Question

Q1.A Company uses 2,500 units during the course of the year, and its usage is relatively constant throughout the year. These units are purchased from a supplier 100 kilometers away for Ksh.15 each, and the lead time is 2 days. The holding cost per unit per year is Ksh.1.50 (or 10% of unit cost) and the ordering cost per order is Ksh.18.75. There are 250 working days per year for this company.
a) What is EOQ?
b) Given the EOQ, what is the average inventory?
c) In minimizing cost, how many orders would be made each year? What would be the annual ordering cost?
d) Given the EOQ, what is the total annual inventory cost ( including purchase cost)
e) What is the time between orders?
f) What is the ROP(re-oder point)? (15mks)

Answer 1

Answer on Question #65630 – Math – Differential Equations

A Company uses 2,500 units during the course of the year, and its usage is relatively constant throughout the year. These units are purchased from a supplier 100 kilometers away for $Ksh. 15$ each, and the lead time is 2 days. The holding cost per unit per year is $Ksh. 1.50$ (or $10\%$ of unit cost) and the ordering cost per order is $Ksh. 18.75$. There are 250 working days per year for this company.

Question

a) What is $EOQ$?

Solution

Economic Order Quantity:

$$Q^* = \sqrt{\frac{2DK}{h}},$$

where $D$ - annual requirement quantity, $K$ - cost per order, $h$ - yearly carrying cost per unit

We have

$$D = 2500; \quad K = 18.75; \quad h = 1.5.$$ $$EOQ = Q^* = \sqrt{\frac{2 \cdot 2500 \cdot 18.75}{1.5}} = 250 \text{ units}$$

Answer: $EOQ = 250$ units

Question

b) Given the $EOQ$, what is the average inventory?

Solution

$$\text{average inventory} = \frac{TC}{N},$$

where

$$N - \text{number of working days per year};$$ $$\begin{array}{l} TC = \text{Total cost} = PD + K \frac{D}{EOQ} + h \frac{EOQ}{2}; \quad P = 15 \text{ is cost per unit}; \\ TC = 15 \cdot 2500 + \frac{18.75 \cdot 2500}{250} + \frac{1.5 \cdot 250}{2} = 37875; \\ \end{array}$$ $$\text{average inventory} = \frac{37875}{250} = 151.5.$$

Answer: average inventory = 151.5.

Question

c) In minimizing cost, how many orders would be made each year? What would be the annual ordering cost?

Solution

Number of orders:

$$N_1 = \frac{D}{EOQ} = \frac{2500}{250} = 10$$

Annual ordering cost:

$$K \cdot N_1 = 18.75 \cdot 10 = 187.5$$

Answer: 10; 187.5.

Question

d) Given the $EOQ$, what is the total annual inventory cost (including purchase cost)

Solution

$$TC = PD + K \frac{D}{EOQ} + h \frac{EOQ}{2} = 15 \cdot 2500 + \frac{18.75 \cdot 2500}{250} + \frac{1.5 \cdot 250}{2} = 37875$$

Answer: 37875.

Question

e) What is the time between orders?

Answer: the lead time = 2 days

Question

f) What is the $ROP$(re-oder point)?

Solution

$ROP = \text{Average daily wage rate} \times \text{Lead time in days}$

$$ROP = \frac{2500}{250} \cdot 2 = 20 \text{ units}$$

When the inventory level reaches 20 units an order should be placed for material.

Answer: 20 units.

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