The purchase inventory model with single discount may be expressed as follows:
Order Quantity  Unit Price (Rs.) 

1 ≤ Q_{1} < b  P_{1} 
b ≤ Q_{2}  P_{2} 
Following are the steps to summarize the approach.
1. Compute the optimal order quantity for the lowest price (highest discount), i.e.,
Q_{2}* =  (2DC_{o})  C_{h}P_{2} 
and compare the value of Q_{2}* with the quantity b which is
required to avail the discount.
If Q_{2}* ≥ b, then place orders
for quantities of size Q_{2}* and obtain discount; otherwise
move to step 2.
2. Compute Q_{1}* for price P_{1} and compare TC(Q_{1}*) with TC(b). The values of TC(Q_{1}*) and TC(b) may be determined as follows:
TC(Q_{1}*) =  DP_{1} + (D/Q_{1}*) X C_{o} + (Q_{1}*/2) X C_{h} X P_{1}  
TC(b) =  DP_{2} + (D/b) X C_{o} + (b/2) X C_{h} X P_{2} 
If TC(Q_{1}*) > TC(b), then place orders for quantities of size b to get the discount.
A big cold drinks company, the Piyo  Pilao Company, buys a large number of pallets every year, which it uses in the warehousing of its bottled products. A local vender has offered the following discount schedule for pallets:
Order Quantity  Unit Price (Rs.) 

Upto 699  10.00 
700 and above  9.00 
The average yearly replacement is 2000 pallets. The carrying costs are 12% of the average inventory and ordering cost per order is Rs. 100.
Solution.
Given
D = 2000 pallets/year, C_{h} = 0.12, C_{o} = Rs. 100,
P_{1} = Rs. 10, P_{2} = Rs. 9.00
The lowest price (highest discount) is RS. 9.00.
Q_{2}* =  (2 X 2000_{} X 100)  0.12 X 9 

= 608.58 pallets/order  
Since Q_{2}* < b (i.e., 608 < 700), Q_{2}* is not feasible.  
Step 2 

Q_{1}* =  (2 X 2000_{} X 100)  0.12 X 10 

= 577.35 pallets/order 
TC(Q_{1}*^{}) = TC(577.35) = 2000 X10 + (2000/577.35)
X 100 + (577.35/2 ) X 0.12 X 10
= Rs. 20692.82
TC(b) = TC (700) = 2000 X 9 + (2000/700) X 100 + (700/2) X 0.12 X 9
= Rs. 18663.71
Since TC(b) < TC(Q_{1}*) and hence the optimal order quantity is the price discount quantity, i.e., 700 units.
Order Quantity  Unit Price (Rs.) 

1 ≤ Q_{1} < b_{1}  P_{1} 
b_{1} ≤ Q_{2} < b_{2}  P_{2} 
b_{2} ≤ Q_{3}  P_{3} 
1. Compute the optimal order quantity for the lowest price (highest discount), i.e., Q_{3}* and compare it with b_{2}
2. Compute Q_{2}* and since Q_{3}* < b_{2}, this implies Q_{2}* is also less than b_{2}. Thus, either Q_{2}* < b_{1} or b_{1} ≤ Q_{2}* < b_{2}
3. Compute Q_{1}* and compare TC(b_{1}), TC(b_{2}) and TC(Q_{1}*) to determine the purchase quantity.
A large dairy firm, the Cow and Buffalo Company, buys bins every year, which it uses in the warehousing of its bottled products. A local vender has offered the following discount schedule for bins:
Order Quantity  Unit Price (Rs.) 

Upto 699  10.00 
700 to 949  9 
950 and above  8 
Solution.
Given
D = 2000 bins/year, C_{h} = 0.12, C_{o} = Rs. 100, P_{1} = Rs. 10, P_{2} = Rs. 9, P_{3} = Rs. 8
The lowest price (highest discount) is Rs. 8. Thus calculating Q_{3}* = corresponding to this range as follows:
Q_{3}* =  (2 X 2000_{} X 100)  0.12 X 8 

= 645.49 bins/order 
Since Q_{3}* < b_{2} (i.e., 645.49 < 950), go to step 2 to determine Q_{2}*
Step 2 

Q_{2}* =  (2 X 2000_{} X 100)  0.12 X 9 

= 608.58 bins/order 
Again, since Q_{2}* < b_{2} and b_{1} (i.e., 608.58 < 950 & 700) go to step 3 to calculate Q_{1}* and compare total inventory cost corresponding to Q_{1}*, b_{1} and b_{2}.
Step 3 

Q_{1}* =  (2 X 2000_{} X 100)  0.12 X 10 

= 577.35 bins/order 
TC(Q_{1}*^{}) = TC(577.35) = 2000 X10 + (2000/577.35)
X 100 + (577.35/2 ) X 0.12 X 10
= Rs. 20692.82
TC(b_{1}) = TC(700) = 2000 X 9 + (2000/700) X 100 + (700/2)
X 0.12 X 9
= Rs. 18663.71
TC(b_{2}) = TC(950) = 2000 X 8 + (2000/950) X 100 + (950/2)
X 0.12 X 8
= Rs. 16666.52
The lowest total inventory cost is TC(b_{2}) = Rs. 16666.52 and hence the optimal order quantity is the price discount quantity of 950 units, i.e., Q* = b_{2} = 950 units.