The average daily demand for component B is 750 units. The average time for waiting during the production process and materials handling is

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?

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Thu Thủy 5 months 2021-08-19T03:52:24+00:00 2 Answers 16 views 0

Answers ( )

    0
    2021-08-19T03:53:36+00:00

    Answer: 4

    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

    0
    2021-08-19T03:53:57+00:00

    Answer:

    4 containers

    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

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