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Operations Research

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Operations Research (MTH601)
72
STEP 1: Calculate usage value by multiplying annual usage of each item with its unit cost and tabulate them and
assign rank by giving rank to the largest usage value as in table below.
Items No.
Annual Usage
Ranking
1
3000
6
2
40500
1
3
300
9
4
5000
4
5
200
10
6
22000
2
7
800
7
8
4000
5
9
9000
3
10
700
8
STEP 2: Arrange items as per ranking and calculate cumulative usage and % cumulative value as in the table below.
% of cumulative
Cumulative %
Rank
Item No.
Annual usage
Cumulative
Annual usage
Annual usage
of item
1
2
40500
40500
47
10
2
6
22000
62500
73
20
3
9
9000
71500
84
30
4
4
5000
76500
91
40
5
8
4000
80500
94.2
50
6
1
3000
83500
98
60
7
7
800
84300
98.6
70
8
10
700
85000
99.4
80
9
3
300
85300
99.8
90
10
5
200
85500
100
100
If you draw the figure, you will see that the curve changes at points (say) X and Y. The items upto X is classified as
class A items and between X and Y as class B and the rest as class C items.
SOME DEFINITIONS
Lead-time: This is defined as the time interval between the placing of the orders and the actual receipt of goods.
Lead-time Demand: This is the lead-time multiplied by demand rate. For example, if the lead-time is 3 weeks and
the demand is at the rate of 50 items per week, then the lead-time demand is 3 x 50 = 150 items.
The lead-time may not be constant. For one batch, a vendor may take 45 days and for the next batch 50 days and so
on. Lead-time itself is therefore a stochastic variable. This complicates the problem of accumulating stock over the
period encompassed by the lead-time. Lead-time may also be forecast exponentially as is done with the demand.
Safety stock or Buffer stock: Lead-time demand is the stock level, which, on the average is sufficient to satisfy the
customer's orders as the stocks are being replenished. "On the average" would mean that during this period of
replenishment 50% of the customer's order can be filled and the remaining 50% may either be refused or back
ordered to be filled later. The reason for this is obvious. Forecast is after all a point estimate only. If the demand is
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Operations Research (MTH601)
73
greater than the forecasts, the customers would not be serviced. If the demand is less then the forecasts,
overstocking would occur. When these two variables, stock level and service to the customer, are summed up over
thousands of stock level and service to the customer, are becomes a major problem for an organisation to find an
acceptable compromise between the two. Sometimes the management in an organisation would like to limit the
disservice to the customer down to 5% or 10% at the cost of extra stocking. This extra stock in excess of the lead-
time demand is called the safety stock. Saftely stock may be expressed as percentages of the lead-time demand. It
may be computed in different ways.
Reorder level: This is defined as the level of the inventory at which the order is placed. It has generally two
components (i) Lead time Demand and (ii) Safety Stock.
Reorder level (ROL) = Lead time demand (LTD) + Safety stock (SS).
Computation of Safety Stock:
As discussed earlier, if the demand exceeds the forecast, the result is bad service to the customer and if the demand
is less than the forecast figure, this results in overstocking. Thus there is a forecast error. This error is assumed to be
normally distributed, with zero mean. If the standard deviation of the forecast error is calculated, then safety stock
may be set with the desired confidence level to result in not more the 5 or 10% shortages etc.
There is another measure of variation, popularly known as mean absolute deviation (MAD), which can be computed
far more easily. It can be routinely smoothed every period for obtaining better estimates of the safety stock.
MAD is related to the standard deviation for a normal distribution as given by.
MAD = 2 SD.
SD = 2MAD
Therefore,
Now safety stock = Z (S.D)
= Z 2MAD
= K MAD
where K is called service factor K = Z 2
The safety stock is for just one period and has to be extended over the lead time. The necessary formula has been
derived by extensive simulation by statisticians and is given below.
MADLT = (0.659+ 0.341 LT) MAD
Safety stock over the lead-time
= K (0.659 + 0.341 LT ) MAD
= Z 2 (0.659 + 0.34 LT )MAD
Example: A company uses annually 50000 units of raw materials at a cost Rs. 1.2 per item. Ordering cost of items is
Rs. 45 per order and item carrying cost is 15% per year of the average inventory.
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74
1.
Find the economic quantity.
2.
Suppose that the company follows the EOQ policy and it operates for 300 days a year, that the
procurement time is 12 days and mum, minimum and average inventories.
Solution:
D = Demand = 50000/year
No. of days = 300/year
C1 = Rs. 1.2 per item
Lead time = 12 days
C2 = Rs. 45 per order
Safety stock = 500
C3 = 15% of Rs. 1.2/item/year.
= Rs. 0.18 per item per year.
2C2D
EOQ =
(1)
C3
24550000
=
0.18
= 25000000
= 5000
(2)
Requirement per day = 50000/300
Lead time demand = 12 x 500/3 = 2000
Safety stock
= 500
Reorder level
= Lead time demand + safety stock
= 2000 + 500 = 2500
Maximum inventory = E.O.Q + Safety stock
= 5000 + 500 = 5500
Minimum inventory = Safety stock = 500
Average inventory
= (Max. Inv. + Min. Inv.)/2
= (5500 + 500)/2 = 3000
Example A scrutiny of past records gives the following distributions for lead time and daily demand during lead time.
Lead time distribution
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Operations Research (MTH601)
75
Lead time (days)
3
4
5
6
7
8
9
10
Frequency
2
3
4
4
2
2
2
1
Demand distribution
Demand/day (units)
0
1
2
3
4
5
6
7
Frequency
2
4
5
5
4
2
1
2
What should be the buffer stock?
Solution:
Computation of average or mean lead time.
(Frequency lead time)
=
Frequency
6+12+20+24+14+6+18+10
=
20
= 6 days
Average demand
(Frequency demand/day)
=
frequency
0+4+10+15+16+10+6+14
=
25
=3
Average lead-time demand
= Average lead time x Average demand/day
= 6 x 3 = 18.
Maximum lead-time demand.
= Max. lead time x Max. demand/day
= 10 x 7 = 70
Buffer stock = Max. lead time demand - Average lead time demand
*Example 7.7.3
For a fixed order quantity system find the EOQ, SS, ROL and average inventory for an item with the
following data.
Demand = 10000 units
Cost of item = Re. 1
Order cost = Rs. 12
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76
Holding cost = 24%
Post lead times = 13 days
Solution:
2C2D
EOQ =
(1)
C3
21210000
=
= 1000 items
0.241
(2)
The Average lead time = (13 + 14 + 15 + 16 + 17)/5
= 15 days ( 30 days time is omitted)
Maximum lead time = 30 days
Safety stock = ( 30 - 15) x 10000/(30 x 12) = 420
(3)
Reorder level = Lead time demand + Safety stock
= 417 + 420 = 837
(4)
Average inventory = (1420 + 420)/2
= 1840/2 = 920.
Example: An airline has determined that 10 spare brake cylinders will give them stock out risk of 30%, whereas 14
will reduce the risk to 15% and 16 to 10%. It takes 3 months to receive items from supplier and the airline has an
average of 4 cylinders per month. At what stock level should they reorder assuming that they wish to maintain an
85% service level.
Solution:
Lead time demand = 3 x 4 = 12 items
Safety stock at 85% service
= 15% disservice
or 15% stock out risk
= 14 items
Reorder level = 12 + 14 = 26 items.
Example: Data on the distribution of lead time for a motor component were collected as shown. Management would
like to set safety stock levels that will limit the stock out to 10%.
Lead time (weeks)
1
2
3
4
5
6
7
8
Frequency of occurrence
10
20
70
40
30
10
10
10
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77
How many weeks of safety stock are required to provide the desired service level?
Solution:
Lead time (weeks)
Frequency
Probability
Cumulative
Probability
1
10
0.05
0.05
2
20
0.10
0.15
3
70
0.35
0.50
4
40
0.20
0.70
5
30
0.15
0.85
6
10
0.05
0.90
7
10
0.05
0.95
8
10
0.05
1.00
200
Average lead time = (lead time x frequency)/ frequency
= (10 + 40 + 210 + 160 + 150 + 60 + 70 + 80)/200
= 770/200 = 3.85 weeks
Upto 90% service level max. lead time is 6 weeks.
Hence 6 - 3.85 = 2.15 weeks of stock would provide the service level of 90%
(Note: If the lead time is given as a continuous time distribution take the mid point.)
Example: Demand for a product during an order period is assumed to be normally distributed with mean of 1000
units and standard deviation of 40 units. What % service can a company expect to provide (i) if it satisfies the
average demand only (ii) if it carries a safety stock of 60 units.
Solution:
1.
If the company provides only average demand, we can expect only 50% service level.
2.
The standard normal variate Z is computed with the following formula.
Z = (Safety stock - 0)/ Standard deviation
= (60 - 0)/40 = 1.5
The area under normal curve for Z = 1.5 is 0.4332.
Service level = 0.50 + 0.4332 = 0.9332
04 93.32%
Example: A manufacturer of water filters purchases components in EOQ's of 850 units/order. Total demand
averages 12000 components per year and MAD = 32 units per month. If the manufacturer carries a safety stock of 80
units, what service level does the this give the firm?
Solution:
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2 MAD = MAD / 0.8 = 32 / 0.8 = 40
Standard deviation =
Z = (S S - 0)/S.D = (80 - 0)/40 = 2
The area under normal curve for Z = 2 = 0.4772.
Service level = 0.9772 or 97.72%
Example: A firm has normally distributed forecast of usage with MAD = 60 units. It desires a service level, which
limits the stock, outs to one order cycle per year.
(1) How much safety stock should be carried if the order quantity is normally a week's supply?
(2) How much safety stock should be carried if the order quantity is weeks supply.
Solution:
No. of orders = 52/year.
1 stock out in 52 weeks means a 98 % = (51/52) values
For 98% area, the value of
Z = 2.05 (from tables)
S D = MAD/0.8 = 60/08
Z = (S S - 0 )/S D= (S S - 0)/60/0.8
2.05 = (S S - 0 ) x 0.8/60
SS = 2.05 x 60/0.8 = 154 units.
(a) Number of orders = 52/5
1 stock out = (52/5 - 1)
Service level = (47/5)/(52/5) = 47/52 = 0.904
Z = 1.285 (for area = 0.404) MAD = 60
1.285 = S S/75
S S = 75 x 1.285 = 96 units.
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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