Inventory 2

160

the length of the time to receive an order, the number of orders per year, and the maximum inventory level. Solution

s = 150MU/order  E=0.75 per unit C=10000m c=10000/311 =32.2m/day Qo=

{(2Cs)/E(1

- c/R)}

=

{(2)(10

000)(150)}

/{0.75(1

- 32.2/150)}

= 2 256.8 m

This value is substituted into the following formula to determine total minimum inventory cost: K e=

2CsE(1

- c/R)

=

2(10

000)(150)(

0.75)(1

- 32.2/150)

= 1 329 MU

or  K e= (C/Q0).s + (Q0/2)E(1-c/R) = (10000/2256.8)150 + (2 256.8/2)(0.75)(1-32.2/150) = 1 329 MU The length of time to receive an order for this type of manufacturing operation is commonly called the length of the  production run. It is computed as follows t1 = Q0/R = 2 256.8/150 = 15.05 days/order  made  No =

The number of orders per year is actually the number of production runs that will be C  Q

=

10000 2256 .8

= 4.43 runs/year 

The maximum inventory level is. Imax= Qo(1- c/R) = 2256.8 (1-32.2/150) = 1 772 m Example 43:

The manager of a bottling plant which bottles soft drinks needs to decide how long a `run` of each type of drink to ask the lines to process. Demand for each type of drink is reasonably constant at 80 000 per month (a month has 160 production hours). The bottling lines fill at a rate of 3000 bottles per hour but take an hour to change over between different drinks. The cost of each changeover (cost of labor and lost production capacity) has been calculated at 100 MU/hour. Stock-holding costs are counted at 0.1 MU/bottle-month. a. How many bottles the companies produce an each run?  b. The staffs who operate the lines have devised a method of reducing the changeover  time from 1 hour to 30 minutes. How would that change the ELS? 160