OPERATIONS MANAGEMENT

A. A. ELIMAM

BUSINESS ANALYSIS AND COMPUTING SYSTEMS

COLLEGE OF BUSINESS

SAN FRANCISCO STATE UNIVERSITY


OPERATIONS MANAGEMENT

It is the systematic direction and control of the processes that transform inputs into finished goods or services.

I. Major Challenges to Operations Managers

Increase the value of the Output relative to the cost of the input, or INCREASE PRODUCTIVITY.

1.1 Decision Making:

An essential Part of Operations Management. Decision making can take place in several areas:

a. Positioning Decisions: Product Planning -Positioning Strategies and Quality Management.

b. Design Decisions: Process Design, Work Force management, capacity, Location and Layout.

c. Operations Decisions: Materials Management, Production Planning and Scheduling, Inventory, Project Scheduling and Quality Control.

1.2. Decision Making Horizons

1. Strategic Planning (5-10 years):

Less certainty- Less detail- Goal-oriented.

2. Operational Planning (3 month- 3 years):

More Certainty--More means-oriented--Better defined

3. Scheduling (weekly-monthly):

More attention to detail

4. Sequencing and Dispatching(hourly-daily):

Exact order and time of implementation

5. Control(hourly-daily)

Feedback on implementation.

II. Product Planning:

Idea Generation-Screening- Development and Testing- Final Product Design

Screening Approaches:

2.1 Preference Matrix: Come up with a weighted score for each product based on selected measures of performance. The product is selected if its total score exceeds a threshold determined by management. (A deficient approach - why?)

2.2. Break-even Analysis (BEA)

In product Screening or selection, BEA is used to determine whether revenues from selling product will offset the cost of making it, Figure 1. So

Total Annual Revenue = Total Annual Cost

Total Annual Revenue = Annual Fixed Cost + Total Variable Cost

P Q = F + C Q [1]

Where

P = Price in $/unit

C = Variable Cost in $/unit

F = Annual Fixed Cost, $/year

Q = Number of units produced and sold per year

So If

1. The Production department determines F and C,

2. The Marketing department estimates the selling price and the amount of units expected to be sold (annual sales volume)

then

the BEA can be used to find out whether the product breaks even at this volume of sales. Clearly, if it does not break-even, then the product must be eliminated from further consideration.

Example 1: Suppose you would like to examine the feasibility of building a furniture plant to manufacture patio furniture sets. The fixed cost to set up the production line is $600,000. The variable cost is $ 50 per set. Your marketing research indicated that you can sell about 15,000 sets at $ 110 per set. Would it be feasible to produce the patio furniture set?

Solution: First we compute the break-even point

Q = F/ (P-c) = 600,000/(110-50) = 10,000 patio furniture sets

Since we can sell 15,000 (more than the BEP), the project is feasible and we should proceed to produce and sell the patio sets.

Example 2: Luxor Inc. began producing Cheese in January 1993. Its 1993 output reached 20,000 lb. at a total cost of $40,000. In 1994 its output increased to 30,000 lb. at a total cost of $50,000.

(a) What must be the variable cost(c) and the fixed cost(F)? Luxor's costs stayed the same during 1993, and 1994.

(b) If Luxor sold the cheese at $ 2.8 and $ 3.20 during 1993 & 1994, respectively, what is Luxor's productivity during these two years.

(c) If Luxor's total cost in 1995, is expected to increase to $ 60,000, how many lb. it should produce and sell in order to maintain the same productivity level of 1994 (selling price is still $ 3.20/lb.)

Solution:

a. Total annual cost = Annual Fixed Cost + Annual Variable cost.

TC = F + c . Q

Since the fixed and variable costs stayed the same during 1993 and 1994 then total cost for each year is given by

40,000 = F + 20,000 c for 1993 [2]

50,000 = F + 30,000 c for 1994 [3]

Subtracting [2] from [3], we get

10,000 =0 + 10,000 c

Therefore c = $ 1 per lb.

Substituting for c in [2], then F = 40,000 - 20,000 (1) = $ 20,000 per year

b. In a give year

Productivity = Output/ Inputs

Productivity = Total Annual Revenues/Total annual costs

or

Productivity = selling price (Q)/ TC [4]

Therefore

Productivity for 1993 = 20,000(2.8)/40,000 = 1.4

Productivity for 1994 = 30,000(3.2)/50,000 = 1.92

which shows that 1994 productivity has improved over that of 1993

c. For 1995, it is required to maintain the same productivity level as that of 1994. We need to find Q in [4] that will ensure that the 1995 productivity will be equal to 1.92.

Hence 1.92 = Q (3.2)/ 60,000, resulting in an amount, Q = 36,000 lb.

BEA can also be used for two more purposes, namely for:

1. Deciding whether to Make-or-buy.

Total Annual Cost of Making = Total Annual Cost of Buying

Fm + Cm Q = Fb + Cb Q [5]

Where the symbols m and b indicate the costs of making and buying respectively.

Once again a decision is made to make or buy depending on whether the number of units needed in a year exceeds the break- even volume, Q.

Example 3: You have a plant to assemble PC’s. In order to make hard disks you need an annual fixed cost of $ 200,000 and a variable cost of $ 50 per unit. You can buy the hard disks you need for $ 130 per unit. Should you make or buy hard disks?

Solution: Using [5], we get

200,000 + 50 Q = 130 Q which yields

Q = 200,000/(130-50) = 2500 units.

Therefore buy the hard disks as long as you need up to 2500 units. Otherwise make the hard disks in your plant

2. Selection Among Two alternatives:

Trying to select between two cars (A and B) for example

Total Annual Cost of Car A = Total Annual Cost of Car B

Fa + Ca Q = Fb + Cb Q [6]

In this case

Ca or Cb provides the cost of operating cars A or B in $/mile and

Q represents the annual miles driven.

The selection is made based on whether the number of miles driven per year exceeds the break-even mileage.

Example 4: You are looking into leasing a new Car and you are considering two models. The lease rates and the running costs for two models are given below:

Annual Costs

HODGE

MISSAN

Lease cost, $

5,000

8,000

Variable cost $/mile

0.3

0.15

(a) Which car should you lease and Why?

(b) Suppose the running cost 0f the HODGE went down to $ 0.25 per mile, would that change your selection ? Why?

Solution: a. Substituting in [6], we get

5000 + 0.3 Q = 8000 + 0.15 Q

which provides Q = 20,000 miles. Therefore, you should lease the HODGE if you expect to drive less than 20,000 miles. Otherwise lease the MISSAN.

b. To answer this question, we need to repeat the above calculation for the new variable cost

5000 + 0.25 Q = 8000 + 0.15 Q

which provides Q = 30,000 miles. Based on the new variable cost per mile for the HODGE, you should lease the HODGE if you expect to drive less than 30,000 miles. Otherwise lease the MISSAN.

The BEA for the selection among alternatives can be expanded to select among locations (two or more). The use of BEA in location will be discussed in detail under chapter 7.

2.3. LP-based Product Selection:

In this approach we formulate a Linear Programming(LP)-based model aimed at selecting product(s) that provides maximum profit while satisfying budget or cost limitations. Specifically, it is required to

Maximize

p1 x1 + p2 x2 + p3 x3 + p4 x4 +.....+ pn xn

Subject to

c1 x1 + c2 x2 + c3 x3 + c4 x4 +.....+ cn xn = B

where

xi = 1 if product I is selected, zero otherwise.

pi = Profit resulting from selecting product I, $

ci = cost of product I, $

B = Budget limitation, $

By solving the above model, we only select the products whose xi = 1 in the solution.

III. Positioning Strategies

It determines whether the production system will be organized by grouping resources around the product or the process.

3.1. Process-Focused strategy:

When to use Process-Focused Strategy?

If we are making low volumes of customized products with intensive customer interactions.

Example Products: aerospace- Buildings-Interior Design

3.2. Product-Focused strategy:

When to use Product-Focused Strategy ?

If we are making high volumes of a standard product without customer interactions.

Example Products: paper Clips-Tires-Floppies

IV. Product Life Cycle

A product passes through five stages, namely: product planning, introduction, growth, maturity, and decline (see figure and example below).

A life-cycle audit is an evaluation of which stage the product is in, based on the changes in sales and profits compared to those that took place in earlier years. Life-cycle audits are used to identify the need for revitalizing or dropping existing products and introducing new ones. Life Cycles vary greatly from product to product. The following table summarizes the features of product life-cycles and how to use it in conducting audits:

 

Stage

Annual Sales

Annual Profits

Action

1. Product planning

  • zero
  • Negative- only spending
  • expedite development once selected
  • 2. Introduction

    • > zero
    • Slow rate of increase (+ slope)
  • starts negative then becomes positive
  • fast rate of increase (+slope)
  • pursue an intensive marketing effort.
  • 3. Growth

    • Steep rate of increase (+ slope)
  • Slow rate of increase (+ slope)
  • Introduce large production volumes
  • 4. Maturity

    • Slow rate of increase (+slope) or decrease (-slope)
    • High Sales
  • very limited rate of change (+ or -)
  • profit is declining
  • Improve efficiency and reduce cost
  • improve marketing operations
  • reduce price
  • 5. Decline

    • (- slope) negative rate of change
  • (- slope) negative rate of change
    • reduce price or enhance marketing activities if still making profit
    • otherwise Drop it

    LINEAR PROGRAMMING

    OUTLINE

    Linear Programming -- a simple example and Motivation

    Characteristics Of LP Models

    Range Of LP Applications

    Problem Formulations

    LP Solution Approaches

    Sensitivity Analysis

     

     

     

    Sensitivity Analysis- Basic Concepts:

    Because of the dynamic nature of real world environments, some of the LP models data (such as the objective function coefficients, the right-hand side of a constraint, et) may change over time. The question is "what happens to the optimal solution if one of these input data parameters changes?" The answer to this question deals with the topic of sensitivity analysis.

    Sensitivity Analysis. Determining how sensitive the optimal solution and the objective function value are with respect to changes in problem data.

    Why Sensitivity Analysis?

    1. to cover for the inaccuracies in the estimated problem parameters.

    2. to answer "What-if" type of questions, particularly to determine:

    or

    and

    3. to include the effect of a technological improvement, hence changing one of the coefficients in the constraint, on the optimal solution.

    This course will only cover items 1 and 2 above.

    1. Sensitivity Analysis of the Objective Function Coefficients

    The objective function coefficients determine its slope, which will change if one of these coefficients (profit for example) is changed. The graphical solution approach for a two variable LP problem provides an excellent illustration of the impact of changes in an objective function coefficient on its slope and accordingly on the optimal solution.

    2. Sensitivity Analysis of the right-hand Side (RHS) Values

    This type of sensitivity analysis is designed to answer the question:" what happens to the optimal objective function value of an LP problem when one of the RHS values is changed and all the other problem data stayed the same?" One way to answer such a question is to solve a new LP problem with the modified RHS value. However, sensitivity analysis can help in answering this question without the need for solving a new LP problem. Once an optimal solution is found, most LP software packages provide, for each resource corresponding to a constraint, a shadow price along with a range within which this price is valid.

    The shadow price determines the change in the optimal value of the objective function that results when one additional unit of this resource is available.

    It is used to determine if it is profitable to acquire additional resources.

    THE TRANSPORTATION PROBLEM

    A special class of LP problems, which is common to many companies, deals with the shipping of finished goods directly from plants to retail outlets at minimum cost. Such a distribution problem is known as the TRANSPORTATION PROBLEM.

    Example. MARS Computers Company has 3 microcomputer assembly plants. The microcomputers are made in three plants: San Francisco, Los Angeles and Phoenix. Each of these plants have their own monthly production capacities. The microcomputers are sold through four retail stores, which require a specific monthly Demand. The costs of shipping one microcomputer from each assembly plant to each of the 4 retail stores, along with the plants' capacities and retail stores' demands are given in the following table.

    Plants

    Shipping cost ( $/unit) from plants to Retail Stores

    Plant Capacity

    , units

    San Diego

    Barstow

    Tucson

    Dallas

    San Francisco

    5

    3

    2

    6

    1700

    Los Angeles

    4

    7

    8

    10

    2000

    Phoenix

    6

    5

    3

    8

    1700

    Demand, units

    1700

    1000

    1500

    1200

     

    Formulate and solve a linear programming model for finding the least cost shipping routes.

    Solution: First we need to define the decision variables.

    Variable Definition. We define a variable for each pair of sources and destinations, as given in the following table:

     

    Decision Variables from plants to Retail Stores Plant

    Plants

    San Diego

    Barstow

    Tucson

    Dallas

    San Francisco

    XSS

    XSB

    XST

    XSD

    Los Angeles

    XLS

    XLB

    XLT

    XLD

    Phoenix

    XPS

    XPB

    XPT

    XPD

    Where XSS = number of computers shipped from San Francisco to San Diego.

    Similarly, the remaining 11 variables defines the number of units shipped from a specific plant to a retail store. Please note that the number of variables will always equal to the product of the number of sources (plants) multiplied by the number of destinations (retail stores)

    Using the above definition the transportation model is presented below

    Minimize (5 XSS + 3 XSB + 2 XST + 6 XSD) +

    (4 XLS + 7 XLB + 8 XLT +10 XLD) +

    (6 XPS + 5 XPB+ 3 XPT + 8 XPD)

    Subject to:

    Capacity Constraints

    XSS + XSB + XST + XSD = 1700 (San Francisco)

    XLS + XLB + XLT + XLD = 2000 (Los Angeles)

    XPS + XPB+ XPT + XPD = 1700 (Phoenix)

    Demand Constraints

    XSS + XLS + XPS = 1700 (San Diego)

    XSB + XLB + XPB = 1000 (Barstow)

    XST + XLT + XPT = 1500 (Dallas)

    XSD + XLD + XPD = 1200 (Dallas)

    Logical ( = 0) Constraints

    XSS, XSB, XST, XSD, XLS, XLB, XLT, XLD, XPS, XPB, XPT, XPD = 0 & integer

    The above formulation is an LP model except for the integer requirements on all variables. However, because of its special structure, a transportation model can be solved as an LP model without the integer requirements and the solution will come out as integer.

    Transportation Models in Finding the least cost Location

    Suppose that the current 4 retail stores demand for MARS microcomputers have grown considerably and accordingly MARS would like to build a plant in a new location. MARS explored 2 potential cities as the location for its plant, namely Seattle or Reno. MARS estimated the cost of production per unit in each of the 2 new cities as well as the cost of shipping from each of these two cities to the four retail stores. The following table provides the data for the new demand of the four retail stores as well as the production and shipping costs for Seattle and Reno.

    Plants

    Shipping cost ($/unit) :Plants to Retail Stores

    Production Cost, $/units

    San Diego

    Barstow

    Tucson

    Dallas

    Seattle

    7

    5

    8

    10

    2

    Reno

    5

    4

    3

    7

    4

    New Demand, units

    2000

    1500

    1700

    1800

     

    MARS plans to build the new plant at 1600 units production capacity.

    In order to solve this problem, we need to develop the updated shipping cost table for the two new cities which is given below

    Plants

    Shipping cost ($/unit) from plants to Retail Stores

    Plant Capacity,

    units

    San Diego

    Barstow

    Tucson

    Dallas

    San Francisco

    5

    3

    6

    9

    1700

    Los Angeles

    4

    7

    8

    10

    2000

    Phoenix

    6

    5

    3

    8

    1700

    Seattle

    9

    7

    10

    12

    1600

    Reno

    9

    8

    7

    11

    1600

    Demand, units

    2000

    1500

    1700

    1800

     

    The above updated table is used to formulate two transportation models:

    1. The first includes the three old cities in addition to Seattle as four sources, while keeping all four retail store destinations. This transportation model is solved for the minimum total shipping & production cost, which we will refer to as TSC(Seattle).

    2. In the second model, Seattle is replaced with Reno and the minimum total shipping & production cost, TSC(Reno), is obtained by solving the resulting transportation model.

    3. If TSC(Seattle) is less than TSC(Reno), then Seattle is selected as the location of the new plant. Otherwise, Reno will be the selected site.

    References

    Lee, Sang M. and Marc J. Schniederjans, 1994," Operations Management ". Houghton Mifflin Company, Princeton, New Jersey.

    Mathur, Kamlesh and Daniel Solow, 1994," Management Science: The Art of Decision Making" Prentice Hall, Englewood Cliffs, New Jersey.

    McClain John O., Thomas, L. Joseph and J. B. Mazzola, 1992," Operations Management: Production of Goods and Services ". third Edition, Prentice Hall, Englewood Cliffs, New Jersey.

    In addition, the following periodicals can provide several Operations Management and Operations Research applications

    1. European Journal of Operational Research

    2. INTERFACES, An International Journal of the Institute of Management Sciences and The Operations Research Society of America

    3. The Institute of Industrial Engineering (IIE) Transactions.