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Example 1 : The demand rate for a particular item is 12000 units/year. The ordering cost is Rs. 100 per order and the holding cost is Rs. 0.80 per item per month. If no shortages are allowed and the replacement is instantaneous, determine:
(1) The economic order quantity. (2) The time between orders. (3) The number of orders per year. (4) The optimum annual cost if the cost of item is Rs. 2 per item.
Solution: Note that the holding cost is given per month and convert the same into cost per year.
C 1 = Rs. 2/item C 2 = Rs. 100/order C 3 = Rs. 0.80/item/month = Rs. 9.6/item/year D = 12000 items/year
a) The economic order quantity
2 3
C D
C
Q
^ ^
500 units
b) The time between orders
t * Q * D 500 / 12000 yr.
500 1000 month
0.5 month
c) The number of orders/year
N D Q * 12000 / 500 24
d) The optimum annual cost
(Number of orders) (Cost of one cycle) 24(500 2) 100 (500 / 2 0.5 0.80) Rs. 28800
Model 2 Purchasing model with shortages
In this model, shortages are allowed and consequently a shortage cost is incurred. Let the shortages be denoted by ‘ S ’ for every cycle and shortage cost by C 4 per item per unit time. This model is illustrated in Fig.
Im Q
s t 1 t 2 t T
Fig. 3
Fig. 3 shows that the back ordering is possible (i.e.) once an order is received, any shortages can be made up as the items are received. Consequently shortage costs are due to being short of stock for a period of time.
The cost per period includes four cost components.
Total cost per period = Item cost + Order cost + Holding cost + Shortage cost
Item cost per period = (item cost) (number of items/period)
C Q 1 (13)
Order cost per period = C 2
Let t 1 be the time period during which only the items are held in stock. Let the maximum inventory be denoted by Im and this is equal to ( Q – S ) or Im = ( Q – S )
From similar triangle concept, the following equations can be obtained, referring to fig 3
t 1 (^) Im t Q (16)
or t 1 (^) t Im Q t Q ( S ) / Q (17)
t (^) 2 S t Q or t 2 t S Q (18)
Since time of one period t Q / D
1 t Q^ S Q Q D
^ (19)
2 t S Q Q D (20)
C (^) S 0
C 3 (^) C S^3^ C S^4^ C 3 S C^ (^3^^ C^4 ) Q Q Q
^ (26)
Solving the equation (26) for S , we get
3 3 4 S C Q (^) C C^ (27)
Substituting the equation (27) into the equation (25), we get
2 3 3 4 3 2 (^2 2 3 ) 0 (^ ) 2 2 ( )
C D C C C C Q Q Q C^ C ^ ^ ^ (28)
2 3 32 (^2 2) 2( 3 4 )
C D C C Q C^ C ^ (29)
Solving equation (29) for Q , we get
2 3 4 3 4 Q * 2 C D^ C^ C C C
^ (30)
which is the economic or optimum order quantity.
2 3 4 3 4 S * 2 C D^ C (^) C C C^ (31)
