Question

Suppose that the annual demand for a component is approximately 60,000 units. The company orders the component from a supplier who has offered the following quantity discount schedule.

Order Quantity	Price per Unit
0-999	$30
1,000-1,999	$29
2,000-3,999	$28
4,000 or more	$27

If the company’s carrying charge is 15 percent of the item’s price and the cost per order is $150, determine the order quantity that would minimize the total related inventory costs for this component.

Answer 1

Annual demand(D) = 60000 units

Ordering cost (S) =$150

Holding cost(H) = 15% of purchase price

Order size Price per unit Holding cost(15% of price per unit)

0-999 30 4.5

1000-1999 29 4.35

2000 or 3999 28 4.2

4000 or more 27 4.05

First find the minimum point for each price starting with the lowest price until feasible minimum point is located.This means until a minimum point falls in the quantity range for its price

Minimum point for price of $27 = Sqrt of (2DS/H)=Sqrt of [(2 x 60000X150)/4.05] = sqrt of (18000000/4.05)= 2108 units.Because an order size of 2108 units will cost $28 rather than $27, it is not a minimum feasible point for $27 per unit.

Minimum point for price of $28 = Sqrt of (2DS/H) =Sqrt of [(2X60000X150)/4. 2]= sqrt of (18000000/4.2) = 2070 units This is feasible as it falls in the $28 per tyre range of 2000-3999

Now the total cost for 2070 units is computed and compared to the total cost of the minimum quantity needed to obtain price of $27 per unit

Total cost for Q=20702 is (Q/2)H + (D/Q)S + (PriceXD)

= [(2070/2)4.2] + [(60000/2070)150] + (28X60000)

= 4347 + 4347.83 + 1680000

= $1688694.83

The minimum quantity needed to obtain a price of $27 is 4000 units So with order quantity(Q) = 4000 units,

Total cost = (Q/2)H + (D/Q)S + (PriceXD)

= [(4000/2)4.05] + [(60000/4000)150] + (27 x 60000)

= 8100 + 2250 + 1620000

= $1630350

So the economic order quantity that would minimize the total related inventory cost is 4000 units as it has the lowest total cost of $1630350