Daily demand for a product is 50 units, with a standard deviation of 30 units. The review period is 5 days and the lead time is 10 days. At the time of review there are 60 units in stock. If 99 percent service probability is desired, how many units should be ordered? (Use Excel's NORMSINV() function to find the correct critical value for the given α-level. Do not round intermediate calculations. Round "z" value to 2 decimal places and final answer to the nearest whole number.) Ordered quantity units

The number of units to be ordered is 961.

Step-by-step explanation:

The optimum demand is given as

From the data

Daily demand is given as =50

Standard deviation is given as

The Lead time is given as L=10 days

The Revier time is given as R=5 days

Initial values given as I=60 units

The confidence level is given as 99% so z is calculated using the NORMSINV() function as NORMSINV(0.99). The value of z is 2.33

Now the standard deviation in terms of the Lead and Review time is given as

Substituting the values in the formula as

So the number of units to be ordered is 961.

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)