As we have already classified the economic order quantity or economic batch size problems into two types, based on the demand position.
Model I
Determination of the optimum quantity ordered (or
produced) and the optimum interval between successive orders (or production) if
the demand is known and uniform with no shortages permitted
An Illustration: Assume
that you are required (as a student) Rs. 100/- to meet your daily
requirements such as food, stay, refreshment etc. In a month, your demand is
Rs. 3000/-. Whenever you are contacting parents, they are in a position to give
your requirements in single stroke on the day itself. To get the money, you
have to go to your native which costs you say Rs. 50/- Instead of keeping the
money in your room, if you keep it in a bank, you may get an interest of
Rs.20/- month. If this is the situation given to you, how much you have to get
from your parents such that your holding/opportunity cost and
ordering/procurement cost is, get balanced?
Here the problem is how much you have to get your
parents every time so that the costs associated with holding the amount is get
balanced? Here the example which may be seems to obvious, instead, if we take a
car manufacturer, whose requirement is say 3000 Gear boxes per month, who is
producing 100 cars in a day, finds that, keeping a gear box in warehouse costs
him Rs. 200/- per month (excluding cost of gear box) and placing the order
would costs Rs. 500 per order. In this situation, how much gear boxes he has to
order, how much should be the order size so that the cost associated
such as ordering and holding get balanced and total costs are minimized?
If we summarize the assumptions, we are approaching
the problems through Economic Order Quantity or Economic Lot Size problems, or
in general known as EOQ problems.
Solution
Let us make the following
assumptions:
1. Let the
demand is known and uniform. Let D be the demand for a period say 1-year.
2.
Shortages
are not permitted. 3.
Let the
replenishment of items be instantaneous. 4.
Lead-time
is zero. 5.
Let Q be
the Economic Order Quantity for every cycle. 6.
Let Cs be
the Set up cost for every cycle 7.
Let C1 be
the inventory holding cost per unit per unit of time. Let us divide the one-year into n
equal parts each of duration‘t’ Therefore n * t = 1 or t = 1/n Let Q be the economic order
quantity for every cycle. The graph of this model is given
by,
Therefore n * Q = D or t = Q/D Inventory for the time period t = Area of the triangle Qot = ½ t* Q Therefore the Average inventory for one unit of
time = [½ t *Q]/ t = Q/2 Since all
the triangles are similar, the average inventory for the whole year = ½ Q Therefore the annual inventory holding cost = ½ Q *
C1 -------- (1)
And
Annual Set up cost | = | | | n * Cs | | | | ---------
(2) |
| = D/Q *
Cs | | | (since
n * Q = D)
| |
Therefore Annual Total Cost Ca = ½ Q * C1 + D/Q * Cs ---- | (3) |
In addition, C_{a} is a function on Q. For
maximization or minimization, the first derivative is found out and equated
with zero. If the second derivative is greater than zero indicates the function
attains its minimum value when the first
derivative is equated with zero.
Moreover, the second derivative
is strictly greater than zero. By substituting the value of Q, which has derived
just now, which is possessing the characteristic of balancing the ordering cost
and holding cost, in the equation for total cost function Ca, we get
Some Special Cases
Case 1 Let us
assume that the year we divided into n parts say, t1, t2, t3, etc., which are
not equal.
The total inventory over time t1 = ½ Q * t1 Total inventory over time t2 = ½ Q * t2 and so on. Therefore
total inventory over 1 year = ½ Q * [t1 + t2 + … +tn] = ½ Q * t And Average inventory = ½Q Therefore Annual inventory holding cost = ½ Q * C1 Since
annual set up cost is the same, we will get the same Ca hence; we will get the
same EOQ and other related issues. Therefore, E.O.Q = [2 * D * Cs / C1] ½ n
= D/Q And t = 1/n Case 2 Let the
Setup cost depend on the # of units that are being ordered or produced. Since the
set up cost for a cycle = Cs + Q * b, where ‘b’ is the cost of ordering one
unit. Therefore Ca = [½ * Q* C1] + D/Q (Cs + Q *b) And d (Ca)/ d Q = 0 implies, d (Ca)/ d Q = ½ C1 - DCs/ Q2 =0 Or D
* Cs / Q2 = ½ C1 Q2 = 2 * D * Cs / C1 Q = [2 * D * Cs / C1] ½ And, Ca =
½ C1 * [2 * D * Cs / C1] ½ + D* Cs *[C1 / 2D Cs] ½ + D* b
On
simplification, we get Ca = [2*D*Cs* C1] ½ + D* b
Note: The only change is addition of D*b term in the cost. Example-1 SMS Limited uses annually 24,000 Paper Boxes, which
costs Rs. 1.25/- per unit. Placing each order cost Rs. 22.50/- and the carrying
cost is 5.4% per year of the average inventory. Find the total cost including
the cost of Boxes. Solution D = 24,000 Cs = Rs. 22.50/- And C1 = 1.25 * 5.4% =
0.0675/Boxes/year. Therefore,
EOQ, Q = [2 * D * Cs / C1] ½ = 4000 Boxes And ‘n’ = D/Q = 24000/ 4000 = 6 i.e.
we have to make 6 orders t = Q/D = 4000/24000 = 0.16666
years or =
0.16666*
12 =
2 months Total Cost Ca, = [2 * D * Cs * C1] ½ =
270/- Total
Cost including cost of raw material = 270 + [24000 * 1.25] = 30270/- Example-2 M/s Shriram Industries has to supply 600 industrial
fans per year. The firm never permitted shortages to occur. Moreover, the
storage cost amounts to Rs. 0.60/ unit/ year. The set up cost per production
run is Rs. 80/-. Find the optimum order quantity, number of orders to place in
a year and average yearly cost. Solution From the given problem, it is clear that, Demand, D = 600 units/year Carrying cost, C1 = 0.60/unit/year Cost of set up, Cs = Rs. 80/-
Example-3Preethi Computers purchases 22,000 silicon chips
every year and each unit cost Rs. 22/-, as they are purchasing in bulk quantity
such a low price is possible. Cost of each order is Rs. 350/-. Its inventory
carrying cost is 18% of average inventory. What should be EOQ. What is the
optimum number of day’s supply for optimum order? What is the annual cost on
inventory including cost of the material? Solution From the given problem, it is clear that, Demand, | | D = 22000 units/year |
Cost of material, | C= Rs. 22 per unit | | |
Carrying cost, | C1 =
18% * (22) = 3.96 | | |
Cost of set up, | | Cs = Rs. 350/-
| | |
Example-4 PRG Engineering Company is a distributor for water
pumps. The company’s sales amount to 50,000 units of Water Pumps per year. The
order receiving/processing and handling cost are Rs. 3 per order while the
trucking cost is Rs. 12 per order. Further, the interest cost is Rs. 0.06 per
unit per year. Deterioration cost is Rs. 0.004 per unit per year. Storage cost
is Rs. 1000/year for 50000 units. Find the EOQ; also find the time between
orders and number of orders to be placed. Solution From the given problem, it is
clear that, Demand, D = 50000 pumps /year Carrying
cost, C1 = 0.006 + 0.004 + 1000/50000
= 0.084 Cost of
set up, Cs = Rs. 3+5= 15 per order
Time between any two successive
orders, t = 1/n = [1/11.83] =
0.08452
years or 1.01424 months =
30.427
days (approx)
Tags : Operations Management - Transportation / Assignment & Inventory Management
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