Question

Q5. A garage requires 100 tires daily with standard deviation of 30 units. Inventories are reviewed and orders are placed daily. lead time for order is 1 day. The holding cost per day is $0.05. The garage wishes to operate with 0.9987 in-stock probability. How many tires are on order ? on hand ?

Answer 1

SOLUTION:

Given data,

A garage requirs 100 tires

standard deviation = 30 units

lead time = 1 day

the holding cost per day = $0.05

0.9987 in- stock probability

find how many tires on order ?

We will initially discover the reorder point,

As we know the formula,

R=d * LT + z*σd*√ LT

where d is a day by day request, LT is lead time, σd is the standard deviation of interest, z will be z-score dependent on the in-stock probability

R=d * LT + z*σd*√ LT

R = 100*1 + z*30*√ 1

(For in-stock probability pf 0.9987, seeing z table we get the z score of 3.02. Z-table is given beneath toward the end for reference)

R = 100*1 + 3.02*30*√ 1

Reorder point, sum available = 190.6 tires

Presently, Amount to order = d*(OI + LT) + z*σd*√ (OI + LT) - (A)

where An is sum available, OI is structure interim (length of time between orders)

OI =1 day, as given

Amount to order = 100*(1+ 1) + 3.02*30*√ (1+1) - (190.6)

Amount to order = 200 + 128.127 - (190.6)

Amount to order = 137.527

Consequently, the Amount to order is 138 (adjusted) and close by is 190 (adjusted)