Question

The average daily demand for component B is 750 units. The average time for waiting during the production process and materials handling is 0.15 days. Processing time per container is 0.1 days. The container capacity is 50 units and the policy variable is​ 5%. What number of containers is required for component​ B?

Answer 1

Answer: 4

Step-by-step explanation:

daily demand(d) = 750 unjts

Policy variable(¢) = 5/100 = 0.05

Waiting time(w) = 0.15 days

Container capacity(c) = 50 units

Processing time per container(p) = 0.1 days

k = number of containers

Using the formula:

k = [d(w + p)(1 + ¢)] ÷ c

k = [750(0.15 + 0.1) × (1 + 0.05)] ÷ 50

k = [750(0.25) × (1.05)] ÷ 50

k = 196.875 ÷ 50

k = 3.9375 = approximately 4 containers

Answer 2

4 containers

Step-by-step explanation:

Given

Average daily demand, d= 750 units

Average time for waiting during production and materials handling, w = 0.15 days

Processing time per container, t = 0.1 days

Container capacity, c = 50 units

Policy variable, α = 5% = 0.05

Number of containers required for component​ B is calculated as follows;

Total Production/Container Capacity.

Total Production = d(w + t)(1 + α)

Total Production = 750 * (0.15 + 0.1) * (1 + 0.05)

Total Production = 196.875

Number of containers = 196.875/50

Number of containers = 3.9375

Number of containers = 4 --- approximated