60/12) = 75 units per container

1).Number of container needed for a one day production = 75*8 / 75 = 8

No. of containers/week = 8 * 5 = 40 containers/week

2). If 5 high stacked containers are there, then number of stacks = 40/5 = 8 stacks

Width of the floor space =60 +2*6 = 72”

Length of the floor space = 8*60 +(8+1)*6 = 534”

Required Floor Space = (534”)*(72”) (1+0.10) = 42292.8 square inch = 293.7 square feet