Optimal Order Quantity and Total Cost Calculation

Questions
Q1
Solution:
 Optimal order quantity: = √ 2 × D × K
H
K= Annual demand=10,000 units
H=Annual holding cost per unit = $3
D=Fixed cost per year= $504
Optimal order quantity= √ 2 ×504 ×10000
3
= 1833 units.
 Total cost = ordering cost + holding cost
= $ 504* number of orders per year( =5.456) +1833 *3
=$ 8248.82
 Number of orders/ frequency of orders = D
EOQ
= 10000
1833
= 5.456
 Time between orders = EOQ/ D = 1833/10000=0.1833 year
= 0.1833 *13
=2.19 months
Q2.

Solutions:
(a)The demands received in the months are shown in the table:
Months 1 2 3 4 5 6 7 8 9 10 11 12
Demand 300 700 800 900 3300 200 600 900 200 300 1000 800
Development of mathematical model:
Assuming:
Yi =1 ; denotes if an order is placed in ith month. (i.e. 0 for order is not placed and 1for order is
placed).
Xi = denotes the ordered quantity in ith month.
Ordering cost per order is: $504
Holding cost per order per month = $ 3
number of months (12)
= $ 1
4
Ii =denotes the ending inventory at the time of ith month.
The holding cost is applied to ending inventory.
The total cost is calculated as = ∑
i=1
n
504 Y i+ 1
4 (I 1+ I 2 + I 3 ... .. ... .+ I 12)
I12 is 0 because the at last inventory gets empty.
Now looking at constraints:
 X1 ≥ D1 means ordered quantity should be greater than the demand for first month.
Also I1 = X1 – D1 ; The orders left are equal to the ending inventory.
 X1 +X2 ≥ D1 + D2 means ordered quantity in first two months should be greater than
cumulative demand in the first two months. Also I2 = X1 + X2 – D1 – D2
Similarly there are 12 constraint equations
From these we can write : ∑
i=1
n
Xi ≥ ∑
i=1
n
Y i
Now there are 12 more constraints arises for relating Xi , Yi ; as
Xi ≤ M Yi (where M is a constant) this will for the order quantity to be zero in month when
there in no order placed;

Now on putting the values of ending inventories which we get from constraint on the total cost
function ;(i.e. I1 , I2 etc)
The total cost is calculated as =
∑
i=1
n
504 Y i+ 1
4 ( ( 12 X1 +11 X2 +10 X3 ... .. ... .+ X12 )− ( 12 D1 +11 D2 +10 D3 ... .. ... .+ D12 ))
The demand term is the constant term which we can remove for linear programming:
So minimize :
∑
i=1
n
504 Y i+ 1
4 ( ( 12 X1 +11 X2 +10 X3 ... .. ... .+ X12 ) )
The excel solver gives the result
Months Demand coefficentsConstraints Xi ≤ M Yi
1 300 1 1000 12 10000
2 700 0 0 11 0
3 800 1 1700 10 10000
4 900 0 0 9 0
5 3300 1 3300 8 10000
6 200 1 1700 7 10000
7 600 0 0 6 0
Yi Xi
(b) 1000 units should be ordered in the first month 1700 units should be ordered in the third
month.,3300 and 1700 should be ordered in the 5th and 6th month, 500 and 1800 should be
ordered in the,9th and 11th month.
The minimize total cost is $ 21249

(c ) When the holding cost is applied to average inventory.
Ending
inventories
I1 I2 I3 .. .. .. .. .. I12
Beginning
inventories
X1 I1+X2 I2+X3 .. .. .. .. .. I12+X1
Average
inventories
(X1+I1)/2 (X2+I1+ I2)/2 (X3+I2+ I3)/2 (X12+I11+ I12)/2
Now the total cost becomes:
total cost = 504*6 +(1000+350+1250+450+3310+1600+450+250+1700+400+1300+400)*(1/4)
=$17810