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__________________________ Decision Science 1

DECISION SCIENCE

MODULE ONE

REVIEW OF MATHEMATICS

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LESSON 1 FUNCTIONS AND GRAPHS OBJECTIVES At the end of the lesson, you should be able to evaluate and graph a function. DISCUSSION We will review the fundamental Mathematics needed to represent functional relationship and to solve systems or sets of these functions. Our aim is not to provide a comprehensive study of Mathematics, but instead to review several fundamental concepts and the theories that represent the foundations of many of the quantitative techniques included in the later lessons ie., those dealing with forecasting, cost, volume, profit analysis and linear programming. 1.1 CONSTANTS AND VARIABLES Mathematical problems involve two kinds of qualities namely: constants and variables. In any given problem, a constant is a fixed quantity that does not varry or is not fixed in value under a given set of conditions. On the other hand, a variable is a quantity that changes in value under a given set of conditions. 2. Functions A variable, such as y in this example is said to be a function of the other variable x, if to each value of x, there corresponds exactly one value of y. A symbol commonly used to denote a function of x is f(x), read “f-function of x”, or f of x. The functional symbol. F(x) indicates substitutes for x. To evaluate a function, simply substitute the specific value for x and find its value by performing indicated operations. Example: Evaluate y = f(x) = 2x – 3 x = 2 and x = -3. Solution: y = f(2) = (2) -3 =4 – 3 = 1 y = f(-3) = 2 (-3) = -6 – 3 = -9




1.3 GRAPH OF FUNCTION The graph of a function, y = f (x) is a set of all points in the coordinate plain whose coordinate (x, y) satisfy the equation. Example: Draw the graph of the function defined by y = 2x-1 Solution: Assign certain values to x and obtain the corresponding value of y, as shown in the Table. The tabulated values help us to plot some points of the graph and draw a curve passing through these points. The graph is a straight line. Table: x -2 -1 0 1 2 y -5 -3 -1 1 3




SELF EVALUATION

1. Construct the graph of the function defined by y = 6 – 2x 2. Evaluate

f(x) = 2x2 – 3x – 2 when x = 3, x = -4 and x = -1

3. a. Express the total cost, ( C ) as a function of the number of units sold, (x) if the fixed cost,

(F) is P 5,000 and the unit variable cost (v) is P4.

b. Determine C when x = 200 units. c. If the selling price, (p) is P 6, find the total revenue (TR).

4. Graph the total cost ( C ) and total revenue (TR) curves in problem no. 3. 5. Graph the function, f(x) = 2x – 6. Is this different from graphing the equation. 2x – y = 6?




LESSON 2 LINEAR EQUATIONS OBJECTIVES At the of the lesson, you should be able to solve linear equations in one and two variables. Likewise, he should be able to graph these equations. DISCUSSION An equation is a mathematical statement denoting the equality of two algebraic expressions. For instance, if the two algebraic expressions: 5x – 4 and 3x – 6 are equal, the statement can be expressed by the equation. 5x – 4 = 3x – 6 In general, there are two basic types of equations namely: conditional equation and identical equation or identity. A conditional equation is an equation which is true only for certain values of the variables or unknowns. For example, the above equations 5x – 4 = 3x – 6 is true if x = -1, that is if x = -1 is substituted to the equation both sides of the equation will be equal to -9. 5x – 4 = 3x - 6 5(-1) – 4 = 3 (-1) – 6 -5 – 4 = -3 – 6 -9 = -9 An equation which is true for all permissible value of variables or unknown is called an identical equation or identity. Consider the equation: 2 (4x + 1) = 8x + 2




For any arbituary value of x, say x = -2, the left-hand side of the equation reduces to – 14. Likewise, is the same value of the right-side of the equation, that is, 2 (4x = 1) = 8x + 2 2 [4(-2) + 1] = 8 (-2) + 2 2 [-8 + 1] = -16 + 2 -16 + 2 = -16 + 2 -14 = -14 and therefore the statement is true for all variables of x. 2.1 LINEAR EQUATION IN ONE VARIABLE OR UNKNOWN

A linear equation in one variable, say x, is one which is of the first degree in x and expressed in the form. ax + b = 0 where a and b are constants and a = 0. The method of solution is obtained by the application of the axioms of the addition, subtraction, multiplication and division successively until the equation x = k is obtained, where the constant, k, is said to be the root or a solution if it satisfies original equation. Example: Consider the equation. 2 (2x – 1) = 6 (x+2) Find the root of the equation and graph the equation. Solution: 4x – 2 = 6x + 12 Expanding both sides of the equation 4x – 6x = 12 + 2 Transposition -2x = 14 Combining similar terms x = -7 Axion of Division




The root of the equation is x = -7. Check: 2[2(-7) – 1] = 6 [(-7) + 2 Substitution 2[-14 – 1] = 6 [-5] 2[-15] = 6 [-5] -30 = -30 The graph of the equation is a vertical line. 2.2 LINEAR EQUATION IN TWO VARIABLES OR UNKNOWS

An equation that can be reduced to the form ax + by + c = 0 where a, b and c are constants and a and b ?� 0, is a linear equation in two variables, x and y. Example: Graph the equation using ordered pairs ( x, y).

2x + 3y = 2




Solution As explained in Section 1.3 – Graph of function, arbitrarily assign certain values to x and solve for the corresponding value of y. Some of the ordered pairs that will be assigned satisfy the equation are: (1,0), (-2, 2), (-5 4) and (-8, 6) The graph of the equation 2x + 3y = 2 is a straight line. 2.3 INTERCEPT METHOD

The easiest and straightforward approach to graph a linear function is by the use of the intercept method. This is done by joining the x - intercept with the y – intercept. The x – intercept is the point where the line crosses the x – axis. This is obtained by setting y = 0, and solve for the value of x. On the other hand, the y – intercept is the point where the line passes the y – axis. This is obtained by setting x = 0, and solve for the value of y. Example: Determine the graph of the equation 2x – y = 8 using the intercept method. Solution: x – intercept 2x – 0 = 8 x 4 0 2x = 8 y 0 -8 x = 4




y – intercept 2(0) –y = 8 y = -8 The graph of the equation using the intercept method.

y




SELF – EVALUATION

1. Find the root of the equation. Graph the equation

a. 3y – 5 = -17 b. -8 = -5x - 3

2. Determine the graph of the equation 2x – y = 4 using:

a. ordered pairs (at least 4 points) b. intercept method

3. Determine whether the point lies on the graph or is outside the graph of the given equation.

a. 3x – y = 5 (0, -5), (1, 2), (1, -2) b. 3x + y = 5 (1, 2), (-1, 8), (1, -2)

4. What is the graph of a linear equation in one variable? In two variables? How many points

are needed to determine the graph of a linear equation? 5. Plot the following linear equations. You may use any of the two methods.

a. 6x + 3y = 36 b. 2x – 3y = 21




LESSON 3 SYSTEM OF LINEAR EQUATIONS OBJECTIVES At the end of the lesson, you should be able to understand the different methods of solving algebraically the system of linear equations in two unknowns. 2. determine the graphical solution of system of linear equations and inequalities A system of linear equations in two variables x and y consists of a pair of linear equations such as: a1x + b1y = c1 a2x + b2y = c2 where a1, b1, c1, a2, b2, and c2, are constants. a1, b1, c1, a2, b2, and c2 = 0 The solution of the system defined by the pair of linear equations is a set of values of x and y that satisfy both equations. In general, there are four different methods of solving the system of equation in two unknown namely:

1. Algebraic solution

1.1 Elimination by addition 1.2 Elimination by subtraction 1.3 Elimination by substitution

2. Graphical solution




1. ALGEBRAIC SOLUTION OF A SYSTEM OF LINEAR EQUATION

1.1 ELIMINATION BY ADDTION Example: Solve the system of linear equation x + y = 9 Eq. 1 x – y = 3 Eq. 2 Solution Using addition: x + y = 9 x – y = 3 2x = 12 x = 6 Then, substitute the value of x in either Eq. 1 or Eq. 2 and solve for the value of the other variable, y. Using Eq. 1 x + y = 9 6 + y = 9 y = 3 Hence, the solution set that satisfies both equations is the ordered pair. ( 6, 3). Check: By substituting the values of x and y in the original equations x + y = 9 Eq. 1 x – y = 3 Eq.2 6 + 3 = 9 6 – 3 = 3 9 = 9 3 = 3




1.2 ELIMINATION BY SUBTRACTION

Example: Solve the system of linear equation x + y = 9 Eq. 1 x – y = 3 Eq. 2 Solution: using subtraction x + y = 9 X: - 1 ( x – y = 3 ) Multiplying by -1. x + y = 9 -x + y = -3 2y = 6 Y = 3 Then, substitute the value of y in either Eq. 1 of Eq. 2, and solve the remaining variable, x. using Eq. 1

x + y = 9 x + 3 = 9 x = 6 Hence, the solution set that satisfies both equations is the ordered pair ( 6, 3). Check: same as shown in Elimination by Addition.




1.3 ELIMINATION BY SUBSTITUTION Example: Solve the system of linear equation x + y = 9 Eq. 1 x – y = 3 Eq. 2 Using substitution: Solve for y in terms of x in either Eq. 1 or Eq. 2. Choosing Eq. 1 x + y = 9 y = 9 – x Substituting the value of y in Eq. 2 and solve for the other variable, x. x – y = 3 x – (9 – x) = 3 x – 9 + x = 3 2x = 3 + 9 2x = 12 2 2 x = 6

Then, substitute the value of x in either Eq. 1 or Eq. 2 and solve for the other variable, y. using Eq. 1. x + y = 9 6 + y = 9 y = 3 Hence, the solution set that satisfies both equations is the ordered pair (6,3). Check: as shown earlier.




2. GRAPHICAL SOLUTION OF A SYSTEM OF LINEAR EQUATION Earlier, we have learned the different approaches of solving system of equation in two unknowns using the algebraic method. We have also studied the graph of the function defined by a linear equation in one variable. Now, we apply the graphical approach in solving the system of linear equation in two variables. Example: Graph the system of linear equation using the intercept method. Solution: x + y = 3 Eq. 1 x – y = 3 Eq. 2 Using the intercept method: x + y = 3 Eq.1 x 0 3 y 3 0 x – y = 3 Eq. 2 x 0 3 y -3 0 Graph the linear equations on the same curtesian plane. Determine the point of intersection of the two straight lines representing the equations x + y = 3 and x – y = 3 using any of the three different approaches of algebraic solution, which is (3, 0). Or, the intersection point can be obtained as shown:




3. GRAPHICAL SOLUTION OF SYSTEM OF INEQUALITIES Inequalities are another type of relationship expressed in mathematical form. Most constraints in Linear Programming are expressed as inequalities, ie., the machine hours used is limited to 75 hours. (< 75 hours), or at least 60 units of ingredients (> 60 units). The solution set for system of inequalities in two variables is a plane. Example: Graph the solution set of the system of inequalities: x + 2y =� 6 x – y =� -6 Solution: Using the intercept method x + 2y =� 6 x 0 6 y 3 0

The solution set is a plane bounded by lines x – 2y = -6 and x + 2y = 6 as shown in the shaded area.




SELF-EVALUATION 1. What is the difference between a linear equation in one variable from linear equation in two

variables? Differentiate the solution set of linear equation in one unknown from the solution set of a linear equation in two unknowns.

2. Solve the system of linear equations using any of the three algebraic methods.

2y + x = 6 Eq. 1 3x – y = 4 Eq. 2

3. Graph the system of linear equations using the intercept method.

4x + 2y = 5 Eq. 1 x – 2y = 3 Eq. 2

4. Solve algebraically the system of equation using elimination by substitution.

4x + 3y = 12 Eq. 1 3x – 4y = 12 Eq. 2

5. Graph the solution set of the following system of inequalities.

2x – 4y =� 16 3x + 9y < 27 Shade their common area that will satisfy both inequalities.

Submit your written reports to your professor.




DECISION SCIENCE

MODULE TWO

BUSINESS FORECASTING




LESSON 1 FORECASTING METHODS OBJECTIVES At the end of the lesson, you should be able to: 1. familiarize yourselves on the different forecasting methods 2. select the most appropriate forecasting method DISCUSSION The term forecasting is used interchangeably with the term prediction. A forecast is a prediction, estimate or determination of some future values based on a certain set of factors. Forecasting deals with expectation of the future, that is, what the manager thinks is likely to happen during the next planning year. When the decision makers are forced to estimate future outcomes such as product demand, number of customers, return on investments, interest rates, market prices and other variables, they are forecasting these outcomes. In general, there are seven different categories of forecasting methods namely: 1. Time –Series Method – makes use of historical data to generate a forecast of the future. Future forecast are based from analysis of past values of variables being forecasted. 2. Casual Method – takes into consideration the factors which are related to that which is being predicted. For example, in forecasting future sales of a firm, competitor’s prices, promotional expenses and other factors that may influence demand are considered.




3. Short-Term Forecasting Method – have a time frame of one day to one month. For instance, the typical routinary business operations i.e., authorization of overtime, granting of sick leaves, manning schedules, etc. 4. Medium – Term Forecasting Method – have a time horizon from one month to one year. For example, in deciding what and how many resources are required ie., number of casual employees to hire to meet the projected increase in demand of a product for the next Christmas season. 5. Long-Term Forecasting Method – have a time frame of over one year. For example, plant and office relocation into a new site, product innovations and automatic of production facilities. Long term forecasting is most vital of all the forecast since it aids management in their strategic planning. However, it is the most difficult, since greater risk and uncertainty is involved. 6. Qualitative Methods – are generally used for long – term, highly uncertain situations in which the use of mathematical model does not seem suitable. For example, in attempting to predict the political development in Mindanao, our economic recovery and technological advancement in science is so filled with approaches that are qualitative rather than quantitative. 7. Quantitative Method – utilizes mathematical methods to indicate the relationship between the variables inorder to make a forecast.




SELF EVALUATION

1. Discuss which forecasting methods are most appropriate with certain time horizon. 2. Under what situations would you recommend the use of qualitative method of forecasting as

against quantitative method. Explain. 3. Suggest an appropriate forecasting method of the following: A. The demand of Christmas lights sets at SM Department Stores this coming Yultide season. B. The expected numbers of customer complaints received next year with Quality Assurance

Department. C. Monthly Compact Disc sales next year at the record outlet in a shopping mall. D. The volatility of the Philippine peso as against the U.S. dollar.




LESSON 2 TIME – SERIES METHOD OBJECTIVES At the end of the lesson, you should be able to use the different Time Series method of Simple Moving Averages, Weighted Moving Averages and Exponential Smoothing in developing forecast of past data for future time periods. DISCUSSION Time series is a historical record or a list of behavior of a variable over time. Time-Series method uses past historical data to generate a prediction of the future. It is implied under time-Series method that what has happened in the past is likely to happen in the future. In making Time-Series analysis, one must look for trend, seasonal, cyclical and random component. The trend component indicates a long – range general movement either in an upward or downward direction over time. Our average population annual growth rate of 2.2% shows an upward trend. The seasonal component reflects changes either up or down occurring with a period of a year or less. As an example, Baguio and Boracay have peak tourist seasons during the summer months. The cyclical components are reflected as a pattern of changes of fixed point in time of duration of more than a year. For example, the El Niño phenomenon hits the Philippines every 5 years resulting into drought and intense heat. Another example is the high consumer spending everytime we have a local and national elections.




The final component of Time-Series analysis is random variation. The ramdom component is what remains after all the other components have been considered. It is referred to as the remaining unexplained noise. The Time-Series analysis is commonly used for both short-term and medium-term forecasting. Included in the Time-Series methods are:

1. Simple Moving Averages 2. Weighted Moving Averages 3. Exponential Smoothing

1. SIMPLE MOVING AVERAGES It is a simple and useful forecasting mode. The average of what has occurred in the past is used to forecast the future. The moving average method use the average of the most recent observation in the time series as forecast in the next period. Everytime a new observation becomes available, the average changes or moves to include the newest observation and drops the oldest observation which was previously used. Mathematically, the moving average calculations is made as follows: N S� xi i = 1 n x1 + x2 + x3 +…. + xn Ft = where Ft = moving average, the forecasted value of x in time period, t. xi = observation or data in period, i. n = number of periods in the changing group

Ft =

n




Example 1 A semi-conductor from wants to forecast demand for its computer chips. Shown below is the demand over the last 6 months. Use a 3 – month moving average to develop forecast for the following. Month Demand (1,000 units) January 127 February 111 March 130 April 142 May 135 June 140 Solution The 3 – month moving average forecast starts for April: 130 + 111 + 127 368 3 3 The 3 - month moving average forecast for July. 140 + 135 + 142 417 3 3 2. WEIGHTED MOVING AVERAGE A weighted moving average is the same as the simple moving average except that it assigns more weight to the recent data to reflect more recent variations in the data in a time series. The formula for the weighted moving average is n S� Wi Xi i=l WFt =

F April = =

F July = = = 139




where: Wi = weight for the period, i Xi = observation or data in the period, i Example 2 Using Example 1, compute the weighted average forecast using weight of 0.40 for the most recent period, 0.30 for the next more recent and 0.20 for the next. Solution The weighted moving average forecast of July: WF = 0.40 (140 + 0.30 (135) + 0.20 ( 142 ) July = 56 + 40.5 + 28.4 = 124.9 3. EXPONENTIAL SMOOTHING Exponential Smoothing is a forecasting technique that uses a weighted average of past time series values to forecast the value of the time series in the next period. The basic exponential smoothing model is as follows: F t+1 = a� Xt + ( 1 – a� ) Ft where: F t+1 = forecast for the next period Xt = actual data in the present period F = forecast for the present period a�t = smoothing constant, with a value between 0 and 1




Example 3 Using Example 1 again, compute the monthly forecast using a smoothing factor, a� = 0.30 Solution The exponential smoothing monthly forecast for period 2 (February) is: a�) F1 F = a� ( X1 ) + ( 1 - 2

F = 0.30 (127) + (1 – 0.30) (127) 2

= 38.1 + 88.9 F = 127 2

The exponential smoothing forecast for period 3 (March) is: a� ) F2 F = a� ( X2 ) + ( 1 - 3

= 0.30 (111) + (0.70) (111) F = 111 3




SELF – EVALUATION 1. Consider the following monthly car sales of dealer:

Period: 1 2 3 4 5 6 7 Month: Jan Feb Mar Apr May June July Sales: 9 11 15 14 16 15 18 (units)

a. Compute a weighted average forecast using a weight of 0.40 for the most recent period, 0.30 for the next most recent 0.20 for the next, and 0.10 for the next.

b. If the actual sales for period 8 is 20 units, forecast sales for period 8 using the same weights

as in part a. 2. A bakeshop observe the following sales (P 000) pattern in the past weeks.

Week 1 2 3 4 5 6 7 Actual 40 45 52 48 53 55 62 sales (P 000)

a. Using a 3 – week moving average forecast, what should be the sales forecast for the 8th

week? b. Compute a 3 – week moving average forecast for the last 7 weeks.

3. Use exponential smoothing to develop a series of forecast for the following sales figure, and

compute the error for each period. ( ERROR = Actual – Forecast)

a. Use a� = 0.20 b. Use a� = 0.30 c. Plot the given data and both sets of forecast on a single graph.




PERIOD MONTH ACTUAL SALES (Units) 1 Jan 25 2 Feb 23 3 Mar 26 4 Apr 28 5 May 25 6 June 30 7 July 32 8 Aug 27 9 Sept 35




LESSON 3 METHODS OF LEAST SQUARE OBJECTIVES At the end of the lesson, you should be able to: 1. determine the equation of a regression line by the Method of Least Square. 2. know how to use the equation in forecasting the value of the dependent variable. DISCUSSION The linear relationship between the two variables is best represented if we plot them. The simplest method to describe the relationship is to locate the line of the best fit on the graph with the combined use of judgement and guesswork. For a given historical data or values, X1, X2, X3, … Xn corresponding to time periods 1, 2, 3 ….. n, we could plot these points ( x, y) on a graph, where the time periods are plotted on the horizontal x – axis and the values on the vertical, y – axis. Fitting a straight line through these given points (x, y) would not exactly pass through every point, or the fitted line would deviate from the plotted points. A mathematical approach in the Time – Series forecasting is the use of regression analysis. Regression analysis is the method of fitting a linear equation to a set of data points in order to obtain the line of best fit. The Method of Least Square is an objective or quantitative approach to locate the linear regression line. The regression line is the line of best fit to a set of data points which describe the relationship between two variables, that is the line that minimize the sum of the squared deviation. The straight line equation of the regression line is of the form: Y = a + bX Eq. 1




where: Y = forecasted value of the dependent variable a = y – intercept of the line of best fit b = slope of the line of best fit x = given value of independent variable, which is the time period for which the forecast is being made. The slope, b is computed by: S�XY - n X ·� Y

S� x2 - n X2 And the intercept, a is given by: a = Y - bX Eq. 3 where: S� = summation sign n = number of paired of observations X, Y = average, value of x and y respectively and is computed as follows: S�x S�y n n

b = Eq. 2

X = Y =




Example: A manufacturing firm reviews its production records for the last 3 years. Annual volume of outputs and production costs for the firm are given as follows: Year Output (1,000 units) Production Cost (P1,000) 1 1 4 2 3 5 3 5 9 Required: 1. Construct a graph depicting the relationship between output and this cost. Locate best fit on the

basis of your judgement. 2. Determine the equation of regression line by the Least – Square method. 3. Construct the graph of the regression line. 4. Find the production cost if the future output is expected to be 8,000 units.




Solution: 1. Draw the line of best fit on the basis of your judgement. 2. Tabulate the given data

Year

Output

(X)

Cost (Y)

XY

X2

Y2

1 2 3

1 3 5

4 5 9

4

15

45

1 9

25

16

25

81

S�X = 9 S�Y = 18 S�XY = 64 S�X2 = 35 S�Y2 = 122




Substituting the preceding summations and n = 3 in Equation 2 and 3, we obtain the following: S�XY - n X Y 64 - 3 ( 3 ) ( 6 ) 9 35 – 3 ( 3 ) 3 18 64 – 54 3 35 - 27 10 8 b = 1.25 a = Y – bX = 6 - 125 ( 3 ) Eq. 1 = 6 - 3.75 a = 2.25 Therefore, the graph of the regression line is: Y = a + bx Y = 2.25 + 1.25x 9 5 3. Draw the graph of the linear regression line, y = 2.25 + 1.25 x, using the Intercept Method.

x 0 -1.8

y 2.25 0

Please return to solution # 1 for the graph 4. Production cost if future output, x = 8,000 units

Y = 2.25 + 1.25 ( 8 ) = 2.25 + 10 Y = 12.25 or P 12,250

S� X2 - n X2 b = = = =

Eq. 2 where:

X = = 3 Y = = 6

4 4 or Y = x




SELF-EVALUATION

1. The following data represents grade that an MBA student obtained on a Decision Science Midterm exam and the number of hours spent preparing for the examination.

Number of Midterm Study time (hrs) Grade 4 65

5 83 6 78 7 84 8 88 9 90 10 95 Required:

1. Develop a mathematical model equation to forecast the midterm grade an MBA student will obtain based on the length of time spent on studying.

2. Plot the given points and draw the line of best fit on the basis of your judgement. 3. what is the expected grade an MBA student will receive if he studies for 6.5 hours. 4. Draw the graph of the regressive line.




2. The following data were collected by an office clerk during a study of business establishments applying

Week: 1

2

3

4

5

6

7

8

Permit 40 Applications:

42

45

38

41

43

45

47

Revenues 55 (Permit Fees)

60

70

50

56

68

71

75

The Electrical Division believes that weekly revenues from Permit Fees is related to the number of electrical permit applications from business establishments. Required: a. Construct a graph showing the linear relationship between applied electrical permits and

revenues from permit fees. Plot the line of best fit on the basis of your judgement. b. Using the Method of Least Square, determine the Regression Line equation that fits the data. c. How much revenues will Electrical Divisions earn if the electrical permit applications are 52? d. Show the graph of the Regression Line.





DECISION SCIENCE

MODULE THREE

COST, VOLUME, PROFIT ANALYSIS




LESSON 1 THE CONCEPT OF CONTRIBUTION OBJECTIVES At the end of the lesson, you should be able to: 1. understand the behavior of cost and revenue. 2. apply the concept of contribution in the cost, profit, volume analysis. DISCUSSION Most basic quantitative models in business applications involve the relationship between volume and cost, revenue and profit. The analysis of these relationship is frequently referred to as cost, profit, volume analysis. • THE BEHAVIOR OR COST

As a starting point of our analysis with basic cost behavior, it is essential to distinguish among fixed cost, unit variable cost and total variable cost. Fixed costs are cost incurred regardless of the number of units produced. They remain constant at different levels of output. Example of fixed costs include insurance, real property taxes, building / office rentals and management’s salaries. Total variable cost is that portion total cost that varies proportionality with the volume of output. Example of total variable cost which increases with the level of output are direct materials, direct labor and selling expenses. Total variable cost (TVC) are obtained by multiplying the unit variable cost (v) by the number of units produced or volume of output. (x)




Total Cost are the sum of fixed costs and total variable costs. Figure 3.1 shows the behavior of these costs in relation to the volume of output. Volume of output (units)

Fig. 3.1 Cost Behavior

Total Cost (TC)

Total Variable Cost (TVC)

Fixed Cost (FC)

Cost (PHP)




Algebraically, the cost behavior can be determined as follows: TC = FC + TVC or TC = FC + vx Where TC = total cost FC = fixed cost TVC = total variable cost v = unit variable cost x = value of output or number of units produced. THE BEHAVIOR OF REVENUE As mentioned earlier, total revenue from sales is related to the volume of output. The more units of goods or services produced and sold, the larger are the total revenues. Like total variable cost, total revenue varies proportionately with output. Total revenue can be determined by multiplying the unit selling price of good or service by the number of units produced or created. As shown in Fig. 3.2, the behavior of total revenue in relation to volume of output.

Fig. 3.2 Total Revenue Behaviour Mathematically, this can be expressed as follows: x TR = p where: TR = total revenue p = unit selling price x = volume of output or number of units produced

Total Revenue (TR)

Volume of output (units)

Revenue (PHP)




THE CONCEPT OF CONTRIBUTION The concept of contribution, the difference between unit selling price and unit variable cost, is a major element in the cost, volume, profit, analysis. It is what remains after deducting variable cost from selling price per unit towards paying off the total fixed costs. Once the contribution equals the total fixed costs, the manufacturer has reached the break-even point or no profit or no loss. When contribution exceeds the total fixed cost, the manufacturer earns a profit. Conversely, had firm contribution fall short of the total fixed cost, then the firm experiences a loss. Example The manufacturer sell its products at P 10 per unit and the product has a unit variable cost of P 8. Determine the contribution to cover the total fixed cost, if the sales volume is 10,000 units and total fixed cost of P 15,000. How much profit, if any will be earned? Solution: TR = P 10 x 10,000 = P 100,000 TVC = P 8 x 10,000 = P 80,000 By Substitution: Total Revenue P 100,000 Total Variable Cost - 80,000 Contribution 20,000 Total Fixed Cost - 15,000 Profit P 5,000 The contribution of P 20,000 exceeds the total fixed cost of P 15,000 and a profit of P 5,000 is realized.




SELF-EVALUTION 1. A sari-sari store owner buys particular canned goods at P 20 per can and sells them P 25 per can.

His fixed cost is P 5,000. Find the number of units to sell to cover the fixed cost. 2. A manufacturing firm sells its product at P 30 per item. It incurred a fixed cost of P 15,000 and a

variable cost of P 20 per item. The projected sales volume is 10,000 items for the year. Determine the following:

a. The contribution per unit b. The contribution towards paying off the total fixed cost c. Profit at the projected sales volume

3. Discuss the effect on contribution to cover the total fixed cost, if the total variable cost is

increased by 10% assuming total revenue as constant. 4. A manufacturer realizes a sales volume of 500,000 units at a selling price of P 2 per unit.

a. If the firm incurs a total fixed cost and total variable cost of 20% and 75% of the total revenue respectively, find the profit earned, if any.

b. If the total fixed cost is increased by an additional 505 what profit will the firm realize?




LESSON 2 COST, VOLUME, PROFIT ANALYSIS OBJECTIVES At the end of the lesson, you should be able to learn the concept of Break-Even Point and its application in business DISCUSSION The firm’s profits (or losses) are determined by the relationship between total revenue and total cost. Accountants deal primarily with cost and revenue data. They have studied relationships which tend to capture the nature of cost and revenue behaviour. The analysis of these accounting – type relationships is referred to as Cost, Volume, Profit Analysis. It represents an important type of quantitative analysis. Cost, volume, profit analysis help the manager in his decision-making in relation to changes in sales and costs to changes in profits. It shows how revenue, variable cost, fixed cost and volume of output or number of units of products made or sole determine the profitability of a firm. The cost, volume, profit relationship as shown in a break-even chart help the manager predict what effects certain decisions would have on profits. We will study both the algebraic and graphical approaches to Cost, Volume, Profit Analysis. The algebraic or mathematical approach provides the accuracy and logic of Mathematics and the graphical approach adds a quick simple understanding with the aid of a chart.




The total revenue ( TR ) is obtained by multiplying the unit selling price ( p ) by the number of units sold or volume ( x ). Total cost (TC) is equal to the sum of fixed cost (FC) and the total variable cost (TVC). Total variable cost is equal to unit variable cost ( v ) times volume ( x ). Fixed Cost (FC) tends to remain constant regardless of the level of volumes or number of unit of products sold. Total variable cost changes in amount as volume or output changes. Since profit (Pr) is total revenue (TR) minus Total Cost (TC), therefore: TR = p.x TC = FC + TVC where: TVC = v.x TC = FC + v.x Hence, Pr = TR – TC Pr = px – [ FC + v x ] Thus, Pr = (p - v) x – FC




ALGEBRAIC METHOD The most popular of the Cost, Volume, Profit Analysis applications is the Break-Even Analysis. We will consider the simplest case of Break-Even analysis, one in which all of the cost and revenues are linear. The break-even point can be derived mathematically. To calculate the Break-Even point, it is necessary to know its fixed costs (FC), its total variable costs (TVC) and its sales volume (x). To determine the Break-Even point in units (BEPx), simply equate the total revenue (TR) to total cost (TC) and solve for the volume of output (x). TR = TC px = FC + vx Solving for x: px - vx = FC x = where: x = BEPx, the Break-Even point in units. Hence, the formulas for Break-Even point in units (BEPx) and in pesos (BEPp) is as follows: BEPx ( units ) = BEPp ( pesos ) = GRAPHICAL METHOD The volume or level of operation at which total revenue (TR) and total cost (TC) are exactly equal is referred to as the break-even point (BEPx). The Break-Even point is reached when TR and TC curves intersect. At this point, the firm neither makes any profit (Pr = 0) or loss money, or simply the firm just break-even. When the volume is greater than break-even point, ie., by selling just one more unit above the break-even point, then the firm realizes a profit.

FC p - v

TFC p - v

TFC 1 – v/p




On the other hand, when the volume is less than break-even point, ie. Selling one unit less, the firm experiences a loss for that period. A typical representation of break-even point is shown in the Break-Even chart below.

Example 1. A toy manufacturers produces and sells its products at P 25 per unit. The product has a variable

costs of P 5 per unit and the fixed cost is P 10,000. Determine each of the following:

a. Total revenue, total cost and profit functions b. Sales volume when profit is P30,000 c. The break-even quantity and revenue (pesos) d. Profit when sales are 550 units e. The amount by which the variable cost per unit has to be increased or decreased in

order to Break-Even at 600 units.




Solution: Let x = volume or number of units to produce and sell.

a. Total revenue function, TR

TR = 25x Total cost function, TC TC = 10,000 + 5x Profit function, Pr Pr = TR – TC = 25x – [ 10,000 + 5x ] Pr = 20x – 10,000

b. If Pr = P 30,000, find x.

30,000 = 20x – 10,000 20x = 30,000 + 10,000 20x = 40,000 x = 2,000 units

c. At break-even point, Pr = O; hence

TR = TC 25x = 10,000 + 5x 25x - 5x = 10,000 20x = 10,000 x = 500 units (BEP quantity)

Since, TR = 25x TR = 25 ( 500 ) TR = P 12,500 (BEP revenue)

d. If x = 550 units, find Pr.

Pr = 20x – 10,000 Pr = 20(550) – 10,000 Pr = 11,000 – 10,000 Pr = P 1,000




e. If BEP quantity = 600 unit, find, the new variable cost per unit

25x – vx = 10,000 25( 600 ) – v ( 600 ) = 10,000 15,000 – 600 v = 10,000 600 v = 5,000

v = P 8.30 / unit; the new variable cost. As compared to the given v = P 5.00 / unit, there is an increased of P 3.30 per unit. GRAPHICAL APPROACH The relationship between revenue, fixed cost, variable cost and volume is determined by use of cost, volume, profit analysis. The break-even chart, a graphical approach to cost, volume, profit analysis is a useful tool in aiding managers estimate the profitability of a firm. Example: Prepare a break-even chart for the toy manufacturer showing the graph of revenue and total cost of functions. Determine the break-even quantity (BEPx) and revenue (BEPp) from the chart Solution Draw the graph of the total cost (TC) and total revenue (TR) functions respectively using the intercept method. TR = 25x x 0 200

TR 0 5,000

TC = 10,000 + 5x x 0 400

TC 10,000 12,000




SELF EVALUATION 1. Snow-White bakeshop is considering producing and marketing a new type of cake. The fixed

cost associated with the production and marketing of the cake are P 500,000 and the per unit variable costs total of P 30 per cake. If the cake is sold to customers at P 80 per unit, what is Snow-White bakeshop break-even point quantity, in units? In pesos revenues?

2. If the unit variable cost is P 40 per cake, what would be the effect on the break-even point

quantity, expressed in units in problems no. 1.? 3. Most firms are forced to reduce their prices, if they want to sell beyond a certain volume. Can a

firm sell any quantity they may want at a given price? The objective of the firm is to operate profitability.

4. PAA, Hoisery, Inc. produces men’s socks at a cost of P 30 per unit and sells them at P 50 each.

The monthly cost of production totals P 10,000. Determine each of the following:

a. The total revenue, total cost and profit functions. b. The break-even point quantity and revenue.

c. If the selling price is raised to P 75 per socks. Find the new break-even point (Fixed

cost and total variable cost remain constant)

d. If the selling prices is lowered t o P 35 per socks, however, an additional overhead cost is incurred, is this decision beneficial to management?




LESSON 3 BREAK-EVEN CHARTS AND DECISION MAKING OBJECTIVES At the end of the lesson, you should be able to expected to use Break-Even charts in managerial decision making. DISCUSSION The Break-Even chart is a useful tool in analyzing managerial problems. Such a chart shows how much business a firm needs to do inorder to break-even financially. It also shows the amount of profit and loss the firm would earn or suffer at different volume of sales or output above and below the Break-Even point. Probably, most manufacturing firms must operate at above 60 percent of their capacity to break-even. Break-Even charts are “if” charts. They say if the volume of sales or output is this, then the costs and profits will be that. Break-Even charts, illustrate the costs and profits as well as losses in search for various volume of sales or output.




The Make-or-Buy Decisions Manufacturing firms making certain components or items which are part of their finish products or of purchasing from our outside sources. In make or buy decision a number of factors are usually considered namely: 1. Available Capacity. If the manufacturer has the available capacity and if it is practical to

produce an item or component, since the additional cost would be relatively smaller than buying the item.

2. The Nature of Demand. When the demand for an item is high and steady, it is better for the

manufacturer to do the work itself. On the other hand, if demand is low and wildly fluctuates, then it is economical to purchase from an outside source.

3. Quality Requirements, Special quality considerations or the ability of a manufacturer to closely

monitor quality may require a firm to perform the work itself. 4. Expertise. If the firm lacks the skills and experience manufacturing the item, purchasing might

be a reasonable option. 5. Cost consideration. Any cost savings obtained from making or buying must be weighed against

available capacity, demand, quality and expertise. Any cost savings generated from the item itself or operational cost savings must be considered.

Break-Even charts or crossover charts can also be used as a tool in the selection from among alternative manufacturing process by comparing the different methods of doing work. For instance, simple machines are usually easy to set but slow and costly to operate. Larger qualities allow for the use of faster machines, which has high set-up cost but once set up are much less costly to operate. Each of the alternative process is the most economical for a certain range of volume. Crossover points can be found by equating the pair of equation, that have a common intersection point.




Example: A manufacturer of motor runs small bushings into one of his production line. At present, he purchases bushing from a supplier at P 24 each. If he plans to produce them, it will cost him an additional P 30,000 annually to his fixed cost, and he estimates his total variable cost per bushing would be P 9.00. Should the manufacture purchase or produce the bushing? Constract the Brea-Even chart.




SELF-EVALUATION 1. The CEO of a medium-size manufacturing firm reviews the break-even chart which his

controller has just submitted and asks: “Is there a way lowering the break-even point?” How should the controller answer the CEO?

2. A manufacturer is making a decision on whether to buy or make a certain product. The

purchasing costs P 3.10 a unit for order up to 10,000 units and P 2.60 a unit for orders exceeding 10,000 units. Making the products is estimated to cost the manufacturer P 7,000 for setting up the machines and the variable costs is P 2 per unit. At what range of volume in the cross-over charts will it be economical for the manufacturer to purchase the product and at what point it will pay to make them. Construct a break-even chart.

3. Recently, the manufacturer of motor has projected his production to reach 3,500 units annually

next year due to an increase in demand of motors in the export market. He can purchase imported bushing at P 30 each. Or, he can use his experience and acquire an automatic machine in producing his own bushing. If he plans to make them, it will cost him an additional P 50,000 to his fixed costs and estimate his variable costs at P 10 per unit.

Required: a. It is advisable for the manufacturer to produce its own bushing to meet the production

requirements of the 3,000 motors? b. At what range is it economical to produce their bushing requirements? c. Construct a break-even chart.

4. A manufacturing firm has three methods to choose from as to the type of machines suitable for

making gears: 1) Lathe machines are general purpose machine that will add P 2,000 annually to his fixed costs but will incur P 4 per unit to cover the costs of machine’s operation, including labor, electricity, depreciation and other costs. Such machines are easy to set up for new jobs, but not very efficient in production; 2) Turret lathes requires more set up time in costs P 8,000 a year and a unit variable costs of P 2.00; 3) Automatic machines take long hours to set up amounting to P 20,000 yearly but will produce a low unit variable costs of only 40 centavos. Determine which machines would be most economical for the manufacturer to operate. Construct a cross over chart.




DECISION SCIENCE

MODULE FOUR

LINEAR PROGRAMMING




LESSON 1 FORMULATION OF LINEAR PROGRAMING MODEL OBJECTIVES At the end of the lesson, you should be able to learn to formulate linear programming models in the mathematical form. DISCUSSION Many problems confronting both personal and business firms can be classified as constrained optimization problems. In these problems, first we have to deal with an objective to be optimized or achieved, and second we are subjected to set of constraints which has to be strictly followed. A technique called Linear Programming is now widely used to solve constrained optimization problem. Linear Programming is a quantitative technique to aid managers for finding the best allocation or use of the firm’s limited resources such as labor, capital or energy under certain restraining conditions. Linear Programming involves a sequence of steps that will lead to an optional solution so long as the solution is feasible. There are two general-purpose approaches in solving linear programming problems namely: graphical method and simplex method. Linear refers to the relationship between two or more variables. Programming is the utilization of certain mathematical techniques to determine the best solution to a problem involving scarce or limited resources. LINEAR PROGRAMMING MODELS Linear programming models are mathematical representations of constrained optimization problems. They have certain common attributes, which could be grouped into two main classifications namely: components and assumptions.




1. Objective Objective is the goal of a linear programming model, which are: maximization and minimization.

A maximization objective includes profit, revenues, efficiency or rate of return. On the other hand, a minimization objective involves cost, time, waste or distance. An objective function is expressed in terms of profit per unit of output, or cost per unit of input.

2. Decision Variables Decision variables are Alternative Courses of Actions ( ACA’s) available to the decision maker

ie., the various combination of inputs to minimize total cost, or combination of output to maximize profits.

3. Constraints Constraints are the limited supply of resources that restrict the alternatives available to the

decision maker. Constraints in linear programming can be expressed as inequalities, ie., greater than or equal to ( =� ), less than or equal to ( =� ) and simply equal to ( = ).

A ( =� ) constraints implies the minimum or the lower limit that must be attained in the final

solution ie., at least 25 kw per hour of electricity generated. A ( =� ) constraint states the maximum or upper limit on the amount of scarce resources ie. 50 hours machine time available. The ( = ) constant specifies exactly what the decision variable should equal. ie., produce 300 units of product x. The number of constraints in linear programming model is unrestricted.

4. Parameters Parameter are mathematical statement of objectives and constraints that consists of symbols, ie.,

x, y that represents the decision variables and numerical constants. There are four basic assumptions that must be satisfied for the effective utilization of linear

programming models namely:

1. Linearity All the mathematical relationships among the variables in a given problem can be described in

terms of first degree equations or inequalities. This applies for both the objective function and the constraints.




2. Divisibility

All the non-integer values of decision variables are considered and accepted. 3. Certainty

All the values of the parameters are given and constant. 4. Non-Negativity

Only positive values of decision variables are accepted as answer to the problem, while

negative values are not acceptable. In summary, linear programming is a popular approach to solving problems in which there are limited resources and constraints. All linear programming problems have an objective function and a set of constraints. The major concern of the decision maker is to achieve the objective while staying within the constraints limits.




SELF-EVALUATION 1. Apply linear programming model in your personal life. Make a statement of objective and

constraints that you have to address to solve your personal problem at home or office. 2. Ang Tibay Corporation manufactures and sells electric ovens and electric irons. The demand for

its products is so large that it can sell all it produces. For each electric oven sold, the contribution to profit is P 250; and for each electric iron, the contribution is P 350. Both products must pass through two processing departments stamping and finishing. Each oven requires 2 hours in the stamping department and 1 hour in the finishing department; each electric iron requires 1 hour in the stamping department and 2 hours in the finishing department. The capacity of stamping department is 10,000 worker-hours, and that of the finishing department is 12,000 worker-hours. Set this problem up as a linear programming model in the mathematical form.

3. Under what conditions can linear programming model be used in the solution of allocation

problem?




LESSON 2 LINEAR PROGRAMMING: GRAPHICAL METHOD OBJECTIVES At the end of the lesson, you should be able to apply the graphical method in solving linear programming problems with two variables. DISCUSSION The graphical solution, as the name implies, makes use of graphs to arrive at the optimal solution. The graphical method of solution is limited in the application for finding optimal solutions to two variables. It is useful, however, to illustrate the area of feasible solution and the manner the objective function is used to determine the optimum solution. All linear programming problems can be formulated in a common format. The general procedure followed using the graphical approach is:

1. Identify the unknowns and represent them using the decision variables, represented by convenient letters.

2. Tabulate the given data and the unknowns. 3. Formulate the objective function and constraints in the mathematical format. 4. Graph the constraints. Draw all the constraints as a graph and find the points, which define

the are of feasible solution. 5. Find the optimum solution. At least one of the intersection points on the extreme boundary

will be optional solution. Hence, evaluate those corner points or vertices of the feasible solution.

6. Substitute the coordinates of vertices forming a convex polygon in the objective function. Make a decision by selecting the largest of maximum value, if it is a maximization problem or the smallest or minimum value, if it is a minimization problem.




Example Therese’s Fashion Company produces and sells two kinds of products: leather bags and wallets. There are 96 units of materials and 72 labor hours available. Each bag requires 12 units of materials and 6 labor hours. Wallet uses 8 units of materials each and requires 12 labor hours per wallet. One leather bag contributes P 120 to profit, while a wallet contributes P 100. Determine the number of leather bags and wallets to produce and sell to maximize the profit. Solution I. Tabulate the given data and the unknowns

Bags ( x )

Wallets ( y )

Available resources

Materials ( units )

12

8

96

Labor Used (hours)

6

12

72

Profit Contribution

P 120/unit

P 100/unit

Let x = number of units of leather bags to be produced and sold. Let y = number of units of wallets to be produced and sold II. Objective Function Pmax = 120x + 100 y III. Subject to Constraints a. Explicit constraints Materials used: 12x + 8y =� 96 Labor used: 6x + 12y =� 72 b. Implicit containts X =� 0, y =� 0




N.B

The “less than or equal to” ( =� ) sign in the explicit constraint means the materials and labor consumed in producing bags and wallets must not exceed the materials and labor available, while the “greater than or equal to” ( =� ) sign in the implicit constraint means that real bags and wallets? IV. Graph the constraints Before graphing, first convert the inequalities to an equation by disregarding the ( =� ) or ( =� ) sign of the explicit constraints, then, plot the equation. The implicit constraints lies on the upper-right hand quadrant or the first quadrant called the “feasible solution” because that is the only quadrant in which the solution maybe found. Eq. 1 12x + 8y = 90 Eq. 2 6x + 12y = 72 x 0 8 x 0 12 y 12 0 y 6 0

N.B The hatched region is the area of feasible solution where the optional solution can be found. The extreme boundary of the feasible region is formed by the lines connecting the four corner points or vertices namely: A(0, 0), (0, 6), C(6, 3) and D(8, 0).




Solve for the coordinate (x, y) of the intersection point, C, of the constraint equation. Using the algebraic method: elimination by addition to solve for y. Eq. 1: 12x + 8y = 96 Eq. 2: x(-2) 6x + 12y = 72: Multiply by – 2. 12x + 8y = 96: Eliminate x -12x -24y = -144 - 16y = -48 48 y = 3 To solve for the other variable, x, substitute the value of the y = 3 in the original equation, Eq. 1 Eq. 1 12x + 8y = 96 12x + 8(3) = 96 12x + 24 = 96 12x = 72 x = 6 Hence, the intersection point is C ( 6, 3 ). V. Finding the Optimum Solution

The list of possible optional solution can be narrowed down to just the four corner points or vertices A, B, C, and D. Test the four vertices of the feasible region accordingly by making another table.

PMAX

120 + 100 y

PA (0,0)

120 (0) + 100 (0)

= 0

PB (0,6)

120 (0) + 100 (0)

= P 600

PC (6,3)

120 (6) + 100 (3)

= 720 + 300 = P 1,020

PD (8,0)

120 (8) + 180 (0)

= P 960

y = 16




The vertex C ( 6, 3 ) yields the largest possible profit of P 1,020.00 Therefore, Therese Fashion Company must produce and sell: X = 6 units of leather bags Y = 3 units of leather wallets Check: By substituting the vertex C (6, 3) in both the explicit constraint equations. Eq. 1 Eq. 2 12x + 8y = 96 6x + 12y = 72 12(6) + 8(3) = 96 6(6) + 12 (3) = 72 72 + 24 = 96 36 + 36 = 72 96 = 96 72 = 72 Since the coordinate C (6, 3) satisfy both equation, therefore, the vertex C (6,3) is the optimal solution.




SELF-EVALUATION 1. Solve the following linear programming problem graphically. Determine

a. The coordinates of the intersection point. b. The hatched region or the area of feasible solution c. The optional solution

Maximize profit = 10x + 4y Constraints: 10x + 6y =� 60 5x + 10y =� 50 X =� 0; y =� 0 2. Will the optional solution of linear programming using the graphical approach always occur at

the vertex or corner point of the feasible region? Why? 3. Consider the following linear programming problem. Maximize Profit, P = 2x + 3y Subject to the following constraints 5x + 3y =� 15 x + 2y =� 6 x, y =� 0

a. Show the feasible region. b. What are the extreme or corner points of the feasible region? c. Find the optimal solution using the graphical procedure.

4. Moda Manika Inc., procedure two molded dolls, Malakas and Maganda. Each of these products

must pass through two manufacturing processes, machine A and machine B. Each unit of Malakas requires one hour in process A and 1 ½ hours on process B. Maganda uses 2 hours in process A and ¾ hours on process B per unit. Machine A is available 100 hours a week and Machine B operates 90 hours a week. Profit per unit runs P 100 for Malakas and P 150 for Maganda dolls respectively. What optimal product mix for Moda Manika, Inc will yield maximum profit?




LESSON 3 LINEAR PROGRAMMING: SIMPLEX METHOD OBJECTIVES At the end of the lesson, you should be able to apply the simplex method in solving complex linear programming problems with more than two decision variables. DISCUSSION The simplex method is a general purpose linear programming procedure which can be used to analyze more complicated linear programming problems regardless of size. We have earlier pointed out that the graphical method is limited to solving the problem with two variables. However, linear programming problem with several variables and or constraints can be solved by using a mathematical procedure or algorithm known as the simplex method. The simplex-method is nothing more than a sophisticated trial-and-error approach to solving linear programming problems with more than two decision variables. As an iterative process, the computational routine is repeated continuously until the optimal solution is reached. Similar to the graphical solution, the simplex method begins at the origin and selects the corner points of the feasible region which gives the greatest improvement in the objective function. Thus, the objective function keeps on improving as you move from one corner point to another. Also, each new solution as indicated in the tableau or table will yield a value of objective function larger than the previous solution. Finally, an optional solution is reached, if all the values of Cj – Zj row of the tableau is zero or negative for simplex maximization problems; and zero or positive for simplex minimization problems. Example: GIC manufacturing firm produces two models of picture frames namely: Model A and Model B. Each of the machine centers. Machine Center 1 has twelve hours while Machine Center 2 has eight hours or excess capacity. Each unit of model A. requires two hours of time on both machine centers. Each unit of model B needs three hours on Machine Center 1 and one hour on Machine Center 2. If the profit is P 6 for each Model A and P 7 for a model B, determine the number of units of Model A and number of units of Model B to produce maximize profit. Solve by Simplex Method.




Solution: I. Tabulate the given data and the unknowns

Model A (x)

Model B (y)

Available Capacity (hrs.)

Machine Center 1

2

3

12

Machine Center 2

2

1

8

Profit Contribution

P 6/units

P 7/unit

Let x = number of units of Model A to be produced y = number of units of Model B to be produced II. Objective Function Pmax = 6x + 7y III. Subject to Constraints Machine Center 1: 2x + 3y =� 12 Machine Center 2: 2x + 1y =� 8 x =� 0, y =� 0




The following steps in solving simplex minimization problem. A. Designing the first or initial tableau Step I. Convert the inequality explicit constraints to equations by adding slack variables. C = 6x + 7y + 0 S A + 0 SB 2x + 3y + 1 SA + 0 SB = 12 2x + 1y + 0 SA + 1 SB = 18 Step II. Enter the variables and coefficients in the First or Initial tableau.

Cj

6

7

0

0

Product MLX

Quantity

x

y

SA

SB

0

SA

12

2

3

1

0

0

SB

8

2

1

0

1

ZJ

0

0

0

0

CJ - ZJ

6

7

0

0

Step III. Compute fa Cj – Zj values a. Solving for Zj: Zj = S� (Cj values x constant coefficient of the constraints) 0 [ 2 3 1 0] = 0 0 0 0 0[ 2 1 0 1] = 0 0 0 0 addition Zj = 0 0 0 0 b. Find the difference, Cj – Zj: Cj: 6 7 0 0 Zj: 0 0 0 0 Cj - Zj: 6 7 0 0

Replaced row

optimum column




N.B In the initial tableau, the contribution to profit (cost) column has zero values as entered in Cj column. Likewise, the slack variables, SA and SB with zero profit (cost) contribution are entered in the variable column or product mix column. B. Revise the current Tableau / Designing the Second Tableau Step IV

a. Identify the optimum column / entering variable, the one with the largest value in the Cj – Zj row.

Answer: They y column b. Identify the replaced row / outgoing variable, the one with the smallest quotient (obtained by

dividing the quantity column and optimum column elements) Answer: 12 = 4 3 8 _____ = 8 1 Since 4 < 8, hence SA row is the replaced row. c. Identify the pivotal element or intersectional element that lies at the “crossroad” of optimum

column and replaced now.

Answer: 3 is the pivotal or intersectional element d. Compute the new values of replacing row of 2nd table by dividing the replaced row by pivotal

element.

12 2 3 1 0 ____ ____ ____ ____ ____

3 3 3 3 3

= 4 2/3 1 1/3 0




Second Tableau

Cj

6

7

0

0

Product Mix

Quantity

x

y

SA

SB

7

y

4

2/3

1

1/3

0

0

SB

4

4/3

0

-1/3

1

Zj

14/3

7

7/3

0

Cj - Zj

4/3

1

-7/3

0

d. Compute the new element of the remaining row by reducing all the elements of the optimum

column to zero. New Old Pivotal New remaining = remaining - element x replacing row row row = 8 - ( 1 x 4 ) = 8 - 4 = 4 = 2 - ( 1 x 2/3 ) = 2 - 2/3 = 4/3 = 1 - ( 1 - 1 ) = 1 - 1 = 0 = 0 - ( 1 - 1/3) = 0 - 1/3 = -1/3 = 1 - ( 1 - 0 ) = 1 - 0 = 1 Solving for Zj: 7 ( 2/3 1 1/3 0) = 14/3 7 7/3 0 0 ( 4/3 0 -1/3 1) = 0 0 0 0 addition hence, Zj = 14/3 7 7/3 0 Solving for Cj - Zj: Cj: 6 7 0 0 Zj: 14/3 7 7/3 0 subtraction hence, Cj - Zj: 4/3 0 -7/3 0

replaced row

replacing row

Entering Column




C. Designing the Third Tableau Continue to computational routine as done in the First and Second Tableau. The final solution is obtained if Cj – Zj row value is zero or positive. Third Tableau:

Cj

6

7

0

0

Product Mix

Quantity

x

y

SA

SB

7

x

2

0

1

½

-1/2

6

y

3

1

0

-1/4

3/4

Zj

32

6

7

2

1

Cj - Zj

0

1

-2

-1

a. Identify the optimum column / entering variable. Answer. the x column b. Identify the replaced row / outgoing variable.

Answer: 4 4/3 4 4/3

Since 3 < 6, hence SB is the replaced row. c. Identify the pivotal or intertional element.

Answer: 4/3 is the pivotal element d. Compute for new value of replacing row of Third Tableau.

4 4/3 0 -1/3 1 4/3 4/3 4/3 4/3 4/3

3 1 0 -1/4 3/4

= 6

= 3




SELF-EVALUATION 1. Using the simplex method, determine the best combination of x and y to produce order to realize

the maximum profit. Compute for Cj – Zj row, then indicate the optimum column and pivotal row.

Maximize: x = 5x + 5y Subject to the following constraints 12x + 8y =� 96 6x + 12y =� 72 x =� 0; y =� 0

2. Tender Loving Care, Inc. manufactures two designs of unbreakable plastic baby bottles, plain and decorated each unit of plain bottles needs 2 hours on process A and 1 hour on process B; each unit of decorated bottles requires ½ hour in process A and 3 hours in process B. Process A has a monthly capacity of 320 hours, and process B monthly capacity is 300 hours. Profit for plain and decorated bottles are P 15 and 25 respectively. Find the best combination of plain and decorated bottles that will maximize profits on a monthly basis? Use the simplex method.

3. A firm produces two styles of products, style A and style B. One machine has 12 hours and a

second machine has 8 hours of excess capacity. Each unit of style A requires two hours of processing time on both machines. Each unit of style B requires three hours of processing time on the first machine and one hour on the second machine. The contribution to profit of each A and style B are P 20 and P 30 respectively. Assuming the objective is to maximize profit, how many units of style A and style B should be produced?

4. Minimize Cost, C = 4x + 3y

Subject to: 2x + y =� 10 x + 2y =� 8 X, y =� 0

Compute for Zj row and Cj – Zj row, then indicate the optimum column and the pivotal row. Lastly, indicate the solution set.





DECISION SCIENCE

MODULE FIVE

INVENTORY CONTROL




LESSON 1 ABC ANALYSIS OBJECTIVES At the end of the lesson, you should be able to learn the ABC Approach for an effective inventory control system. DISCUSSION An inventory for small or large-scale companies, whether it’s a manufacturing plant or a fastfood chain may range from a simple item ie, paperclips, rubber bands amounting only a few centavos to expensive spare motors. Under such situations, the firm must devise an inventory management technique on all items for them to operate profitably. To determine the degree to which the inventory of a given item should be controlled, firms must design their own inventory classification system. This means that a decision has to be made as to which inventory items are little things and which need more careful control. The Pareto’s vital few and trivial many concept, the 20-80 principle applies here. The inventory controller should look over the stock records, item by item and classify them into groups. One basis of classifying inventories according to their peso value or unit price of their respective item is the ABC system or ABC analysis. The class A represent a small portion of the high-priced items that account for only 10 percent of the total number of inventory items but 70 percent of the total pesos invested in inventory. The class B items are the intermediate items that account for 30 percent of the total items stocked for 20 percent of the peso invested in inventory. Lastly, the class C items represent a large volume of low-priced items that account for 60 percent of the total inventory items, but only 10 percent of total inventory investment. In this regard, it is suggested that close control should be exercised over class A items, moderate control to the class B items and loose control over class C items.




Example: MYRA novelty store carries 11 items in one of its outlet, as follows:

Item no.

Unit cost (P)

Number of items (Quantity)

Amount ( P )

01

60

50

3,000

02

100

20

2,000

03

35

100

3,500

04

5,000

3

15,000

05

45

60

2,700

06

3,500

5

17,500

07

200

10

2,000

08

70

30

2,100

09

7,000

2

14,000

10

90

12

1,080

11

75

18

1,350 Total: 310 64,230 Conduct an ABC analysis on these stems:




Solution Consider all the 11 items listed in inventory. Three of the items, namely items number 04, 06 and 09 have a unit cost not lower than P 3,500, while they represent a single-digit number of inventory items, 3, 5, and 2 units respectively. Collectively the three high-value items totaling only 10 units amounts to P 46,500, representing about 75% of the whole amount of inventory belongs to class A items, the vital few. On the other hand, the class C items, the low-priced items number 03 and 05 totaling 160 units or 51.6% of the total inventory stock, but amounts to only P 6,200 or 9.6% of total inventory cost. Hence, the firm will end up ahead with loose control on item C. Likewise, the remaining items or the class B items should get middle treatment or moderate control. The firm should focused more attention on class A items and implement strict control on them.




SELF-EVALUATION

1. How do the ABC method of inventory classification work? How does a firm which recognizes this concept, benefit as compared to one which does not.

2. Perform an ABC analysis on the inventory analysis items listed below.

Item Number

Unit Price (PHP)

Number of units

Amount (P)

11

10

8,000

12

80

1,500

13

750

500

14

4,000

200

15

500

400

16

55

2,500

17

6,000

150

18

20

4,500

19

70

1,000

3. Explain the differences in inventory control techniques which should be used for A, B, and C items.




LESSON 2 ECONOMIC ORDER QUANTITY OBJECTIVES At the end of the lesson, you should be able to determine the economic order quantity and its use as a tool in controlling inventory. DISCUSSION Inventory refers to idle materials or goods that are kept by business firms for future use. Our way to classify inventory systems is by their relation to the overall sequence of production operations namely: supplies, materials, in-process and finished goods. Inventory serves as buffer against fluctuating usage and keeps a supply of items available and needed by the firm or its customers. FUNCTIONS OF INVENTORIES 1. Improved Customer Service. By maintaining an inventory of finished goods (or buffer stocks), customer demand can be fulfilled from stocks during varying lead time. Lead time (L) is the time between placing an order to replenish an inventory and receiving an order from suppliers. 2. Production Smoothing. Instead of varying rates of production and size of work-force during lean and peak sales seasons, production rate can be maintained constant approximately by building inventory during lean months and reducing inventory during peak months of sales. Cost savings may be realized by avoiding overtime work, subcontracting, hiring and laying off workers, loss sales and varying production rates.




3. Coordination of Operations. Allowing an inventory of work in process materials to accumulate as buffer in front of the next work station simplifies coordinating the different rates opf the operation of the two work stations in an intermittent processing system. 4. Economies of Scale. By letting inventories to accumulate, the fewer are the number of orders placed per year. Also, large quantities of materials maybe purchased at volume discounts. The amount of inventory on hand behaves in a cyclical manner. Beginning at a high level, the quantity is reduced as units are consumed. Once the inventory level gets low, an order is placed which, when received, lifts the inventory back up and the cycle is repeated. The amount of inventory is controlled by the timing and quantity of teach order. Inventory models having independent demand are considered in which the demand for the item does not depend on the demand for the other items. In this regard, the two basic inventory management decisions have to made namely:

1. How much to order as replenishment. 2. when to order.

The question of how much to order is determined by using the economic order quantity

(EOQ) model. The purpose of using the EOQ model is to obtain the optimal order quantity which minimizes the total inventory costs. INVENTORY COSTS In designing an inventory system, two costs are relevant-the ordering costs and carrying (or holding) cost. 1. Ordering Costs. These are the cost incurred each time an order is placed to replenish inventory. It is a fixed cost, independent of the size of the order placed. The order cost consists of paperwork, typing, stationery, telephone calls and time spent in receiving and expecting the items. The behavior of order cost is shown in Fig. 5.1 with the total cost per order remaining constants, the total ordering costs rise linearly with the number of orders. PHP 30 20 10 1 2 3 Number of Orders Fig. 5.1 Ordering Costs ie., Co = P 10/order

Total Ordering Cost




2. Carrying Costs. Also called holding costs, are the cost of maintaining an item in inventory. This consists of cist capital (or opportunity cost), insurances, taxes, spoilages or damages, pilferages and product obsolescence. Annual holding costs are 1q5 to 45 percent of the average inventory value. Total holding cost is the product of holding cost per unit times average inventory level. As the number of orders increases, average inventory declines, thereby reducing total holding costs. As the number of orders is increased until there is one order for every item demanded, a point is reached wherein there would be no inventory at all. The behavior of carrying costs is shown on Fig. 5.2.

Fig. 5.2 Carrying Cost. i.e., Ch = P 10 / unit / year The remaining costs associated with inventory are purchased costs and shortage cost. Purchase Costs. These are costs to buy an item. This includes the price of the item plus taxes and Transportation costs. Shortage Costs. Refers to costs incurred when an item is out-of-stock resulting into lost sales and goodwill and an unsatisfied customer. EOQ: A Fixed-Order Quantity Model The economic order quantity (EOQ) is a fixed-order quantity model in which demand is met out of inventory on hand. When the inventory declines to an established reorder point, a replenishment order is placed for the same amount. To simplify the inventory analysis, the following consumptions will be considered:

1. Demand is constant and continuous (uniform) 2. Lead time and all costs are constant 3. Supply arrives by batches, not piece meal (lump sum)




Demand is uniform and occurring at the rate of D units per time ie., year. Supply is received in batch sizes of Q units per order. The inventory level starts at a peak of Q units at time 0 (zero) and gradually declines to the reorder point (R), at which time a new order Q is placed. When the order arrives, inventory level instantly increases from zero to Q units and the cycle is repeated. The behaviour of inventory level as a function of time is shown in a saw-tooth curves in Fig. 5.3

Fig. 5.3 EOQ: Fixed-Order Quantity System COMPUTING THE EOQ: ALGEBRAIC METHOD Only two inventory costs are considered namely: the ordering cost and the carrying (or holding) cost. Thus: Total Inventory Cost = Total Ordering Cost + Total Carrying Cost For inventory analysis, one year of total operation is commonly used. Although any length of time could be used. (day, week, month). First, consider the total annual ordering cost. Total annual ordering cost (TOC) is expressed as: D TOC =

Q Co




where: D = annual demand rate (units/year) Q = order size or quantity (units/order) D/Q = optimum number of orders per year (order/year) Co = fixed cost to place an order (p / order) Next, consider the total annual carrying (or holding) cost. Total annual carrying cost (THC) would be: Q Q THC = Pr = Ch 2 2 Where: P = product cost or unit cost of SKU* (P / unit) r = carrying Ch = cost to carry one unit of item in inventory for one year ( P/unit/year) * Each item that is stocked in inventory is called Stockkeeping Unit (SKU) ie., red Sock, blue tie, belt, etc. Now, the total annual inventory cost (TIC) can be expressed as: D Q TIC = Co + Pr Q 2 Note that the purchase cost of the items are not included. At EOQ, the total annually ordering cost (TOC) equal total annual holding cost (THC): Total annual ordering cost = Total annual holding costs or, by substitution: D Q Co = P r Q 2




For any given values of D, Co, P and r, we could derive a a value for Q to minimize the total cost of inventory. The formula for economic order quantity (EOQ) can be expressed as follows: EOQ = 2DCo P r This formula gives the optimum quantity to order each time an order is placed. Example: A manufacturing plant needs 2,000 small parts for next year. The parts cost P 50 each. The ordering cost is P 30 per order. The average inventory carrying cost is 40% of unit cost. Determine the following:

a. Economic order quantity b. Total annual ordering cost c. Total annual carrying cost d. Total annual inventory cost e. How many orders would there be in 1 year? f. How many calendar days would there be between orders? (Assume 365 days for a year) Given: D = 2,000 units / year Co = P 30 / order P = P 50 / unit R = 0.40 Solution: a. Economic Order Quantity (EOQ)

EOQ = 2 D Co P r

EOQ = 2 (2000) (30) 0.40 (50)

= 77.4 units




b. Total annual ordering cost (TOC)

D TOC = Co Q

2000 77.4 c. Total annual carrying cost (THC)

Q Q

THC = Pr = Ch 77.4 2

d. Total annual inventory cost (TIC)

D Q TIC = Co + P r

TIC = 775 + 775 = P 1.550 e. Number of orders per year f. Days between orders or days supply (or order period)

365 25.8

TOC = (30) = P 775

2 2

THC = 0.40 x 50 = P 775

Q 2

order period = = 14 days




SELF-EVALAUTION

1. The following information for an inventory item is given:

Demand = 10,000 units per year Average carrying cost = 25% of unit cost Ordering cost = P 50 per order Unit cost = P 200 per item

Find the optimal order quantity which minimizes total inventory cost.

2. How effective is the economic orderquantity as a tool in inventory control? 3. If the annual demand for an item is uniform at P9,000 units, the unit cost is P 250 per item,

the ordering cost is P 125 per order, and the average carrying cost is 10% of the unit cost.

a. Determine the economic order quantity b. What is the total annual inventory cost?

c. How many orders would there be in one year?

d. How many days would there be between orders?

4. In actual practices, the EOQ’s are by no means always used. Why not?




LESSON 3 REORDER POINT OBJECTIVES At the end of the lesson, you should be able to apply Fixed-Order Quantity model when shortage cost is unknown in determining the Recorder Point. DISCUSSION One of the two basic inventory management decisions mentioned in Lesson 2 is When to Order or, simply, we are ask to determine the Reorder Point. For an EOQ model, the demand is uniform and lead time is assumed to be constant. The reorder point (R) is equal to the demand which will occur during the lead time period as shown in Fig. 5.1. EOQ: Fixed-Order Quantity system. Mathematically, R = DL 365 where: R = reorder period (units) D = annual demand (units/ year) L = lead time (days) 365 days / year




Example 1 A manufacturing plant needs, 2,000 small parts for next year. The parts cost P 50 each. The ordering cost is P 30 per order. The average inventory carrying cost is 40% of unit cost. The small parts are locally available with a one-week lead time. Determine the following

a. The Reorder Point (RP) b. Make a statement of an inventory policy for ordering small parts.

Given: D = 2,000 units / year L = 1 week = 7 days Solution:

a. Reorder Point (RP)

DL RP = 365

b. Inventory policy statement

The inventory policy, then, is to order 2,000 units whenever t he inventory of small parts declines to 28 units.

FIXED-ORDER QUANTITY MODEL WHEN SHORTAGE COST IS UNKNOWN Under the fixed-order quantity model as shown in Fig. 5.4, demand (D) and lead time (L) are allowed to vary randomly, while supply is assumed to arrive in batches. When the inventory level declines to reorder point (RP), an order for fixed quantity (Q) is placed. Likewise, the amount on hand at the time of receipt of the order also varies. Shortages (or stockout) or being out-of-stock may occur, as shown during L2 period. As a remedy, raising the reorder point reduces the possibility of shortages, but increases the carrying or holding costs. The objective is to determine the best trade-off between the risk of being out-off-stock and higher carrying cost.




Fig. 5.4 EOQ: Fixed-Order Quantity System when Shortage or Stockout Cost is Unknown a. Determining Fixed Order Quantity To find the Fixed Order Quantity (Q), simply use the basic EOQ model from Lesson 2. Q = EOQ = 2DCo CH where: D = annual demand rate (units/year) Co = cost of one order or order cost (P / order) CH = holding or carrying cost (P/unit/year)




DETERMINING TH REORDER POINT (R) To find the optimum reader point (R), we need to know the distribution of demand during the lead-time period called the Lead-Time Demand. The probability of being out-of-stock exist only during lead-time period as shown in Fig. 5.4. Whenever the inventory level falls to the reorder point, an order is placed and the lead time-period begins. It is only during these periods (L1 and L2) is there any probability of being out-of-stock. The distribution of Lead-Time demand is shown in Fig. 5.5 which shows a normal probability distribution centered on the average lead-time demand, (Dd L). If reorder point (R) is set equal to average lead-time demand (Dd L). if reorder point (R) is set equal to average lead-time demand (Dd L), the stock on hand at time receiving an order on the average will be zero (0). However, 50 percent of time, the stock will be more than zero (0) and half of the time less than zero (0) or, out-of-stock, which is too high. Hence, some safety stock (Ss) must be added to avoid shortage. To avoid stockout, buffer or satefy stock (SS) must be added as shown in Fig. 5.6. As a result, the reorder point (R) is raised to provide added protection against stockouts during lead-tme period.




The formula for reorder point (R) can be expressed as follows: Rp = Dd L + Ss where: Dd = average daily demand (units) L = average lead time (days) Ss = safety (or buffer) stock (units) The amount of safety stock is based on management decision about service level, which is the probability of having an item in-stock when needed. Service lead ranges from 80 to 99 percent or, the chance of out-of-stock is from 20 to a low of 1 percent. The safety (or buffer) stock (Ss) is given by: Ss = Z ?� d where: S = standard deviation of lead-time demand Z = value is obtained from standard normal probability Table in Appendix A The Z value corresponds to the desired service level. Example The average daily demand for an item is 20 units. The lead time has an average of 4 days. The standard deviation for lead-time demand is 30 units. The owner wants a 95 percent service level of the item. Determine the following aspects of the inventory policy.

a. Safety Stock (Ss) b. Reorder point (RP)

Given: Dd = 20 units/day L = 4 days d = 30 units




Solution:

a. Safety Stock (Ss)

Ss = zd where:

Z = 1.64 (at 95 percent Service Level) Ss = (1.64) (30) = 49 units

b. Reorder Point (Rp)

Rp = Dd L + S = (20) (4) + 49 = 80 + 49 Rp = 129 units

Determining the reorder point as shown below in the probability distribution of lead-time demand.




SELF-EVALUATION 1. Suppose the demand for a product averages 20 units per day (for a 300 day year). Product cost is

P 60. Fixed ordering cost is P 125, and the carrying cost rate is 20 percent. The lead time varies slightly; with an average of 5 days. The demand has a normal probability distribution with a mean of 100 units and a standard deviation of 20 units. Identify the following aspects of the inventory policy.

a. Economic Order Quantity. b. Find the safety stock level for a 95 percent service level.

c. Calculate the number of orders per year.

d. Find the time between placing orders

e. What is the annual cost of carrying this safety (or buffer) stock?

f. Determine the reorder point

g. Specify the optimal inventory policy for placing an order (Include the total annual

inventory costs) 2. We don’t use reorder points because of our sales turnover we are often out-of-stock before the

replishment order is received”. Is this statement a reasonable one? Explain. 3. A manufacturer of computers buys a component unit for P 250. Forecasted demand for next year

is 10,000 units, depending on sales. The CEO believes demand will be normally distributed at an average of 10,000 units. (for a 300 working days). The lead-time has a average of 9 days. The demand has a standard deviation of 17.3 Order cost are P 500 per order while carrying costs are P 50 per day.

Determine the following: a. The optimal order quantity which minimizes total inventory cost. b. Safety stock level for a 99 percent. c. Reorder Point d. A statement of optimal inventory policy.





DECISION SCIENCE

MODULE SIX

SIMULATION




LESSON 1 THE SIMULATION PROCESS OBJECTIVES At the end of the lesson, you should be able to develop simulation models in solving complex business problems. DISCUSSION In the preceding lesson, the emphasis was on the formulation and analysis of mathematical models. In these models, the relevant elements from the real-word system and their inter-relationship were expressed in mathematical formulas and equations. Then, these models will be analyzed and solved. Sometimes problems are very complex and are impractical to solve mathematically. When deciding on a new product, for example, the marketing manager must consider the price, brand name, quality, promotion, distribution, customer reaction, competition and other factors. A fast food store manager is staffing check out counters, must consider the hourly and daily variation in customer traffic, crew attendance, level of skills and experience, as well as the impact of special promotional sales. These problems involve random events, many variable having complex relationships and dynamic phenomena. No known equation or optimizing model will work for these problems because of the behavior of the variables and the complexity of their interactions. Under such conditions, the only practical alternative is to simulate the system, or to conduct an experiment. Simulation is a technique for experimentation using logical and mathematical models. We will focus our attention on probabilistic simulations, and see how the random are treated. When a simulation is developed, the emphasis is first directed at the behavior of the individual elements of the system as described by a probability distribution.




Stimulation is a quantitative technique by which a look-a-like model is developed to perform like the “real world” represents. Then, we can vary the conditions in the simulation model to perform as if the conditions were changed in real world. Under this approach, the decision maker can evaluate the behavior of the model over a range of future conditions. THE SIMULATION PROCESS Simulation can be implemented in a sequence of five basic steps namely:

1. Definition of the problem and setting the objective. A simulation can be conducted to facilitate the understanding of an existing system or in the design of proposed system. It is necessary in the objective understanding of an existing system or in the design of a proposed system. It is necessary that the objectives are clearly defined to determine which variables may be included or not in the model. 2. Formulation of the Model

The decision marker identifies the mathematical expressions and logical relationships used in the computation of the output given the input values. 3. Testing the Model The validation phase is done by comparing the model with the behavior of the real world. 4. Conducting the experiment By actual running of the model, the decision maker learns about the behaviour of the system. 5. Evaluation of the outcomes Under probabilistic model, experimental runs or observations must be conducted for better results. 6. Return the simulation

The Decision maker analyzes and interprets the results depending on the extent the simulation model approximates the real world.




ADVANTAGES OF SIMULATION As a decision-making tool, there are a number of reasons on extensive use of simulation in business namely:

1. Highly complex conditions make it impossible to develop mathematical solutions. In contrast, simulation models can provide a better picture to depict the situation without sacrificing simplicity.

2. Simulation maybe the sole method available due to its difficulty in observing the actual

environment.

3. Actual observation of a system maybe too expensive. 4. Time constraints as to the extensive operation of the system.

5. Availability of user – friendly computer software packages even for sophisticated simulation

models. SHORTCOMINGS OF USING SIMULATION Applying simulation involves certain disadvantage that must be weighed against its advantages. Some of the disadvantages are these:

1. Simulation doest not provide an optimum solution. Solutions obtained from simulation are only approximate. It merely indicates an approximate behavior for a given set of inputs due to inherent randomness, or the present of random numbers in simulations are based on models, and models are only approximation of reality.

2. It is too expensive to implement large-scale simulation models. The expenses includes

model design process, data collection about the real process, writing and debugging the computer programs, validating the model, and running the computer program on a computer system.

3. In more complex situations, it is necessary to employ specialized statistical techniques to

ensure accuracy of the simulation results.




SELF-EVALUTION

1. Visit a bank that has a number of teller windows in your locality. Describe how you will design a simulation model representing the teller windows with customer arriving at random during the day. Determine if the present number of bank tellers will be adequate to serve their clients. What proposals would you recommend to management in providing quality service to their bank’s customers?

2. Somebody with a P 100,000 cash ask you advice on what investment proposals would you

suggest under the present economic conditions. Prepare a simulation model representing the interests that will be earned while considering the riskiness of the investment proposals.

3. Consider a fund-raising project for your parish community to rebuilt the dilapidated church.

You are tasked by the parish priest to propose at least two possible options to raise the money. Explain how a simulation model might be used to evaluate the alternative options and help identify the best possible solution.




LESSON 2 MONTE CARLO SIMULATION OBJECTIVES At the end of the lesson, you should be able to apply the Monte Carlo Simulation technique in solving complex business problems and the use of Random Number Table in generating values for the variables. DISCUSSION Problems that are two complex for reasonable mathematical solution can be handled quite satisfactorily by the use of Monte Carlo simulate ion. Monte Carlo simulation does not use formulas. It is an attempt to imitate a business situation so that the decision maker can test and evaluate various observations. This is usually done by a computer program to act like the system being studied. Simulation allows decision makers to investigate changes in systems and results without incurring the costs of manipulating real systems. Monte Carlo Simulation is a probabilistic type of simulate ion technique that can be used when a process has a random component. Probabilistic simulation is a form of random sampling with each run representing one observation. The larger the degree of variability inherent in the simulation results, the greater the number of simulation runs needed, or the larger sample size will be used to attain a reasonable level of confidence. A more formal method and commonly practiced in Monte Carlo Simulation is the use of a Table of Random Number. (Table 1) It is possible to implement this simulation without using a computer. Or, an intuitive method ie., numbers-in-a-hat method maybe used to simulate the process.




Table 1 Random Number Example:

1. A newly – opened 24 hours convenience store has experienced the following historical daily sales for a particular item for the last 50 days.

Sales (units) Frequency (days) 0 3 1 5 2 13 3 11 4 8 5 7 6 3 Required:

a. Construct a cumulative frequency distribution. b. Use a Random numbers ( Table 1); simulate daily demand of the item for a 5 – day period.

Using the last or fifth column, starting from the last two digits of the sixth row.

c. Determine the mean number of sales for a 5-day simulation period.

42336 49230 87444 57565 41094 70551 16834 82722 78239 35710 42027 96061 96834 62664 80971 69097 42016 38177 33948 43368 78037 34949 35350 32095 06267 10632 36604 75115 39045 93913 68378 62644 80983 54443 53328 48096 63346 65726 32923 27640 13002 85275 07796 26772 31961 75767 04053 17150 96371 04372




Solution a. Construct a cumulative frequently distribution. Step 1: Convert the given frequencies into relative frequency or probability by dividing the

individual frequency by the total frequencies. 3 5 50 50 Step 2: Obtain the cumulative frequency distribution by successively adding the relative

frequencies. Sales Frequency Relative Cumulative Frequency Frequency 0 3 0.06 0.06 1 5 0.10 0.16 2 13 0.26 0.42 3 10 0.20 0.62 4 8 0.16 0.78 5 7 0.14 0.92 6 4 0.08 1.00 b. Simulate the daily demand of the item for 5 days. Step 1: Assign Random Number intervals to correspond with the cumulative frequency for daily

sales. Sales Frequency Relative Cumulative Corresponding 0 3 0.06 0.06 00-05 1 5 0.10 0.16 06-15 2 13 0.26 0.42 16-41 3 10 0.20 0.62 42-61 4 8 0.16 0.78 62-77 5 7 0.14 0.92 78-91 6 4 0.08 1.00 92-99

= 0.06 = 0.06




Step 2: Using the 5th or last column, starting from the last two digits of the sixth row, we have. 13 28 40 61 72 Step 3: Convert the random number into number of sales 13 is found in the interval 06-15 and corresponds therefore, to one sale on day 1. 28 is obtained in the interval 16-41 and corresponds therefore, to two sales on day 2. 40 is found in the interval 16-41 and corresponds therefore, to two sales on day 3. 61 is obtained in the interval 42-61 and correspond therefore, to three sales on day 4. 72 is obtained in the interval 62-77 and corresponds therefore, to four sales on day 5. Step 4: Summarized all the result of a 5-day simulation in a tabulated form. Frequency Frequency Random Number Day Random Number Stimulated Number Sales Number 1 13 1 2 28 2 3 40 2 4 61 3 5 72 4 c. The average number of sales this 5 – day simulation period is 12/5 = 2.40 / day. As compared to

the expected number of items based on the past data, we have: = 0 ( 0.06 ) + 1(0.10) + 2(0.26) + 3(0.20) + 4(0.16) + 5(0.14) + 6(0.08) = 0 + 0.10 + 0.52 + 0.06 + 0.64 + 0.07 + 0.48 = 2.41 / day It is noteworthy to consider that simulation should be viewed as a sample. An additional runs of 5 numbers, would result into a different average values. However, due to inherent variability in the results of small sample size, it is not correct to base our conclusion solely from them. In actual experiment, much larger samples are recommended.




SELF-EVALUATION 1. GIC Rent-a-Car Services, has recorded the demand of its utility vans over the past 25 days as

shown below:

a. Use the following five random numbers to generate 5 – days of rental demand for the firm.

18 51 75 55 92 b. Determine the average number of utility vans rented out for 5 days. c. What is the expected daily demand for utility vans?

2. The following data were taken on the number of patients arrival at emergency room that occur

each hour at Marie Therese’s Medical Center during the grave shift from midnight to 4:00 AM. Using Monte Carlo techniques, simulate 10 hours of emergency room activity

No. of Emergency Calls Frequency 0 2 1 3 2 6 3 7 4 15 5 17 6 20 7 14 8 12 9 4

Use the second column, starting from the first two digits of the first row moving downwards of the Random Number Table I.




3. Values of the random variable and probability are tabulated below:

Variable

Probability

D E G I N O R S T U W

Space

0.03 0.03 0.07 0.12 0.15 0.20 0.12 0.10 0.03 0.06 0.04 0.05

4. A Table of Simulation of an investment proposal is shown for your guidance.

Trial Random Number

Revenues RN Manufacturing Expenses

Profit Benefit/ Cost Ratio

Required:

a. Assign random number intervals for the variables b. List the values of the variable for the following sequence of random numbers:

14, 30, 96, 9, 51, 2, 97, 92, 4, 99, 83 66, 87, 75, 84




LESSON 3 SIMULATION APPLICATIONS OBJECTIVE At the end of the lesson, you should be able to apply Simulation as a decision making tool in solving management problems related to queuing systems, inventory systems and investment proposals. DISCUSSION Simulation differ from the previous topics in this module in an important respect – it is far more general. There are no defined principles or theorems of simulation. It is not limited to any one category of problem, such as inventory control., or to any method of solution which does not exist as those of human imagination, computer resources, and time. In general, the simulation technique is used either because the assumptions are not available or because required by an assumption, optimizing technique are not personally satisfied in a given situation. Usually, the decision maker will turn to simulation as a more reasonable option for obtaining descriptive information about the system study. Simulation has various applications across a wide range of operations management problems, both simple and complex ones. In this lesson, we will apply simulations in management problems related to queing or waiting line systems, inventory systems and investment proposals.




SELF-EVALUATION 1. A sari-sari store has encountered the following daily demand for a particular item for the last 300

days. Simulate for 10 days, using the third column, starting from the last two digits of the first row.

Demand / day Frequency 50 15 51 30 52 55 53 90 54 70 55 40

2. Making use of given data in Problem #1, assume that particular item on sale is the newspaper

tabloid at the newsstand. The owner buys the paper at P 5 each and sells the paper for P 6. Unsold papers at the end of the day are returned for a P 2 credit. If a paper is demanded but not available, goodwill cost is P 2 each.

The owner at the newsstand follows on ordering policy: the quantity ordered each day is equal to the quantity sold the previous day plus the number of lost sales the previous day. Simulate 15 days operation inorder to determine the average profit per day from the sale of newspaper tabloids. Assume the demand for the previous day was 50 papers, and 5 lost sales was incurred. Use the first column of the Random Number Table, starting from the last two of the first row moving downwards. Continue the second column, starting again from the last two digits of the first row going down.




3. GIC is contemplating on whether to add a new product to its production lines. The investment in machinery and equipments to produce the new product is estimated at P 250,000. Assume that the project will have an expected life of one year. The marketing department has estimated that there is a 20 percent probability that the projected first year sales ill be P 400,000; a 30 percent probability that the first year sales will be P 500,000; a 40 percent probability that it will be P 600,000; and a 10 percent probability that will be 700,000.

The machine capacity is P 15,000 units annually. The operation manager consider that there is 10 percent probability that manufacturing expenses would be kept as low as P 200,000; an 80 percent probability that it could be P 300,000; and a 10 percent probability that it could be as high as P 400,000. Simulate the investment proposal using 20 trials and estimate the firm’s probability. Using the benefit-cost ratio obtained by dividing the benefits ( or revenues less manufacturing expenses) derived from the project by its investment cost. Use the fourth row of Random Number table starting from the first two digits of the first row moving downwards. Continue on the fifth column starting again from the first two digits of first row moving downward.




4. BURGERLICIOUS - is a fast-food company which offers dine-in and take-out services to its customers. Top management is considering the installation of an additional drive-thru window at one of its outlets. The existing single drive-thru window is not adequate in meeting the demand and has resulted into long waiting lines which could cause traffic into the busy street. Unfortunately, there is inadequate space to construct two drive-thrus. The CEO has requested that a study be made to determine if this problem really exists. Arrival patterns at a nearby fast food outlet which maintains two drive-thru windows were taken to closely approximate that of the new system presented in Table 6.1. Table 6.1 Time between arrivals at nearby food outlet Drive-Thru Window Table between Frequency Probability of Time arrivals, mins between arrivals 1 10 2 20 3 50 4 10 5 5 6 5 100 1.00 Likewise, the service pattern for this proposal drive-thru outlet was also approximated at the same nearby Makati outlet by observing one of its Drive-thrus shown in Table 6.2. Table 6.2 Service Times at nearby outlet Drive-Thru Window Service Number of Probability time, mins observations Service-Time 1 20 2 30 3 30 4 10 5 6 6 4 100 1.00




Develop a simulation model. Simulate the first hour of operation. The Drive-Thru outlet opens at 7:00 a.m. Use the second column of Random Number Table 1, starting from the first two digit of the first row moving downwards. Continue on the third column starting again from the first two digits of the first row moving downwards. A Table of Simulation of Drive-Thru outlet as guide with the first row completely field. Random Number

Time between Arrivals (mins.)

Clock time at arrival

Random Number

Service Time (mins.)

Time at which service begin

Time at which service ends

Waiting Time (mins.)

49

3 6

7:03 a.m.

7:09 a.m

16

42

1

7:04 a.m

7:09 a.m

7:00 a.m

7:11 a.m

0 0

Determine the following: a. The total time waiting time (minutes) of customer or the total time the customer has to wait in

the first hour of operation. b. The maximum time (minutes), the customer has to wait. c. The average waiting time for all customers.

Submit your written reports to your professors.




BIBLIOGRAPHY

Victoriano, Praxedes S., Quantitative Technique for Business Management, 2nd edition. Rex Book

Store., 1990 Del Rosario, Asuncion C., Quantitative Techniques for Business Decision, 1999 Committee University of the East, Quantitative Technique for Business, Rex Book Store, 1999 Andenson, David R; Sweeney, Dennis, J and Williams, Thomas, W., Quantitative Methods for

Business 7e, South-Western College Publishing, 1995 Render, Barry, Stair Ralph M. Jr., Quantitative Analysis for Management 6th ed. Prentice Hall, 1997 Lawrence, John A. Jr., Pastermack Barry A., Applied Management Science-A Computer-Integrated

Approach for Decision Making. John Wiley and Sons, Inc., 1998.



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