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Operations Research (MTH601)
68
ORDER QUANTITY WITH PRICE-BREAK
The concept of Economic Order Quantity fails in certain cases where there is a discount offered when purchases are
made in large quantities. Certain manufacturers offer reduced rate for items when a larger quantity is ordered. It may
appear that the inventory holding cost may increase if large quantities of items are ordered. But if the discount offered
is so attractive that it even outweighs the holding cost, the probably the order at levels other than the EOQ would be
economical. An illustration is given in the following example and the rationale is explained.
Example: A company uses 12000 items per year supplied ordinarily at a price of Rs. 3.00 per item. Carrying costs
are 16% of the value of the average inventory and the ordering costs are Rs. 20 per order. The supplier however
offers discounts as per the table below:
Order size
Price per item
Less than 2000
Rs. 3.00
2000 to 3999
Rs. 290
4000 or more
Rs. 2.85
Compute the economic order size.
2 12000 20
EOQ =
= 1000
3.00 0.16
EOQ at Rs. 2.90 per item
2 12000 20
=
= 1017
2.90 0.16
EOQ at Rs. 2.85 per item
2 12000 20
=
= 1026
2.85 0.16
The EOQ at Rs. 2.90 per item = 1017. But the price per item is Rs. 2.9 only if the items are ordered in the range of
2000 to 3999. This is therefore an infeasible solution. Similarly the EOQ at Rs. 2.85 per item is 1026. This price is
valid only for items ordered in the range 4000 or more. This is also an infeasible solution.
We follow the routine procedure and calculate the cost for various order sizes: 1000, 2000, 4000,
Order size
1000
2000
4000
Item cost (Rs.)
36000
34800
34200
Order cost (Rs.)
240
120
60
Holding cost (Rs.)
240
464
912
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Operations Research (MTH601)
69
Exercises
1.  Assume the following price structure
Units
Units Price
0-199
Rs. 10.00
200-399
9.75
400-599
9.50
600
9.25
Purchase cost per order
= Rs. 25
Cost of the item
= Rs. 10
Annual demand
= 950 Units
Carrying cost
= Rs. 2/Unit/year
2.  Find the optimal order quantity for a product for which the price-breaks are as follows:
Items
q
Price/Unit
0
Rs. 20
q < 100
100
Rs. 18
q < 200
200
Rs. 16
q
The monthly demand for the product is 600 units. The storage cost is 15% of unit cost and the cost
of ordering is Rs. 30 per order.
DYNAMIC ORDER QUANTITY
The basic assumption in the derivation of Economic Order Quantity models discussed previously is that the
demand is uniform. But in certain situations the demand is not uniform. It may rise and fall, depending on seasonal
influences. A general method is discussed below that can be applied to any pattern of varying demand due to
seasonal or irregular variations.
Consider the following example to illustrate how a varying demand problem can be tackled. This is known as
Dynamic Order Quantity model.
Example: The requirements for 12 months are given below:
Month
1
2
3
4
5
6
7
8
9
10
11
12
Requirement
20
40
10
10
10
2
40
30
40
40
10
20
Set up cost
= Rs. 20
Unit price
= Rs. 5 per item.
Interest
= 24% per year
or 2% per month.
Solution: To calculate the dynamic order quantity we can adopt the following procedure.
69
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Operations Research (MTH601)
70
The first month's requirement has to be ordered in the first month itself at a procurement cost of Rs. 20. Now we
have to decide whether the second month's requirement can also be ordered along with first month's requirement.
This involves additional carrying cost, but this will result in saving an extra set up. Hence if the saving on set up costs
outweighs the carrying costs, then we include the second month's requirements along with the first month. Similarly a
decision can be taken whether to include the third month's requirement in the first month itself. This procedure is
followed until the procurement costs and carrying costs are balanced.
Let n represent the month, n = 1, 2, ..., 12. Let Rn be the requirement during nth month and Rn+1 be the
requirement during (n + 1)th month. If the (n + 1)th month requirement namely Rn+1 is absorbed in nth month itself,
the additional procurement cost is saved.
Hence the saving in procurement cost is (C2/n). But the additional carrying cost is C3 n (Rn+1). If the
additional carrying cost is less than the procurement cost, then the (n+1)th month requirement is to be ordered also
with nth month.
Hence we have to check whether
n Rn + 1 C3 < C2/n
n2 Rn + 1 < C2/C3
If the answer to the above inequality is yes, then, the future month's requirements are included in the first month itself.
If the answer is 'no', then that requirement is to be ordered afresh and this is treated as month
n = 1. In this
example C2/C3 = 20/0.01 = 200. All the eleven information can be represented in the table as shown.
n2Rn+1.
n
Is
Month.
Requirement. Rn
2
n Rn+1<200
Action.
1
20
1
40
Yes
include 40 in month 1
2
40
2
40
Yes
include 10 in month 1
3
10
3
90
Yes
include 10 in month 1
4
10
4
160
Yes
include 10 in month 1
5
10
5
50
Yes
include 2 in month 1
6
2
6
1440
No
set up again in month 7
Total
92
7
40
1
30
Yes
include 10 in 7th month.
8
30
2
160
Yes
include 40 in 7th month.
9
40
3
360
No
set up again in month 10
Total
110
10
40
1
10
Yes
include 10 in month 10
11
10
2
80
Yes
include 20 in month 10
12
20
Total
70
Hence we order three times a year in the first month, seventh month and tenth month, the batch sizes being 92, 110
and 70 respectively.
Exercise: Compute the dyanamic EOQ is for the following requirements.
Month
1
2
3
4
5
6
7
8
9
10
11
12
Requirement.
50
100
10
170
150
180
1
260
100
80
150
200
70
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Operations Research (MTH601)
71
ABC ANALYSIS
The ABC analysis is the analysis that attracts management on those items where the greatest savings can be
expected. This is a simple but powerful tool of statistical sampling in the area of inventory control or materials
management.
In this analysis, the items are classified or categorized into three classes, A, B and C by their usage value. The usage
value is defined as,
The ABC concept is based on Pareto's law that few high usage value items constitute a major part of th capital
invested in inventories whereas a large number of items having low usage value constitute an insignificant part of the
capital. It too much inventory is kept, the ABC analysis can be performed on a sample. After obtaining the random
sample the following steps are carried out for the ABC analysis.
STEP 1: Compute the annual usage value for every item in the sample by multiplying the annual requirements by
the cost per unit.
STEP 2: Arrange the items in decending order of the usage value calculated above.
STEP 3: Make a cumulative total of the number of items and the usage value.
STEP 4: Convert the cumulative total of number of items and usage values into a percentage of their grand totals.
STEP 5: Draw a graph connecting cumulative % items and cumulative % usage value. The graph is divided
approximately into three segments, where the curve sharply changes its shape. This indicates the three segments A,
B and C.
The class A items whose usage values are higher are to be carefully watched and are under the strict and continued
scrutiny of the senior inventory control staff. These items should be issued only an indents sanctioned by the staff.
The class C items on the other extreme can be placed on the shop floor and the personnel can help themselves
without placing a formal requisition. The class B items fall in between A and C.
ABC concept conforms to the consideration implied in the EOQ model. 'A' items have high inventory carrying
costs and should therefore be placed with EOQ concept. The 'C' items require very little capital and have therefore
low inventory carrying costs. Hence, they can be purchased in bigger lots. 'B' items are usually placed under
statistical stock control.
Example: Perform ABC analysis on the following sample of 10 items from an inventory.
Items
1
2
3
4
5
6
7
8
9
10
Annual
Usage
300
2700
30
1000
50
220
160
800
600
70
(Unit)
Unit
Cost
10
15
10
5
4
100
5
5
15
10
(Rs.)
Solution:
71
Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION