chapter 1 introduction - university of agriculture ... like an economic balance between the cost of...

Chapter – 1

INTRODUCTION

1Prof. Dr. Muhammad Iqbal, Department of

Farm Machinery & Power, University of Agriculture, Faisalabad

Introduction

• Before the advent of industrial revolution in Europe, business organization and industrial enterprises were very small in size.

• There was no management problem of considerable magnitude as we face them today.

• Each enterprise had a single boss who could direct all the activities of business single – handed.

• The boss used to do purchasing, plan and supervise production, sell the product, hire and fire the personnel etc.



• The rapid scientific invention led to the mechanization of the production with the result that pace of growth in magnitude of industrial enterprise was so great that it became impossible for one man to perform all the managerial functions.

• A division of management function with a reasonable head of each division took place.

• The ever – increasing spiral of continued mechanization supported by automation resulted in still faster industrial growth.

• Faster industrial growth necessitated further decentralization of operation and division of the management function.



• The decentralization process resulted in “Executive

Type Problems”.

• The special type characteristics of “Executive Type

Problems” needed organized activity.

• For smooth and efficient working of an organization,

each function unit (division, department) had to

perform its function as a single unit.

• Each part had to perform its functions efficiently in a

manner so as to accomplish the overall objectives of

the entire organization.

• Each functional units had to develop objectives of its

own.



• These objectives are not often consistent. Frequently they

come into direct conflict with one another.

• A policy, which is most favorable to one department, may

not be favorable for the other.

• The questions then arise, what is the best policy for the

whole organization as one single unit?

This is the “executive type problem” having the

following characteristics:



Executive type problem characteristics:

• To divide and sub – divide functions for the

effectiveness of the organization as a whole.

• Solve conflicting interest of the functional units of the

organization.

A scientific base for solving problems involving the

interaction of components of the organization in the

best interest of the organization as a whole was

required. The development of “Operations

Research” has greatly solved this problem.



Historical background & development• Operations research (OR) is derived from that of

Operational Radar

• The optional research had its origin in the military activities

of the 2nd world war.

• There was an urgent need to allocate scares resources to the

various military operational activities within each operation

in an effective manner.

• The birth of the OR began with the assignment to the

teachers of university educated personal in United Kingdom

from all areas of science to assist the military in solving the

management problems of using their limited resources.



Problems facing the English people were:

1. How to face German air raids with the limited equipment

2. How to fight the U – boat attacks

3. How to deploy the armed forces in the most efficient manner

• Great faith was placed in the scientific method and the research approach.

• Different teams were constituted. The most prominent team was headed by Professor P.M.S Blcakett. The work of the team dealt essentially with the problems of antiaircraft defense, especially in the London area. The studies of the group were devoted to the operational use of Radar, the activity then came to be known as “OPERATIONAL RESEARCH”.



• The early work of United Kingdom operational researchers was later supported by U.S.A, who latter joined the war.

• Armed forces were successful in using the operations research personal in such activities:

– Developing strategies for mining operations

– Improving bombing accuracy

– Determining the best research patterns for submarines.

• After the war, operations research groups were maintained in all branches of the U.S. forces. Interest in applying the operations research approach in military activities grew tremendously.



The brief history in developments of OR:

• 1950 – Operations research approach began to be taken

seriously by American industry. Numerous industrial

and business applications have been made since then.

• 1953 – American society of operational research

(ASOR) came into being.

• The success of OR in military and non – military

activities caused not only the formation of the American

Society of Operations Research (ASOR), but also the

development of educational program in many

universities throughout the United States (in

engineering colleges, through their industrial and

systems engineering departments)



• 1957 – The American Management Association

conducted a study among 631 companies about their

involvement with OR.

324 were then using OR techniques

144 out were considering OR adoption in the future

The left over did not still think about OR



• The OR societies have been developed in Argentina,

Australia, Belgium, Brazil, Canada, Denmark,

France, Germany, Greece, Ireland, Israel, Italy, Japan,

Mexico, The Netherlands, Norway, Spain Sweden,

Switzerland, and Thailand.

• 1959 – Industrial federations of operations research

societies (IFORS) came into being. Its main objective

was to develop operational research as a unified

science and to advance it in all nations of the world.



What is operations research?

• According to he professor Arnold Kaufmanns OR is The

body of tools and methods which makes possible strong

confidence upon the rational determination of the most

efficient and economical solution in policy – decision

problems concerning the management of economic or

human phenomenon, drawing upon statistical,

mathematical procedure which sometimes require the

use of high – speed computers.



• Operations research can also be defined as: “the

application of scientific methods to deal with

management so as to provide executives in

controlling system with a sound quantitative basis for

making decision regarding such problems.

• Operations research is research on problems for

optimality



Application of the course

• Problems may be considered from two aspects, “Type” and “Contents”.

– Executives, in general, tend to look at problems according to the area of the business in which the problems occur, that is, their content.

– On the other hand the type to which problems belong has to do with their underlying logical structure. Almost all the decision problems that a manager faces will appear as one of a few different types.



Types of problems

• Management decision problems that can be analyzed by operations Research methodology include the following six types:1. Allocation

2. Inventory

3. Replacement

4. Waiting line

5. Competition

6. Sequencing



1. Allocation Problems

• A larger number of important decision problems

involve the allocation of scare resources to various

activities in such a way so as to optimize the

objective function. They can arise whenever one has

to select the level of certain activities that must

complete for certain scarce resources necessary to

perform those activities.

• Techniques usually used in solving allocation

problem include; Linear Programming (LP), Dynamic

Programming (DP) and other kinds of Non – Linear

Programming.



2. Inventory problems

• Inventory analysis deals with the control and

maintenance of quantities of useable but idle resources

like; Material, Manpower, Money, or Water in a

reservoir.

• The decision problem concerned with the frequency of

useful addition and quantity acquired so that minimum

total cost can be attained.

– The total cost is basically the sum of holding cost,

ordering cost, stock cost, and shortage costs

• Optimum decision rules for ordering policies may be

obtained by such methods as calculus, probability

theory, and dynamic programming.



3. Replacement problems

• Replacement of aging facilities (for example old tractor

from the Agriculture department). There are generally two

types of such facilities; items that deteriorate during use

and items that suddenly fail, for example; 12-hp Honda

tractor failed in Pakistan.

• Replacing after economic use (machinery and vehicles).

The calculus method and dynamic programming helps in

solving such problems.

• Replacing after sudden failure (Tires, Electronic

Computers, and Light Bulb in a football stadium).

Statistical sampling techniques are usually used to solve

this type of replacement problem.19

Prof. Dr. Muhammad Iqbal, Department of Farm Machinery & Power, University of

Agriculture, Faisalabad

4. Waiting Line Problems

• QUEUING theory problems. Problems of planning service facilities to

meet the demand of the customer that arrive in a random manner.

• This situation is a characteristic of sales customers, doctor’s

offices, Maintenance & Repair of machines and similar

problem areas.

• If the service is inadequate, congestion is unavoidable.

Associated with it are certain costs, such as the one of losing

customers. However, an increase of service facilities usually

increases the idle time of servers, and thus incurs a cost of

idleness.

• The decision problem is to determine the optimal number of service

facilities like an economic balance between the cost of service and the

cost of waiting is obtained.

• Probability theory and differential difference equations are extensively

used in the analysis of waiting line problems.



5. Competition problems

Problems of making decisions between two or more individuals or

Organizations in situation that involve a conflict of interest.

1. Decision making under certainty (DMUC)- The competitors

course of action is known in advance with certainty.

2. Decision making under risk (Intuitive approach)- The

competitors choice is not known with certainty but can be

predicted subject to error.

3. Decision making under uncertainty (DMUU)- Nothing is known

in advance concerning the alternatives of competitors but previous

sales data can help for probability evaluations.

The application of statistical decision theory has proved to be

successful in most situations, both simple and complex.



6- Sequencing problems

• Determining the optimal sequence to perform a set of jobs

or activities in order to minimize the time required. In

general, the number of alternative to be considered is so

high that it will not be practical to examine them

completely. In view of such time consuming matter, most

sequencing problems are approached by the Simulation

Method.

• To solve such problems Program Evaluation Review

Technique (PERT) and Critical Path Method (CPM) are

used.



The Methodology of Operations Research

Analyze the Situation / observations

Specify the Problem

Formulate a Hypothesis by a Model

Evaluate the Model

Test the Model by Experimentation

Compare the Model Result with the Real System



Chapter-2

DECISION THEORY

1Prof. Dr. Muhammad Iqbal, department of


Decision Making Under Uncertainty

• Every human being / business man is uncertain about

his future and yet he plans on certain assumptions and

expectations.

• Every body knows that business organizations do not

operate under conditions of certainty but most of the

decisions are made with the help of past experience

and subjective judgment of the decision maker along

with the analysis of quantitative data with in and

outside the business organization.

• Mathematical theory of the probability is of great

help in arriving at a suitable



Objective and Subjective Probabilities

Objective probability-Take the example of a coin. If

a coin is tossed the probability of a head is p=0.5 and

the probability of a tails is also q=0.5, as there can be

only two possible events. The assumption in this case

is that the coin is fair.

Subjective probability- A quite different situation

may arise when the historical information is not

available. Personal experience and subjective

judgment are than made the basis of probability

assignments. Such probabilities are called subjective

probabilities



Conditional and expected values

Assume that expected monetary value (EMV) is the suitable criterion

for decision making.

Example:

• Suppose a grocer has to decide for how many crates of tomatoes

should be ordered in order to meet tomorrows demand. The selling

and purchase prices per crate are Rs. 15/- and Rs. 12/- respectively.

Since the product is perishable, therefore, it is assumed that unsold

crates do not bring any salvage value. The future demand and the

corresponding action to meet the demand are both uncertain. The

grocer has maintained the past record and the historical data about the

demand in the past as in given in the following table-1



Table 1. Historical record of sales of tomato

Events (crates) Frequency (days) Probability

15 30 30/300 = 0.1

16 90 90/300 = 0.3

17 120 120/300 = 0.4

18 60 60/300 = 0.2

Total 300 300/300 = 1.0



Conditional values/profit

• The grower has a free choice of purchasing 15, 16, 17, or

18 crates. Each choice is denoted as an action and each

demand an event.

• Corresponding to each action there is a possibility of

occurrence of any one of the events resulting in certain

conditional +ve or –ve profit.

• The profits are conditional because corresponding to any

specific purchase action any sale event may occur

resulting in a profit or loss conditional upon the

occurrence of that sales event. All the possible

combination of events and actions are shown in the

following Table-2 6

Prof. Dr. Muhammad Iqbal, department of Farm Machinery & Power, University of


Table – 2 Conditional values/profit

Event

(Demand)

Action (Crate)

Stock 15 Stock 16 Stock 17 Stock 18

15 crate 45 33 21 9

16 crate 45 48 36 24

17 crate 45 48 51 39

18 crate 45 48 51 54



Table – 3 conditional opportunity and unsold loss

Event

(Demand)

crates

Opportunity and unsold loss (Rs.)

Action (Crate)

Stock 15 Stock 16 Stock 17 Stock 18

150

(16 – 15) * 12

= Rs. 12

(17 – 15) *

12 = Rs. 24

(18 – 15) *

12 = Rs. 36

16 (15 – 12) * 1

= Rs. 30

(17 – 16) *

12 = Rs. 12

(18 – 16) *

12 = Rs. 24

17 (15 – 12) * 2

= Rs. 6

(15 – 12) * 1

= Rs. 30

(18 – 17) *

12 = Rs. 12

18 (15 – 12) * 3

= Rs. 9

(15 – 12) * 2

= Rs. 6

(15 – 12) * 1

= Rs. 3

0



Table – 4 Expected Monetary Values (EMV)

Expected Monetary Values (EMV) of each action

Event (p) Act. 15 Act. 16 Act. 17 Act. 18

CV EMV CV EMV CV EMV CV EMV

15 0.1 45 4.5 33 3.3 21 2.1 9 0.9

16 0.3 45 13.5 48 14.4 36 10.8 24 7.2

17 0.4 45 18.0 48 19.2 51 20.4 39 15.6

18 0.2 45 9.0 48 9.6 51 10.2 54 10.8

Total EMV 45.0 46.5 43.5 34.5



Table 5. Expected opportunity and unsold loss of

each action

Expected opportunity and unsold loss of each action

Eve

nt

(p) Act. 15 Act. 16 Act. 17 Act. 18

CL EOL CL EOL CL EOL CL EOL

15 0.1 0 0 12 1.2 24 2.4 36 3.6

16 0.3 3 0.9 0 0 12 3.6 24 7.2

17 0.4 6 2.4 3 1.2 0 0 12 4.8

18 0.2 9 1.8 6 1.2 3 0.6 0 0

Total EOL 5.1 3.6 6.6 15.610



Steps in brief in solving such problems:

1. Prepare a conditional value table

2. Assign respective probabilities to each event

3. Calculate EMV for each decision by p * CV and add

them up

4. Select the action which gives the highest EMV



Table 6. Expected profit under uncertainty

(EPUC)

Event P Conditional

profit under

certainty

Expected profit

under certainty

15 0.1 Rs. 3 * 15 =

Rs. 45

0.1 * 45 = Rs. 45.0

16 0.3 Rs. 3 * 16 =

Rs. 48

0.3 * 48 = Rs. 14.9

17 0.4 Rs. 3 * 17 =

Rs. 51

0.4 * 51 = Rs. 20.4

18 0.2 Rs. 3 * 18 =

Rs. 54

0.2 * 54 = Rs. 10.8

Total expected profit under certainty = 50.1 12Prof. Dr. Muhammad Iqbal, department of


Table 7. Evaluation of Actions

Decision Stock 15 Stock 16 Stock 17 Stock 18

EMV 45.0 46.5 43.5 34.5

EOL 5.1 3.6 6.6 15.6

EPUC 50.1 50.1 50.1 50.1



Expected value of perfect information (EVPI)

• The best action was to stock 16 and earn an expected profit of

Rs. 46.5. But if perfect information is available the expected

earning could be increased to Rs. 50.1. Therefore, the

maximum that the grocer can pay for the perfect information is

EOL = Rs. 3.6, Hence EVPI = Rs. 3.6

• Thus EVPI = EVUC – EMV of optimum action

• Or EVUC = EVPI + EMV of optimum action.

= EOL + EMV of optimum action.

Conclusion: the grocer would prefer to get the perfect

information so long as the cost of obtaining the perfect

information is less than Rs. 3.6



Marginal profit approach

• The effect of adding one more unit on profit is the basis of

marginal approach analysis. So long as the profit is increasing,

more and more units are added till we reach turning point

where addition of more unit decreases the profit. This is the

point of maximum profit which is selected the best solution of

the problem.

• The marginal unit is, therefore, the last additional unit bought.

There are two possibilities if one additional unit is purchased.

It may be sold or it may not be sold but held in stock. The sum

of probabilities of these two events must be equal to 1.0



• Let p = probability of selling this additional unit (Success)

• Then 1-p = probability of not selling it (Failure)

• If the additional unit is sold the profit earned from it will

be the marginal profit (MP) and if it not sold the loss will

be the marginal loss (ML).

• Therefore expected marginal profit from selling the

additional unit, EMP = P * (MP)

• And expected marginal loss from not selling this unit,

EML = (1 – p ) * (ML)

• The businessman will, therefore, continue to add more

units in the stock so long as the expected increase in profit

is greater than expected loss due to units unsold.



• Or mathematically,

EMP > EML

• Or p * (MP) > (1 – p) * (ML)

• Businessman will continue adding one additional unit

till the additional profit is exactly equal to loss if it is

not sold. Hence net profit will be maximum, when

• p * (MP) = (1 – p) * (ML)

• or p * (MP) = ML – p * (ML)

• or p (MP + ML) = ML

• or Pc = ML/ (ML + MP), critical probability



Rules to achieve the optimum point:

• So long as the probability of selling one or more units

is greater than the Pc, we should order that unit.

• When the probability of selling the additional unit is

equal to Pc, we should be indifferent to include or not

to include the additional unit.

• If the probability of selling the additional unit is less

than Pc the additional unit should not be ordered.



Example:

• Let in previous example each unsold crate can be sold for

Rs. 10/-

• So the data become like this:

• Purchase price per crate = Rs. 12/-

• Sale price per crate = Rs. 15/-

• Salvage value per crate = Rs. 10/-

• Applying the principle of marginal approach we see that:

• MP = Rs. 15 – 12 = Rs. 3.0 per unit

• ML = Rs. 12 - 10 = Rs. 2/- per unit

• Pc = ML/(ML+MP)= 2/(2+3) = 2/5 = 0.4



Table. Probability, cumulative probability,

and absolute certainty of different Events

Demand

(Crates)

Probability Cumulative

probability

Absolute

Certainty

15 0.1 1.0 100%

16 0.3 0.9 90%

17 0.4 0.6 60%

18 0.2 0.2 20%



• As pc = 0.4 we can maximize profit by purchasing an additional unit so

long as we are sure by 40% (Absolute certainty = 40%) or above of selling

that unit. The cumulative probability column shows that the optional

solution is 17 crates. Let us look at the following calculations:

• Expected monetary value on selling 15 crates:

= cumulative probability * (MP) * 15 Crates

= 1.0 * (Rs. 15-12) * 15 = Rs. 45/-

• Increasing from 15 to 16 crates (adding 1-unit) we have

E (MP) = cumulative probability of 16 crates ‘p’ * (MP) * 1 crate

= 0.9 * (Rs. 15-12) = Rs. 2.7/-

E (ML) = (1 – p) * (ML) * 1 crate

= (1-0.9) * (Rs. 12-10) = RS. 0.2/-

• Therefore,

Net expected marginal profit on 16th crate = Rs. 2.7-0.2 = Rs. 2.5/-

Total EMV of 16 crates = Rs. 45 + Rs. 2.5 = Rs. 47.5/-





= 0.6 * (Rs. 15-12) = Rs. 1.8/-

E (ML) = (1 – p) * (ML) * 1 crate

= (1-0.6) * (Rs. 12-10) = RS. 0.8/-

• Therefore,

Net expected marginal profit on 17th crate = Rs. 1.8-0.8 = Rs. 1.0/-

Total EMV of 17 crates = Rs. 47.5 + Rs. 1.0 = Rs. 48.5/-





= 0.2 * (Rs. 15-12) = Rs. 0.6/-

E (ML) = (1 – p) * (ML) * 1 crate

= (1-0.2) * (Rs. 12-10) = RS. 1.6/-

• Therefore,

Net expected marginal profit on 18th crate = Rs. 0.6-1.6 = - Rs.

1.0/-

Total EMV of 18 crates = Rs. 48.5 - Rs. 1.0 = Rs. 47.5/-

The addition of 18th crate has reduced the net

expected monetary value; therefore, the grocer

should stock 17 crates for optimum profit.



Using the Standard Normal Probability Distribution

• We can use the idea of the standard normal probability distribution to solve a decision theory problem employing a continuous distribution.

• Assume that a manager sells an article having normally distributed sales with a mean of 50 units daily and a standard deviation in daily sales of 15 units. The manager purchases this article for Rs.4 per unit and sells it for Rs.9 per unit. If the article is not sold on the selling day, it is worth nothing. Using the marginal method of calculating optimum inventory purchases levels, we can calculate our required p:


Agriculture, Faisalabad24

This means that the manager must be 0.44 sure of selling at least an additional unit before it would pay to stock that unit. Let us reproduce the curve of past sales and determine how to incorporate the marginal method with continuous distributions of past daily sales.

Now refer to figure. If we erect a vertical line b at 50 units, the area under the curve to the right of this line is one half the total area. This tells us that the probability of selling 50 or more units is 0.5. The area to the right of any such vertical line represents the probability of selling that quantity or more. As the area to the right of any vertical line decreases, so does the probability that we will sell that quantity or more.



MLMP

ML

44.04.5.

4.

RsRs

RsP =

μ=50

• Suppose that manager considers stocking 25 units, line a. Most of the entire area under the curve lies to the right of the vertical line drawn at 25; thus, the probability is greater that the manager will sell 25 units or more. If he considers stocking 50 units (the mean), one half the entire area under the curve lies to the right of vertical line b; thus he is 0.5 sure of selling the 50 units or more. Now, say he considers stocking 65 units. Only a small portion of the entire area under the curve lies to the right of line c; thus the probability of selling 65 or more units is quite small. The 0.44 probability that must exist before it pays our manager to stock another unit. He will stock additional units until he reaches point Q. If he stocks a larger quantity, the shaded area under the curve drops below the 0.44 and the probability of selling another unit or more falls below the required 0.44. How can we locate the point Q?



We can use Appendix Table 1 to determine how many standard deviations it takes to include any portion of the area under the curve measuring from the mean to any points such as Q. In this particular case, since we know that the shaded area must be 0.44 of the total area, and then the area from the mean to point Q must be 0.06 (the total area from the mean to the right tail is 0.50).



• Looking in the body of the table, we find that 0.06 of the area under the curve is located between the mean and a point 0.15 standard deviations to the right of the mean. Thus we know that point Q is 0.15 standard deviations to the right of the mean (50). We have been given the information that 1 standard deviation for this distribution is 15 units; so 0.15 times this would be 2.25 units. Since point Q is 2.25 units to the right of the mean (50), it must be at about 52 units. This is the optimum order for the manager to place; 52 units per day.

•

•



Q1.Highway road construction in Islamabad is concentrated in the months from May through September. To provide some measure of protection to the crews at work on the highways, the Department of Transportation (DOT) requires that large, orange MEN WORKING signs be placed well in advance of any construction. Because of vandalism, wear and tear, and theft, each year DOT purchases a number of new signs to be put into service. The signs are actually made under the auspices of the Department of Corrections, but because of interdepartmental budgeting and accounting, DOT is charged a price equivalent to one it might pay were it to buy the sign from an outside source. The interdepartmental charge for the signs is Rs.110 if more than 30 of the same kind are ordered. Otherwise, the cost per sign is Rs.150. Because of budget pressures, DOT attempts to minimize its costs both by not buying too many signs and by attempting to buy in sufficiently large quantity to get the Rs.110 price. In recent years, the department has averaged purchases of 84 signs per year with a standard deviation of 11. Determine the number of signs DOT should purchase.



Q2. Nestle Milk Pack Lahore, is preparing for the celebration of the 11th Annual Milk and Dairy Day. As a fund-raising device, the management once again plans to sell souvenir T-shirts. The T-shirts, printed in 6 colors, will have a picture of a cow and the words, “11th Annual Milk and Dairy Day,” on the front. The management purchases heat transfer from a supplier for Rs.75 and plain white cotton T-shirts for Rs.150. A local merchant supplies the appropriate heating device and also purchases all unsold white cotton T-shirts. The management plans to set up a booth on Main Street and sell the shirts for Rs.325. The transfer of the color to the shirt will be completed when the sale is made. In the past year, similar shirt sales have averaged 200 with a deviation of 34. The management knows that there will be no market for the patches after the celebration. How many patches should the management buy?



Q3. Bike wholesale parts was established in 1970s in response to demands of several small and newly established bicycle shops that needed access to a wide variety of inventory but were not able to finance it themselves. The company carries a wide variety of replacement parts and accessories but does not maintain any stock of completed bicycles. Management is preparing to order 27” x ” rims from the BECO company in anticipation of a business upturn expected in about 2 months. BECO makes a superior product, but the lead time required necessitates that wholesalers make only one order, which must last through the critical summer months. In the past, Bike Wholesale Parts have sold an average of 120 rims per summer with a standard deviation of 28. The company expects that its stock of rims will be depleted by the time the new order arrives. Bike Wholesale Parts have been quite successful and plans to move its operations to a larger plant during the winter. Management feels that the combined cost of moving some items such as rims and the existing cost of financing them is at least equal to the firm’s purchase cost of Rs. 730. Accepting the hypothesis that any unsold rim at the end of the summer season are permanently unsold; determine the number of rims the company should order if the selling price is Rs.810.



Q4. The B & G cafeteria features barbecued chicken each Thursday. The special has become a popular item, and cafeteria manager Amjad wants to ensure that the cafeteria will make money on the special in the long run. Including labor, each portion of chicken costs the cafeteria Rs. 95 to prepare and sell. The customers pay Rs. 120 and consider it a good deal. Data taken from last year indicate the barbecued chicken sales are normally distributed, with mean 160 and standard deviation 23. If B & G Cafeteria converts one chicken into two portions of barbecued chicken, how many chickens should be prepared each Thursday?



Q5.Paige’s Tire Service stocks two types of radial tires: polyester-belted and steel-belted. The polyester-belted radials cost the company Rs.3000 each and sell for Rs.3500. The steel belted radials cost the company Rs.4500 and sell for Rs.6000. For various reasons, Paige’s Tire Service will not be able to reorder any radials from the factory this year, so they must order just once to satisfy customer’s demand for the entire year. At the end of the year owing to new tire models, Paige will have to sell all its inventory of radials for scrap rubber at Rs.500 each. The annual sales of both types of radial tires are normally distributed with mean and standard deviation indicated below.

• How many polyester- belted radials should be ordered?

• How many steel-belted radials should be ordered?



Annual

Mean Sales

Standard

Deviation

MP ML

Polyester – Belted 300 50 500 2500

Steel – Belted 200 20 1500 4000

Q6.Plain Games, a toy and hobby shop, is getting ready to make its final order of toys and games, which will arrive exactly 4 weeks before Eid-Ul-Fitar. The manager is most concerned about a line of toys produced by the North Pole Company--- they are decorated to be very Eid’s-oriented and will be worth almost nothing to the store after Eid. Each such toy sells for Rs.600, but will be sold to a discount retailer for only Rs.100 after the holiday season. Plain Games pays the factory Rs.400 for each toy. The manager knows from past experience that the demand for North Pole toys during the 4 weeks before Eid is normal distributed, with mean 600 and standard deviation 120. How many North Pole toys should be stocked for the last 4 weeks of the Eid buying season?



INVENTORY MANAGEMENT



Inventory?A stock or store of goods (Investments in business)

Business type Examples of inventory

Manufacturing firm

Raw material, purchased parts,Partially finished goods,finished goods, spare parts of machines, tools etc.

Departmental store

Clothing, furniture, stationary,gifts cards, toys, cosmetics etc.

Hospital Drugs, surgical supplies, life monitoring equipment, sheets, pillows etc.

Supermarket Fresh, canned and frozen goods, Magazines, dairy products etc.



Types of Inventories ?• Raw materials & purchased parts

• Partially completed goods called work in progress

• Finished-goods inventories– (manufacturing firms)

or merchandise (retail stores)



Types of Inventories (Cont’d)

• Replacement parts, tools, & supplies

• Goods-in-transit to warehouses or customers



Functions of Inventory

1.To meet anticipated demand of customers

A customer can be---a person who walks in off the market to buy his wants

2.To smooth production requirements

Build up inventories during off season period to meet high demands during certain seasonal periods

3.To protect against stock-outs (Keep safety stock)

Delayed deliveries/unexpected increase in demand increases the risk of shortage.

Delays could be due to—bad weather, stock outs, deliveries of wrong materials, low quality or nonstandard supplies



Functions of Inventory (Cont’d)

4.To help hedge against price increases or to take advantage of quantity discounts

5.To permit operations by keeping buffer stock

Equipment breakdown can cause temporary shutdown of operations. So buffer stock permits other operations to continue in production systems



Objectives of inventory control

• Inventory management concerns:1. level of customer service

keep right goodskeep enough quantitieskeep goods at the right placegoods be available at the right time

2. costs of goods/stock• Objectives:

To achieve satisfactory levels of customer service while keeping inventory cost within reasonable bounds



Good Inventory Management

Successful operation of Business

Poor inventory management

hampered operations

customer un-satisfaction

increased operating cost



• A system to keep track of inventory

• A reliable forecast of demand

• Knowledge of lead times

• Reasonable estimates of

– Holding costs

– Ordering costs

– Stock cost

– Shortage costs

Effective Inventory Management



Inventory Counting Systems (tracking system)

• Periodic SystemPhysical count of items made at periodic intervals (weekly, monthly) which help in deciding when and how much to order?

• Perpetual Inventory System System that keeps track of removals from inventory continuously (by UPC),

thus monitoringcurrent levels of each item.



Inventory Counting Systems (Cont’d)

• Two-Bin System - Two containers of inventory; reorder when the first is empty

• Universal Bar Code - Bar code printed on a label that hasinformation about the item to which it is attached

(UPC).0

214800 232087768



Demand forecast & lead time information

• Reliable estimates of time & amount of demand

• Reliable estimates of time required for order replenishment (Lead time)

• Electronic point of sale (EPOS) system: (this records sales electronically)



Reasonable estimates of inventory costs

1.Holding cost/carrying cost:interest, insurance, taxes, depreciation, obsolescence, deterioration, warehouse costs (rent, light, heat, security etc.

2.Ordering cost:-costs of preparing invoices (clerical,administrative costs)-shipping costs, transportation costs-inspection costs on arrival at point of sale



Reasonable estimates of inventory costs (cont’d)

3.Shortage cost/stock-out cost(cost when demand exceeds the supply)This may result in loss of customer goodwill

4.Stock costbuying in prices direct cost of production(discounts available for bulk purchase)



ABC Classification System

Classifying inventory according to some measure of importance and allocating control efforts accordingly.

A - very important

B – moderatelyimportant

C - least important

Figure 13-1

Annual

$ volume

of items

A

B

C

High

Low

Few Many

Number of Items



1.Demand: the amount required for sales, production (units/time)

2.Lead time: time interval between ordering and receiving the order (weeks, months)

3.Economic order quantity (EOQ): the calculated ordering quantity that minimizes the total cost

Inventory Terminologies



Key Inventory Terms (cont’d)4.Buffer stock (safety stock): stock kept to cover

errors in forecasting the lead time or the demand during the lead time

5.Reorder level: the level at which a further replenishment order should be placed

6. Reorder quantity: the quantity of the replenishment order



• Economic order quantity (EOQ) model

• Economic production quantity model

• Quantity discount model

Economic Order Quantity Models



Assumptions

• Only one product is involved

• Annual demand requirements known

• Demand is even throughout the year

• Lead time does not vary

• Each order is received in a single delivery

• There are no quantity discounts

Economic order quantity (EOQ) model



The Inventory Cycle

Profile of Inventory Level Over Time

Quantity

on hand

Q

Receive

order

Place

orderReceive

order

Place

order

Receive

order

Lead time

Reorder

point

Usage

rate

Time

Figure 13-2



Total Cost (EOQ model)

Annualcarryingcost

Annualorderingcost

Total cost = +

Q2

HDQ

STC = +



Cost Minimization Goal (EOQ model)

Order Quantity

(Q)

The Total-Cost Curve is U-Shaped

Ordering Costs

QO

An

nu

al C

os

t

(optimal order quantity)

TCQ

HD

QS

2

Figure 13-4



Deriving the EOQ (EOQ model)

Using calculus, we take the derivative of the total cost function and set the derivative (slope) equal to zero and solve for Q.

Q = 2DS

H =

2(Annual Demand )(Order or Setup Cost )

Annual Holding CostOPT



Total Costs with Purchasing Cost (EOQ

model)

Annualcarryingcost

PurchasingcostTC = +

Q2

HDQ

STC = +

+Annualorderingcost

PD+



Total Costs with PD (EOQ model)

Co

st

EOQ

TC with PD

TC without PD

PD

0 Quantity

Adding Purchasing cost

doesn’t change EOQ

Figure 13-7



Quantity discount model

Total Cost with Constant Carrying Costs

OC

EOQ Quantity

To

tal C

ost

TCa

TCc

TCbDecreasing

Price

CC a,b,c

Figure 13-9



When to Reorder with EOQ Ordering

• Reorder Point - When the quantity on hand of an item drops to this amount, the item is reordered

• Safety Stock - Stock that is held in excess of expected demand due to variable demand rate and/or lead time.

• Service Level - Probability that demand will not exceed supply during lead time.



Safety Stock

LT Time

Expected demand

during lead time

Maximum probable demand

during lead time

ROP

Qu

an

tity

Safety stock

Figure 13-12



Reorder Point

ROP

Risk of

a stockout

Service level

Probability of

no stock out

Expected

demand Safety

stock

0 z

Quantity

z-scale

Figure 13-13



• Orders are placed at fixed time intervals

• Order quantity for next interval?

• Suppliers might encourage fixed intervals

• May require only periodic checks of inventory levels

Fixed-Order-Interval Model



• Tight control of type A items

• Items from same supplier may yield savings in:– Ordering

– Packing

– Shipping costs

• May be practical when inventories cannot be closely monitored

Fixed-Interval Benefits



• Requires a larger safety stock

• Increases carrying cost

• Costs of periodic reviews

Fixed-Interval Disadvantages



• Too much inventory– Tends to hide problems

– Easier to live with problems than to eliminate them

– Costly to maintain

• Wise strategy– Reduce lot sizes

– Reduce safety stock

Operations Strategy



Linear Programming

• Linear programming (LP) techniques consist of a sequence of steps that will lead to an optimal solution to problems, in cases where an optimum exists

Linear Programming

Linear Programming Model

• Objective: the goal of an LP model is maximization or minimization

• Decision variables: amounts of either inputs or outputs

• Feasible solution space: the set of all feasible combinations of decision variables as defined by the constraints

• Constraints: limitations that restrict the available alternatives

• Parameters: numerical values

Linear Programming Assumptions

• Linearity: the impact of decision variables is

linear in constraints and objective function

• Divisibility: non-integer values of decision

variables are acceptable

• Certainty: values of parameters are known and

constant

• Non-negativity: negative values of decision

variables are unacceptable

Graphical Linear Programming

• Set up objective function and constraints in mathematical format

• Plot the constraints

• Identify the feasible solution space

• Plot the objective function

• Determine the optimum solution

Linear Programming Example

0

2

4

6

8

10

12

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Plot

Constraint 1

X1 + 3X2 = 12


0

2

4

6

8

10

12

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Add

Constraint 2

4X1 + 3X2 = 24

Constraint 1

X1 + 3X2 = 12

Solution space


0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Z x x

x x

x x

x x

4 5

3 12

4 3 24

0

1 2

1 2

1 2

1 2,

Z = 60

Z = 40

Z = 20

X1

X2

• Redundant constraint: a constraint that does not form a unique boundary of the feasible solution space

• Binding constraint: a constraint that forms the optimal corner point of the feasible solution space

Constraints

Slack and Surplus

• Surplus: when the optimal values of decision variables are substituted into a greater than or equal to constraint and the resulting value exceeds the right side value

• Slack: when the optimal values of decision variables are substituted into a less than or equal to constraint and the resulting value is less than the right side value

Simplex Method

• Simplex: a linear-programming algorithm that can solve problems having more than two decision variables

• Tableau: One in a series of solutions in tabular form, each corresponding to a corner point of the feasible solution space

Sensitivity Analysis

• Range of optimality: the range of values for which the solution quantities of the decision variables remains the same

• Range of feasibility: the range of values for the fight-hand side of a constraint over which the shadow price remains the same

• Shadow prices: negative values indicating how much a one-unit decrease in the original amount of a constraint would decrease the final value of the objective function

1

Linear Programming Big M Method

Artificial Starting variables Artificial Variable takes the role of slack variables at the 1

st iteration, only to be disposed of at the later

iteration. Artificial Variable are extraneous to LP Model, a penalty is assigned them in the objective

function to force them to zero at a later iteration of the Simplex Algorithm.

Given M is a large positive value, the variable Ri is penalized in the objective function using:

-MRi in Maximization problem

+MRi in Minimization problem

Example:

Minimize Z= 4X1 + X2

Subject to:

1. 3X1 + X2 = 3

2. 4X1 + 3X2 >= 6

3. X1 + 2X2 <= 4

Where, X1 and X2 >=0

Standard LP form by adding slack or subtracting slack will be

Minimize Z= 4X1 + X2

Subject to:

1. 3X1 + X2 = 3

2. 4X1 + 3X2 – X3 = 6

3. X1 + 2X2 + X4 = 4


Use Ri in 1st and 2

nd constraint and penalize the objective function

Minimize Z= 4X1 + X2 + MR1 +MR2

Subject to:

1. 3X1 + X2 + R1 = 3

2. 4X1 + 3X2 – X3 + R2 = 6

3. X1 + 2X2 + X4 = 4


Row# BV Z X1 X2 X3 R1 R2 X4 RHS

Row0 Z 1 -4 -1 0 -M -M2 0 0

Row1 R1 0 3 1 0 1 0 0 3

Row2 R2 0 4 3 -1 0 1 0 6

Row3 X4 0 1 2 0 0 0 1 4

Develop new Z-Row to eliminate Ri

Action Z X1 X2 X3 R1 R2 X4 RHS

Old Row0 1 -4 -1 0 -M -M2 0 0

+M(Row1) 0 3M 1M 0 1M 0 0 3M

+M(Row2) 0 4M 3M -1M 0 1M 0 6M

2

New Row0 1 -4+7M -1+4M -M 0 0 1M 9M

Iteration 0

Row# BV Z X1 X2 X3 R1 R2 X4 RHS Ratio

Row0 Z 1 -4+7M -1+4M -M 0 0 1M 9M

Row1 R1 0 3 1 0 1 0 0 3 1

Row2 R2 0 4 3 -1 0 1 0 6 1.5

Row3 X4 0 1 2 0 0 0 1 4 4

Mark the key column (most +ve in minimization problem), key row (lowest ratio) and find pivot element.

Divide the key row by pivot element to make the key column the part of identity matrix.

Now X1 will become the BV and R1 will become NBV. Apply the Gauss Elimination Method.

Iteration 1


Row0 Z 1 0 (1+5M)/3 -M (4-7M)/3 0 0 4+2M

Row1 X1 0 1 1/3 0 1/3 0 0 1 3

Row2 R2 0 0 5/3 -1 -4/3 1 0 2 6/5

Row3 X4 0 0 5/3 0 -1/3 0 1 3 5

Mark the key column (most +ve in minimization problem), key row (lowest ratio) and find pivot

element. Divide the key row by pivot element to make the key column the part of identity matrix.

Now X2 will become the BV and R2 will become NBV. Apply the Gauss Elimination Method.

Iteration 2


Row0 Z 1 0 0 1/5 (24-15M)/15 (-1-5M)/5 0 18/5

Row1 X1 0 1 0 1/5 9/15 -1/5 0 3/5 3

Row2 X2 0 0 1 -3/5 -4/5 3/5 0 6/5 -2

Row3 X4 0 0 0 1 -4/3 1 1 1 1

Mark the key column (most +ve in minimization problem), key row (lowest ratio) and find pivot element.

Divide the key row by pivot element to make the key column the part of identity matrix.

Now X3 will become the BV and X4 will become NBV. Apply the Gauss Elimination Method

Iteration 3

Row# BV Z X1 X2 X3 R1 R2 X4 RHS

Row0 Z 1 0 0 0 (19-15M)/15 -M 0 17/5

Row1 X1 0 1 0 0 2/15 0 0 2/5

Row2 X2 0 0 1 0 -1/5 0 0 9/5

Row3 X3 0 0 0 1 -4/3 1 1 1

This is optimum solution with Z=17/5, X1=2/5, X2=9/5, and X3=1

The Transportation Model



Requirements for Transportation Model

• List of origins and each one’s capacity

• List of destinations and each one’s demand

• Unit cost of shipping



Transportation Model Assumptions

• Items to be shipped are homogeneous

• Shipping cost per unit is the same

• Only one route between origin and destination



The Transportation Problem

D

(demand)

D

(demand)

D

(demand)

D

(demand)

S

(supply)

S

(supply)

S

(supply)



A Transportation Table

Warehouse

4 7 7 1

100

12 3 8 8

200

8 10 16 5150

450

45080 90 120 160

A B C D

1

2

3

Factory Factory 1

can

supply

100

units per

period

Total

supply

capacity

per

period

Total demand

per period

Demand

Warehouse B can use 90 units per period

Table 8S-1



Special Problems

• Unequal supply and demand

• Dummy: Imaginary number added equal to the difference between supply and demand when these are unequal



Summary of Procedure

• Make certain that supply and demand are equal

• Develop an initial solution using LCM, NWC, or VAM methods

• Check that completed cells = R+C-1

• Evaluate each empty cell

• Repeat until all cells are zero or positive



Excel Template



Forecasting



Introduction• A statement about the future• Used to help managers

a. To plan the system--long-range planning about:

O--type of products & services to offerO--type of facilities and equipment and where

to locate themb. To plan the use of the system

--short range and intermediate range planning:O--planning inventory and work forceO--planning purchasing and productionO--budgeting and scheduling



• Farmers rely on weather forecast for seed bed preparing, planting, spraying, harvesting crop etc.

• Businessmen rely on forecasts for:o budgeting and planning for capacityo product sales o production and inventoryo manpowero purchasing etc.



Accounting Cost/profit estimates

Finance Cash flow and funding

Human Resources Hiring/recruiting/training

Marketing Pricing, promotion, strategy

MIS IT/IS systems, services

Operations Schedules, MRP, workloads

Product/service design New products and services

Uses of Forecasts

Features common to all forecasts• Assumes causal system

past ==> future

• Forecasts rarely perfect because of

randomness (actual results usually

differ from predicted values)

• Forecasts more accurate for

groups vs. individuals

• Forecast accuracy decreases

as time horizon increases

I see that you will

get an A this semester.



Elements of a Good Forecast

Timely

AccurateReliable

Written




(cont’d)1. Forecast be timely: give time to implement the forecast e.g. capacity

can't be increased overnight, inventory increase needs time

2. Forecast be accurate: degree of accuracy be stated. This will help in

comparing different alternative forecasts3. Forecast be reliable forecast should work consistently




(cont’d)4. Forecast be in meaningful units financial planners need to know how many dollars

needed production planners need to know how many

products needed schedulers need to know what machines & skills

involved5. Forecasts be in writing a written forecast permits an objective basis for

evaluating the forecast once actual results are in6. Forecast be simple to understand and easy to use

properly people lack confidence in the use of sophisticated

forecasts



Steps in the Forecasting Process

Step 1 Determine purpose of forecast

Step 2 Establish a time horizon

Step 3 Select a forecasting technique

Step 4 Gather and analyze data

Step 5 Prepare the forecast

Step 6 Monitor the forecast

“The forecast”



Types of Forecasts

1. Judgmental - uses subjective inputs

2. Time series - uses historical data

assuming the future will be like the past

3. Associative models - uses explanatory

variables to predict the future



1. Judgmental & opinion Forecasts

i. Executive opinionsA small group of upper level managers (e.g. marketing, operations, and finance) meet and collectively develop forecastthis approach is used as a part of long range planning and new product development

ii. Sales force / customer service staff opinions

a good source of information because of their direct contact with consumers



1. Judgmental & opinion

Forecasts (cont’d)iii. Consumer surveys

excellent approach for recording demand and preferences of consumers

take a samples of potential consumers for getting their opinions

iv. Delphi method

managers and staff complete a series of questionnaires, each developed from the previous one, to achieve a consensus forecast



2. Time Series Forecasts

• Trend – A long-term upward or downward

movement in data

• Seasonality – A short-term regular

variations in data

• Irregular variations - caused by unusual

circumstances

• Random variations - caused by chance



2. Time Series Forecasts (cont’d)

Trend

Irregular

variation

Cycles

Seasonal variations

90

89

88

Figure 3-1



Techniques for averaging

CHARACTERISTICS• Smoothen the fluctuations in a time series due to

individual highs and lows of data points

• FC based on an average tends to exhibit less variability than the individual data values

• Responding to changes in expected demand often entails considerable cost (e.g. changes in---production rate, in the size of work force, and in inventory level), so it is desirable to avoid reacting to minor variations



Techniques for averaging (cont’d)

• Naïve forecast

• Moving average

• Exponential smoothening



N.F: The forecast for any period equals the previous period’s actual value

with a stable series (variations around an average) the last data point becomes the forecast for the next period

with seasonal variations, the forecast for current season is equal to the value of the series last “season”

for trend variations, the forecast is equal to the last value of the series plus or minus the difference between the last two values of the series (example at next page)

Naïve Forecasts (N.F)



Naïve Forecasts (cont’d)

period actual Change from

previous

value

forecast

t-1 50

t 53 +3

t+1 53+3=56

Naïve Forecasts (cont’d)Benefits• simple to use• virtually no cost• data analysis is nonexistent• easily understandable• cannot provide high accuracy• can serve as a standard of comparison

to judge the cost and accuracy of other techniques

Drawbacks:• Traces the actual data

• Does not smooth the curve at all



Moving average

• This method uses a number of most recent actual data values in generating a forecast

Man= (ΞAi/n)

where, i=most recent period

n=number of periods, data points

A=actual value with age i

MA= forecast



Example of moving average (cont’d)

period age demand

1 5 42

2 4 40

3 3 43

4 2 40 Three most

recent demands

5 1 41

Example of moving average (cont’d)

• MA3= (41+40+43)/3 = 41.33

If actual demand in 6th period turns out to be 39, moving average forecast for 7th

period would be

MA3= (39+41+40)/3 = 40

“as new actual becomes available, the forecast is updated by adding the newest value and dropping the oldest and then re-computing the average”



Simple Moving AverageFigure 3-4

MAn =n

Aii = 1

n

35

37

39

41

43

45

47

1 2 3 4 5 6 7 8 9 10 11 12

Actual

MA3

MA5



Moving average (cont’d)+VE POINTS:• Easy to compute and understand• FC updated by adding newest value and dropping the

oldest value. So FC moves by reflecting the most recent value

• Fewer the data points in an average, the more responsive the avg. tends to be real changes

• More the data points, smooth the curve will be but less responsive to real changes

-VE POINTS:All the data points in an avg. are weighted equally



Weighted average technique:

Weighted average: dt= w1(d1)+ w2(d2)+ w3(d3)+ w4(d4)

Where,dt =demand forecast for next period (t)w1=weightage for most recent demand value (d1)w2=weightage for next demand value (d2)w3=weightage for next after demand value (d3)w4=weightage for next next after demand value (d4)



Example of weighted average technique (cont’d)• Let w1=0.4, w2=0.3,w3=0.2, w4=0.1

Period 1 2 3 4 5 6

demand 42 40 43 40 41 39

Demand forecast for 6th, 7th, and 7th period as below:

d5=0.4(40)+0.3(43)+0.2(40)+0.1(42)

d6=0.4(41)+0.3(40)+0.2(43)+0.1(40)=41

d7=0.4(39)+0.3(41)+0.2(40)+0.1(43)=40.2

Advantage: W.A. is the most reflective of the most recent occurrences than the simple moving average



Exponential Smoothing

Premise--The most recent observations might have the highest predictive value.

– Therefore, we should give more weight to the more recent time periods when forecasting.

Ft = Ft-1 + (At-1 - Ft-1)Ft = demand FC for period “t”

Ft-1 = demand FC for period “t-1”

= smoothing constant, usually varies 0.0 to 0.5

At-1 = actual demand for period “t-1”

(At-1 - Ft-1) = percentage of previous occurrence



Exponential Smoothing (cont’d)• Let

Ft-1 = 42 units = 0.1At-1 = 40 unitsthen, Ft = Ft-1 + (At-1 - Ft-1)

=42+0.1(40-42)=41.8 unitsNote:• Larger the “” value, greater the response and less the

smoothing• Smaller the “” value (close to “0”), smaller the FC will

be to adjust to forecast error (greater the smoothing)



Period Actual Alpha = 0.1 Error Alpha = 0.4 Error

1 42

2 40 42 -2.00 42 -2

3 43 41.8 1.20 41.2 1.8

4 40 41.92 -1.92 41.92 -1.92

5 41 41.73 -0.73 41.15 -0.15

6 39 41.66 -2.66 41.09 -2.09

7 46 41.39 4.61 40.25 5.75

8 44 41.85 2.15 42.55 1.45

9 45 42.07 2.93 43.13 1.87

10 38 42.36 -4.36 43.88 -5.88

11 40 41.92 -1.92 41.53 -1.53

12 41.73 40.92

Example of Exponential Smoothing



Picking a Smoothing Constant

35

40

45

50

1 2 3 4 5 6 7 8 9 10 11 12

Period

De

ma

nd = .1

= .4

Actual



Common Nonlinear Trends

Parabolic

Exponential

Growth

Figure 3-5



Linear Trend Equation

• “b” is similar to the slope. However, since it is

calculated with the variability of the data in

mind, its formulation is not as straight-forward

as our usual notion of slope.

Yt = a + bt

0 1 2 3 4 5 t

Y



Calculating a and b

b =n (ty) - t y

n t 2 - ( t) 2

a =y - b t

n



Linear Trend Equation Example

t y

Week t2

Sales ty

1 1 150 150

2 4 157 314

3 9 162 486

4 16 166 664

5 25 177 885

t = 15 t2 = 55 y = 812 ty = 2499

( t)2 = 225



Linear Trend Calculation

y = 143.5 + 6.3t

a =812 - 6.3(15)

5=

b =5 (2499) - 15(812)

5(55) - 225=

12495-12180

275 -225= 6.3

143.5



Associative Forecasting Techniques(PP-110)

The essence of “AT” is the development of an equation that summarizes the effects of predictor variables (X)

• Predictor variables - used to predict values of variable of interest

• Regression - technique for fitting a line to a set of points

• Least squares line - minimizes sum of squared deviations around the line



Linear Model example-8, pp-111

Healthy Hamburgers has a chain of 12 stores in northern Illinois. Sales figure and profits for the stores are given in the following table. Obtain a regression line for the data, and predict profit for a store assuming sales of $10 million

Sales (X) 7 2 6 4 14 15 16 12 14 20 15 7

Profit (Y) 15 10 13 15 25 27 24 20 27 44 34 17



Calculations for regression parameters Ex-8• X Y XY X2 Y2

7 0.15 1.05 49 0.0225

2 0.10 0.20 4 0.01

- - - - -

132 2.71 35.39 1796 0.7159



Calculations for regression parameters Ex-8

b = 12(35.29) - 132(2.71) = 0.01593

12(1796) - 132(132)

a = 2.71-0.01593(132) = 0.0506

12

Y = 0.0506 + 0.01593 X



Linear Model Seems Reasonable

EXAMPLE-8 (PP-111)

0

10

20

30

40

50

0 5 10 15 20 25

X Y

7 15

2 10

6 13

4 15

14 25

15 27

16 24

12 20

14 27

20 44

15 34

7 17

Computed

relationship



Comments on using Regression Analysis

1. Variations around the line are random. So no pattern such as cycles or trends should be apparent when the line and data are plotted

2. Deviations around the line should be normally distributed

3. Prediction are being made only within the range of observed values



FORECAST ACCURACY (FA)

• Forecast accuracy works as a basis of comparison among different techniques to be used

• For making periodic forecasts (Ft) {e.g. weekly} monitor forecast error to determine if the errors are within reasonable bounds. If they are not then take necessary action.

• Error - difference between actual value and predicted value, Error, et = At-Ft)

this should be within reasonable bounds• Forecast errors influence decisions in two ways:

1. Making a choice among various FC alternatives2. Evaluating the success or failure of a technique in use



Forecast Accuracy

• Mean absolute deviation (MAD)

-Average absolute error

• Mean squared error (MSE)

-Average of squared errors

• Tracking signal

-Ratio of cumulative error and MAD



Mean absolute deviation (MAD) and

Mean squared error (MSE)

MAD = Actual forecast

n

MSE =Actual forecast)

-1

2

n

(



Example of Forecast Accuracy, FA

Compute MAD and MSE

Period Actual Forecast Error [error] [error]2

1 217 215 2 2 4

2 213 216 -3 3 9

3 216 215 1 1 1

4 210 214 -4 4 4

5 213 211 2 2 2

6 219 214 5 5 5

7 216 217 -1 1 1

8 212 216 -4 4 4

-2 22 76

Example of Forecast Accuracy, FA• Calculations

– MAD = 22/8 =2.75

– MSE = 76/(8-1) =76/7 =10.86

APPLICATIONS OF MAD and MSE– MAD weighs all errors evenly– MSE weighs all errors according to their squared values

– To compare alternative forecasting methods, using either MAD or MSE, a manager can compare the results of exponential smoothing with values of 0.1, 0.2, 0.3 and select the one that yields the lowest MAD or MSE for a given set of data



Controlling the forecast• Monitor the FC errors to ensure that the FC is performing adequately by comparing FC errors to predetermined values or action limits (Figure below). Errors outside of either limit (+ or -) signal that corrective action is needed

+ Need for Upper limit corrective action

Error 0 range ofacceptable variation

- Lower limit Time _



CONTROLLING THE FORECAST (CONT’D)

SOURCES OF FC ERRORS1. The model may be inadequate due to• Omission of an important variable• Sudden appearance of trend or cycle• Appearance of a new variable (new competitor)2. Irregular variations may occur due to severe

weather, temporary shortages, or breakdowns etc.3. FC technique incorrectly used4. There are always random variations in data.

Randomness is the inherent variation that remains.



Controlling the forecast (cont’d)(Tracking signal)

FC can be monitored using either tracking signals or control charts

Tracking signal: the ratio of cumulative FC error to the corresponding value of MAD used to monitor a FC

Cumulative forecast error, Ξ(At-Ft): It reflects bias in forecast, which is the persistent tendency for forecasts to be greater or less than the actual values of time series



Tracking Signal

Tracking signal =(Actual -forecast)

MAD

Tracking signal =(Actual -forecast)

Actual -forecast

n



Controlling the forecast (cont’d)(Tracking signal)

TRACKING SIGNAL (TS): TS values = ±3 to ±8use ±4 which are comparable to ±3 σ

CONTROL CHART (CC): This involves setting upperand lower limits for individual FC errors. The limits are multiples of the square root of MSE. This method assumes the followings:1. FC errors are randomly distributed around a mean2. Errors follow normal distribution curve (take µ = 0)

µ± 1σ = 0.6628, µ± 2σ = 0.95, µ± 3σ = 0.9987



Linear Trend Equation (PP-129)T3-3



Simple Linear RegressionT3-5



Project Management



Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.

Build A

A Done

Build B

B Done

Build C

C Done

Build D

Ship

JAN FEB MAR APR MAY JUN

On time!

Projects



• Deciding which projects to implement

• Selecting a project manager

• Selecting a project team

• Planning and designing the project

• Managing and controlling project resources

• Deciding if and when a project should be terminated

Key Decisions



Project Manager

Responsible for:

Work Quality

Human Resources Time

Communications Costs



Project Life Cycle

Concept

Feasibility

Planning

Execution

Termination

Man

ag

em

en

t



• Risk: occurrence of events that have undesirable consequences

– Delays

– Increased costs

– Inability to meet specifications

– Project termination

Project Risk



• Identify potential risks

• Analyze and assess risks

• Work to minimize occurrence of risk

• Establish contingency plans

Risk Management



Work Breakdown Structure

Project X

Level 1

Level 2

Level 3

Level 4

Figure 18-3



MAR APR MAY JUN JUL AUG SEP OCT NOV DEC

Locate new

facilities

Interview staff

Hire and train staff

Select and order

furniture

Remodel and install

phones

Move in/startup

Planning and Scheduling

Gantt Chart



PERT and CPM

PERT: Program Evaluation and Review Technique

CPM: Critical Path Method

• Graphically displays project activities• Estimates how long the project will take• Indicates most critical activities• Show where delays will not affect project



Project Network – Activity on Arrow

1

2

3

4

5 6

Locate

facilities

Order

furniture

Furniture

setup

InterviewHire and

train

Remodel

Move

in



Network Conventions

a

b

c ab

c

a

b

c

d

a

b

c

Dummy

activity



Example 1

1

2

3

4

5 6

8 weeks

6 weeks

3 weeks

4 weeks9 weeks

11 weeks

1 week

Move

in



Example 1 Solution

Path Length(weeks)

Slack

1-2-3-4-5-61-2-5-61-3-5-6

182014

206

Critical Path

• Network activities– ES: early start– EF: early finish– LS: late start– LF: late finish

• Used to determine– Expected project duration– Slack time– Critical path

Computing Algorithm



Probabilistic Estimates

Activity

start

Optimistic

time

Most likely

time (mode)

Pessimistic

time

o pm te



TOTAL QUALITY MANAGEMET

(TQM)



Total Quality Management

A philosophy that involves everyone in an organization in a continual effort to improve quality and achieve customer satisfaction.



1. Find out what the customer wantsThis involves; surveys, focus groups, interviews of customers etc.

2. Design a product or service that meets or exceeds customer wants

3. Design production processes that facilitates doing the job right the first time

4. Keep track of results and use those to guide improvement in the system. Never stop trying to improve.

5. Extend these concepts to suppliers

The TQM Approach



Important Elements of TQM1.Continual improvement- philosophy that seeks to

make never-ending improvements to the process of converting inputs into outputs (equipment, methods, materials, and people). Japanese say this term of continual improvement as Kaizen.

2.Competitive benchmarking - identify companies or organizations that art best at something and study how they do it and learn how to improve your operation (the company need not be in the same line of business as yours).

3.Employee empowerment – giving workers the responsibility for improvements and authority to make changes. Motivate them for the sake of existence of organization with integrity. For doing so give them incentives.



Important Elements of TQM (cont’d)4.Team approach-

- use team approach for solving problems - promote a spirit of cooperation and shared values among employees

5.Decisions based on facts rather on opinions-- management gathers and analyzes data as a

basis for decision making6.Knowledge of tools – employees and managers

are trained in the use of quality tools7.Supplier quality- ensure high quality material supply

by suppliers8.Quality at the Source- The philosophy of making each

worker responsible for the quality of his or her work.



1.Lack of Company-wide definition of quality-efforts are not coordinated, people are working at cross purposes, addressing different issues, using different measures of success.

2.Lack of Strategic plan for change-It lessens the chance of success.

3.Lack of Customer focus-without there is a risk of customer satisfaction

4.Lack of real employee empowerment-this gives the impression of not trusting employees

5.Lack of strong motivation-managers need to make sure employees are motivated

Obstacles to Implementing TQM



6.Lack of time to devote to quality initiatives-don’t add more work without adding additional resources

7.Lack of leadership- managers need to be good leaders

8. Poor inter-organizational communication-the left hand does not know what the right hand is doing

9. View of quality as a “quick fix”-Needs to be a long term, continuing effort

10.Emphasis on short-term financial results-

Obstacles to Implementing TQM (cont’d)



Basic Steps in Problem SolvingProblem solving is one of the basic procedures of TQM

• Define the problem and establish an improvement goal

• Collect data

• Analyze the problem

• Generate potential solutions- methods include brainstorming, interviewing, and surveying

• Choose an economically best solution

• Implement the solution- keep every one informed

• Monitor the solution to see if it accomplishes the goal- if not modify the solution accordingly



The Plan-Do-Study-Act (PDSA) CyclePlan

Do

Study

Act



The PDSA) Cycle (cont’d)• It is a frame work for problem solving and

improvement activities. There are 4-steps in the cycle

1.Plan - begin with studying the current process- document that process- collect data on that process / problem- analyze the data and develop a plan for

improvement- specify measures for evaluating the plan

2. Do - implement the plan, on a small scale if possible

- document any changes made during thisphase

- collect data systematically for evaluation10

Prof. Dr. Muhammad Iqbal, Department of Farm Machinery & Power, University of


The Plan-Do-Study-Act (PDSA) Cycle

3.Study - evaluate the data collection during the DO phase

- check how closely the results match - the original goals of the plan phase

4. Act - if the results are successful, standardize the new method and communicate it to all the concerns associated with the process

- implement training for the new method- if not successful results then revise the

plan and repeat the process or cease this project



Process Improvement• Process Improvement is a systematic approach to

improving a processFor improving the functioning of process, it involves:

»documentation»measurement»analysis

Typical goals of process improvement include:» increasing customer satisfaction»achieving higher quality»reducing waste» increasing productivity»speeding up the process



The Process Improvement Cycle(another version of PDAS cycle)

Implement the

Improved process

Select a

process

Study/document

Seek ways to

Improve it

Design an

Improved process

Evaluate

Document



Overview of Process Improvement (cont’d)

A. Process mapping-1. Collect information about the process,

identify each step in the processfor each step, determine:

»The inputs and outputs»The people involved»Document such measures as time, cost

space used, working conditions revenues, profits

2. Prepare the flow chart and accurately depict the process



Overview of Process Improvement (cont’d)B. Analyze the process

1. ask the following about the process•is the flow logical•are any activities missing•are any there duplication

2. ask the followings about each step• is the step necessary• does the step add value• does any waste occur at this step

C.Redesign the process-

• using the results, redesign the process if needed



Process Improvement Tools

– There are a number of tools that can be used for problem solving and process improvement

– Tools aid in data collection and interpretation, and provide the basis for decision making



Process Improvement Tools (cont’d)

1.Flowcharts- a diagram of the steps in process

2.Check sheets- a tool for recording and recognizing data to identify a problem; a tally of

problems or other events by category

3.Histograms- a chart that shows an empirical frequency distribution

4.Pareto Charts-

• a diagram that arranges categories from highest to lowest frequency of occurrence

• Pareto analysis is a technique for classifying problem areas according to degree of importance, and focusing on the most important



Process Improvement Tools (cont’d)

5.Scatter diagrams-

a graph that shows the degree and direction of relationship between two variables

6.Control charts-

a statistical chart of time-ordered values of a sample statistic

it is used to monitor a process

it can indicate when a problem occurred and give insight into what caused the problem



Process Improvement Tools (cont’d)7.Cause-and-effect diagrams

– a structured approach used to search for causes of a problem; also called fish bone diagram

– this tool is used after brainstorming sessions, to recognize the ideas generated

8.Run chart

a tool for tracking results over a period of time



Check Sheet

Billing Errors

Wrong Account

Wrong Amount

A/R Errors

Wrong Account

Wrong Amount

Monday



Pareto Analysis

80% of the problems may be attributed to 20% of the

causes.

Smearedprint

Nu

mb

er

of

def

ects

Offcenter

Missinglabel

Loose Other



Control Chart

970

980

990

1000

1010

1020

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

UCL

LCL

Figure11-9



Cause-and-Effect Diagram

Figure11-11

Air ticketerrors

Materials(ticket stock)

Method (printing)

machinepersonnel

ageMaintenancefrequency

type

training

carbonspeed

supervision

agequality

Attentionto detail

paper



Run Chart

Time (Hours)

Dia

met

er



Control chart for Tracking Improvements (cont’d)

UCL

LCL

LCLLCL

UCLUCL

Process not centered

and not stable

Process centered

and stable

Additional improvements

made to the process

Figure11-15



Methods for Generating Ideas

1.Brainstorming

2.Quality circles

3.Interviewing

4.Benchmarking

5.5W2H



1. Brainstorming• It is a technique for generating a free flow of

ideas in a group of people to identify problems, and finding causes, solutions, and ways to implement solutions

• A group of people share thoughts and ideas on problems in a relaxed atmosphere that encourages un-restrained collective thinking



• Quality circles are groups of workers who meet periodically to discuss ways of improving products or processes

• A team approach works best : This may involve one or more of the following methods– List reduction- it is applied to a list of problems /

solutions. Its purposes is to clarify items and reduce the list of items by solving the problems

– Balance sheet- it lists pros and cons of each item and focuses discussion on important issues

– Paired comparisons- in this, each item on a list is compared with every other item, two at a time, and selected the preferred item (applicable for five or fewer items only)

2. Quality Circles



3. Interviewing

It is technique for identifying problems and collecting information

internal problems may require interviewing employees

external problems may require interviewing external customers



It is the process of measuring performance against the best in the same or another industry

• Identify a critical process that needs improving

• Identify an organization that excels in this process

• Contact that organization

• Analyze the data

• Improve the critical process

4. Benchmarking Process



5. 5W2H

It is a method of asking questions about a current process that includes what, why, where, when, who, how, and how much

(5W2H = 5 ‘w’ words and 2 ‘h’ words)

This leads to important insights about why the current process isn’t working as well as it could, as well as potential ways to improve it



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