Question

Daily demand for a product is 100 units, with a standard deviation of 25 units. The review period 10 days and the lead time is 6 days. At the time of review there are 50 units in stock. If 98 percent service probability is desired, how many units should be ordered.

Answer 1

Answer:

1755 units are ordered

Explanation:

given data

Daily demand = 100 units

standard deviation = 25 units

review period = 10 days

lead time = 6 days

stock = 50 units

service probability = 98 percent

to find out

how many units should be ordered

solution

order quantity is calculated in fix time period formula is express as

q = $\bar{d}(L+R) + z \sigma_{L+R} - I$        .........................a

here L is lead time and R is review time and σ is standard deviation and I is stock and d is Daily demand

so first we find here standard deviation that is

$\sigma_{L+R} = \sqrt{L} * \sigma$   ...................1

$\sigma_{L+R} = \sqrt{25} * 25$

$\sigma_{L+R} =100$

so the value of z is for 98 % service probability is  2.05

so put here value in equation 1

q = 100 × ( 6 +10) +(2.05) × 100 - 50

q = 1755 units

so 1755 units are ordered