1 / 25

ACC423: Management Accounting II Module4 (Week10) Covering: Regression Analysis EOQ Analysis Linear Programming

ACC423: Management Accounting II Module4 (Week10) Covering: Regression Analysis EOQ Analysis Linear Programming. By: E. P. Enyi, Ph.D, MBA, ACA, FAAFM, RFS, MFP, FIIA Head, Dept of Accounting, Covenant University, Ota, Nigeria. Regression Analysis.

thyra
Download Presentation

ACC423: Management Accounting II Module4 (Week10) Covering: Regression Analysis EOQ Analysis Linear Programming

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ACC423: Management Accounting II Module4 (Week10) Covering: Regression Analysis EOQ Analysis Linear Programming By: E. P. Enyi, Ph.D, MBA, ACA, FAAFM, RFS, MFP, FIIA Head, Dept of Accounting, Covenant University, Ota, Nigeria

  2. Regression Analysis • Regression analysis is also called CURVE FITTING • It is a statistical technique used for short or medium term activity forecasts like next season’s projected sales, activity costs etc. • It is also used in management accounting to establish the relationship between two or more variables and to separate cumulated costs into fixed and variable elements. • There are primarily two main methods of finding the line of best fit; which is the linear relationship between the variables under consideration. • It is a SIMPLE regression analysis when it involves only two variables but it becomes a MULTIPLE regression analysis when three or more variables are involved. • To find the line of best fit, it is necessary to calculate a line which minimizes the total of the squared deviations of the actual observations from the calculated. This is known as the LEAST SQUARES.

  3. The LEAST SQUARES method of linear regression is given as y = a + bx Where, a and b are constants and where a represents the fixed element and b represents the slope of the curve; y in this case is the dependent variable e.g. total sales while x is the independent variable e.g. selling price per unit. The following graph illustrates this:

  4. Before you can find the value of y and x, you must first find the value of a and b by solving the following two equations simultaneously: ∑y = an + b∑x (1) ∑xy = a∑x + b∑x2 (2) where n = number of observations or pairs of past data being used for the analysis. Example: TMC Ltd has collected data for the past seven years on the sales of its prime product named HODAC which is presented as follows: No of Colors 1 2 3 4 5 6 7 Sales(‘000) 14 17 15 23 18 22 27 The management wish to know whether there is any relationship between the number of colors included in the product and the total sales volume and what would be the sales volume if 9 colors were used.

  5. Solution: As a matter of fact, you can solve this problem using the visual fit method but the solution will be less accurate than the more scientific method of least squares. The outline of the least square solution is as follows: Colors(x) Sales(y) xy x2 1 14 14 1 2 17 34 4 3 15 45 9 4 23 92 16 5 18 90 25 6 22 132 36 7 27 189 49 Totals 28 136 596 140

  6. Here, n = 7; ∑x = 28; ∑y = 136; ∑xy = 596; ∑x2 = 140 Such that: 136 = 7a + 28b …….(1) 596 = 28a + 140b …….(2) Solving the equations simultaneously, we have: a = 12 b = 1.86 Therefore, the regression line or linear relationship between the number of colors in HODAC and the sales volume achieved is given by the equation: y = 12 + 1.86x i.e. sales volume (‘000) = 12 + 1.86(No of colors used) If 9 colors are used, then the expected sales volume will be: y = 12 + (1.86 * 9) = 28.74 or 28,740 units of HODAC

  7. For Class Work: The following data were extracted from the production logbook of Azundu Chemicals Ltd: Batch Units ((x) ‘000) Total Costs ((y) N’000) a. 9 116 b. 16 211 c. 14 152 d. 38 410 e. 21 256 f. 25 298 g. 20 220 h. 15 180 i. 15 185 j. 11 129 Required: Separate the total costs into its fixed and variable elements. If the company is to manufacture 30 units, how much will it costs? How many units will a cost of N500,000 produce?

  8. Economic Order Quantity (EOQ) EOQ is a stock control tool. It is used in conjunction with other stock control tools such as Re-Order Level, Minimum Stock Level, Maximum Stock Level etc. EOQ is defined as the optimal number of items to be placed in a single stock replenishment order in order to achieve the least total over-all stock holding costs per annum. EOQ can be determined in two ways:- • By constructing a graph • By the use of a formula. ASSUMPTION UNDERLYING THE USE OF EOQ • That there is a known constant stock holding cost; • That there is a known constant stock ordering cost; • That the rates of demand are known; • That the price per unit is known and constant; • That replenishment is made instantaneously, i.e. batch is delivered whole

  9. GRAPHICAL APPROACH To calculate EOQ graphically, the following ingredients are necessary: • Total costs per annum (i.e. ordering cost plus carrying costs) • Number of orders required per annum (i.e. Annual DD / Order Qty) • Average stock (i.e. Order Qty / 2) Example: A company uses 50000 bottles per annum which costs N10 each to purchase. The ordering and handling costs are N150 per order and carrying costs are 15% of the purchase cost per annum. Required: Determine the Economic Order Quantity that will minimize the total stock holding cost for the coming year; assuming all other parameters remain as projected. Solution: Graphical solution is implemented by trying a combination of assumed order quantities, e.g. 1000, 2000 etc. It is the quantity of each assumed order batch size that will determine the number of orders per annum.

  10. A Table of Assumed Order Levels: • (b) (c) (d) (e) (f) Order Average Annual Average Stock Total Qty No of Ordering Stock Holding Cost Orders Costs Cost p.a. ---------- ---------- ----------- ------------ ------------ -------- Assumed Dd/(a) (b)*150 (a)/2 (d)*1.50 (c)+(e) ---------- ---------- ------------ ------------ ------------ --------- 1000 50 7,500 500 750 8250 2000 25 3,750 1,000 1,500 5250 3000 16.67 2,500 1,500 2,250 4750 4000 12.5 1,875 2,000 3,000 4875 5000 10 1,500 2,500 3,750 5250 6000 8.5 1,250 3,000 4,500 5,750 ------------------------------------------------------------------------------------------------------- From the above table the graph can be prepared:

  11. From the above graph, the EOQ is approximately 3,160 units as indicated by the downward arrow.

  12. EOQ BY FORMULA The scientific formula for EOQ determination is given as: EOQ = √(2*Co*D)/Cc Where, Co = Cost of order per batch Cc = Carrying cost per item per annum D = Annual Demand Solution: EOQ = √(2*150*50000)/1.5 = √10000000 = 3,162 units There are other measurements of EOQ which are of interest to the management accountant. These include: • EOQ With Gradual Replenishment – where there is internal arrangement for stock usage replenishment; and • EOQ Where STOCK-OUTS Occur – where stocks can be completely allowed to be used up before replenishment (this will involve additional stock-out costs)

  13. EOQ with Gradual Replenishment = √(2*Co*D)/(Cc(1-(D/R))) Where, 1-(D/R) = rate of stock usage R = rate of production per annum Example: Assume the company decides to make its own bottles with a machine that can produce up to 250,000 units per annum, with other data remaining the same, the EOQ with gradual replenishment will be: EOQ = √(2*150*50000)/(1.5(1-(50000/250000))) = √12500000 = 3,536 units per order EOQ where STOCK-OUTS Occur = [√(2*Co*D)/Cc] * [√(Cc+Cs)/Cs] Where, Cs = Cost of stock-out per item per annum. CLASS WORK: Assuming that a courier service charges N0.75 to make a quick delivery per item in addition to an admin cost of N0.25 per item for every stock-out orders, what is the new EOQ?

  14. Linear Programming Techniques LP is a mathematical technique concerned with the scientific allocation of scarce resources to attain optimum value in decision making. For LP to be effectively deployed: • The problem must be capable of numerical translation; • All factors involved in the problem must have linear relationship; • The problem must provide alternative causes of action; • There must be one or more restrictions on the factors involved. USING THE LP SOLUTION TECHNIQUE There are two stages involved in using the LP technique. The first is to formulate the model and the second is to solve the problem using the model formulated. It is important that the model formulation is accurately done otherwise it will be a case of garbage in and garbage out. To formulate the model, you first determine the objective function, then you determine the main body of the LP problem.

  15. Methods: There are two main methods of solving a LP problem: • By graph – this is only possible when the resources are limited to two. The graphical method cannot solve complex situations. Even so, the solution provided will not be as accurate as the more scientific simplex method. • By Simplex Algorithm – this is a more scientific method and can handle complex situations such as those involving more than two resource combinations. Maximization – is used when it is necessary to work out the solution (mix) that gives the maximum benefit e.g. optimal profit. Minimization – is used when it is necessary to work out the solution (mix) that produces the least cost. Example: A firm makes two products fork and knife. Fork has a contribution of N3 per unit and knife N4 per unit. The firm wishes to establish weekly production plan which maximizes its total contribution. The production data are as follows:

  16. PER UNIT Machining (Hrs) Labour (Hrs) Material (kgs) Fork 4 4 1 Knife 2 6 1 Total Available Per Week 100 180 40 Because of a trade agreement, sales of Fork are limited to a weekly maximum of 20 units and to honor an agreement with an old customer, at least 10 units of Knife must be sold per week. Step 1: Formulate the LP model; this is given thus: Maximize 3f + 4k SUBJECT TO: A 4f + 2k ≤ 100 (Machining Hr Constraint) B 4f + 6k ≤ 180 (Labour Hr Constraint) C f + k ≤ 40 (Materials Constraint) D f ≤ 20 (Fork sales Constraint) E k ≥ 10 (Knife sales Constraint)

  17. f, k ≥ 0 where, f = number of units of fork k = number of units of knife As this is a problem with only two unknowns (f and k) it can be solved graphically. Step 2: Draw the axis of the graph ensuring that you calibrate as appropriate: f 0 k

  18. Step 3: Remove the inequalities from the equations and solve for each constraint as follows: Constraint Equation A 4f + 2k = 100 B 4f + 6k = 180 C f + k = 40 D f = 20 E k = 10 In A, where f=0, then k=100/2=50; where k=0, f=100/4=25 In B, where f=0, then k=180/6=30; where k=0, f=180/4=45 In C, where f=0, then k=40; where k=0, f=40 In D, f=20 In E, k=10 These are then plotted in the given graph below to produce the following image:

  19. Fork Knife LP Model Graph Showing the FEASIBLE REGION

  20. THE FEASIBLE REGION The feasible region is the portion of the graph which falls within the boundary of possible solutions. It is the only area of the graph which satisfied all the conditions imposed by the various constraints. It is usually shaded to distinguish it from other non-feasible area. It is from this region that a set of possible solutions are outlined. In this graph, there are 4 possible outcomes from which we choose the one with the optimal contribution as follows: Point Fork(N3) Knife(N4) Total I 20 * 3 10 * 4 100 II 15 * 3 20 * 4 125 (Choice) III 0 * 3 30 * 4 120 IV 0 * 3 10 * 4 40 From the above analysis, we can see that the combination which produced the highest benefit is point II, i.e. the one obtained at the rightmost upward projection of the feasible area on the graph. This is translated to mean that a production combination of 15 forks and 20 knives per week should be pursued by the firm as this is the mix that will give maximum contribution.

  21. MINIMIZATION Provided that there are two unknowns (i.e. two products), we can also use the graphical method to solve minimization problem, but, with the following differences: • In minimization problem the limitations are of the greater than or equal to type (≥) rather than the less than or equal to type (≤); • The feasible region is found above all or most of the limitations rather than below them; • The normal objective is to minimize cost so that the objective function line represent cost rather than profit or contribution and because the objective is to minimize cost, the optimum point will be found from the cost line furthest to the left which still touches the feasible region (the opposite of the maximization method). Example: A firm is to market a new fertilizer which is to be a mixture of two ingredients A and B. The properties of the two ingredients are given in the table below:

  22. INGREDIENT ANALYSIS BONE NITR- PHOS COST MEALOGENLIMEPHATE /KG Ingredient A 20% 30% 40% 10% N12 Ingredient B 40% 10% 45% 5% N8 It has been decided that: • The fertilizer will be sold in bags containing a minimum of 100kg • It must contain at least 15% nitrogen, 8% phosphate and 25% bone meal. The firm wishes to meet the above requirements at the minimum cost possible SOLUTION Set the objective function as follows: MINIMIZE 12a + 8b Subject to: A a + b ≥ 100 (Weight Constraint) B 0.3a + 0.1b ≥ 15 (Nitrogen Constraint) C 0.1a + 0.05b ≥ 8 (Phosphate Constraint) D 0.2a + 0.4b ≥ 25 (Bone Meal Constraint) a ≥ 0 and b ≥ 0

  23. Where, a = kgs of ingredient A b = kgs of ingredient B Remove the inequalities and solve thus: In A, if b=0 then a=100; if a=0 then b=100 In B, if b=0 then a=15/0.3=50; if a=0 then b=15/0.1=150 In C, if b=0 then a=8/0.1=80; if a=0 then b=8/0.05=160 In D, if b=0 then a=25/0.2=125; if a=0 then b=25/0.4=62.5 The above computations are then graphed as follows:

  24. Ingredient A Feasible Region Ingredient B

  25. Reading from the graph: PointIng A (N12)Ing B (N8)Total (N) I 125 * 12 0 * 8 1500 II 75 * 12 30 * 8 1140 III 60 * 12 40 * 8 1040 (Lowest = Optimal) IV 0 * 12 160 * 8 1280 The above implies that the best combination of ingredients that would minimize total cost (i.e. result in the least cost) is 60 kg of A and 40 kg of B. SIMPLEX ALGORITHM METHOD This is a step-by-step arithmetic technique whereby one moves progressively from a zero position to the maximum possible contribution. It utilizes the solution method of matrix algebra. The steps involved in the simplex algorithm method are numerous and the underlying logic is complex and cannot be adequately covered within the time scheduled for this lecture. The student is advised to study this topic independently.

More Related