1. F2:Management Accounting

2. 2 Designed to give you knowledge and application of: Section F: Short–term decision–making techniques F1. Cost –Volume-Profit (CVP) Analysis F2. Relevant Costing F3. Limiting Factors

3. 3 Learning outcomes  Identify a single limiting factor. [1]  Determine the optimal production plan where an organisation is restricted by a single limiting factor. [2]  Formulate a linear programming problem involving two variables. [1]  Determine the optimal solution to a linear programming problem using a graphical approach. [1]  Use simultaneous equations, where appropriate, in the solution of a linear programming problem. [1] Study Guide F3 : Limiting Factors

4. 4 Limiting factors (LF) Limiting factor Restricting production Scarce factors Optimal utilisation essential to attain the highest profits Difficulty in the short run Identify a single limiting factor Example Pinocchio Plc makes and sells puppets to toy shops. It can supply 400 different types of puppets in the coming season. However, the demand for puppets has gone down and the company now has an order for only 250 puppets. The demand, in this case, is a constraint or the limiting factor since Pinocchio Plc can produce 400 puppets but cannot sell as much.

5. 5 Limiting factors under Considerations while devising a product mix Profit contribution Market demand LF units consumed Scarce resources used in the production of a single product  Devise an optimal product mix  Produce the product with maximum contribution per LF Multi- products manufacturing Single product manufacturing

6. 6 Determine the optimal production plan where an organisation is restricted by a single limiting factor. Refer to Example on page F2.3 Optimal production plans (OPP) Determine contribution per unit of LF Rank products in order of highest contribution Draw up a production schedule with maximum hours / units of LF for first ranked products Use remaining hours / units of LF in order of ranking Designing OPPs

7. 7 Example Hammer Plc manufactures hardware tools of three types: screwdrivers, pipes and spanners. It wishes to analyse the sales and production budget for the coming quarter. The information available is given below: 1,0001,200700Required hours 200400350Estimated sales demand 532Machine hours required per unit $14$8$10Contribution SpannerPipe Screwdriver Continued…

8. 8 231 Ranking $2.8 ($14/5) $2.67 ($8/3) $5 ($10/2) Contribution per machine hour 532 Machine hours required per unit $14$8$10 Contribution SpannerPipeScrewdriver Example continued The machine hours available are only 2,000, so are insufficient to meet the demand. The machine hours are the limiting factor in this case. In order to arrive at the optimal production plan, the contribution per unit of limiting factor is calculated first. The product that gives the highest contribution per unit of limiting factor will be ranked as first in the order of production. Thereafter, the products will be ranked according to the contribution per machine hour. Once the products have been ranked in order of highest contribution per unit of limiting factor, the schedule of production can be decided. This means using the hours available for the maximum demand of the first ranked product. The remaining hours are used in the order of ranking. Continued…

9. 9 Statement of production plan Calculation of hours required Machine hours Total machine hours available2,000 Hours used in the manufacture of the maximum demand of screwdrivers 2 hours per unit x 350 screwdrivers 700 Balance hours left1,300 Hours used to manufacture spanners 5 hours per unit x 200 spanners 1,000 Balance hours left300 Remaining hours to be used for manufacture of pipes. The demand for pipes is 1,200 units which require 1,200 x 3 hours per unit = 3,600 hours. However, there are only 300 hours left, so we will utilise them for the manufacture of pipes. Units manufactured will be 300 / 3 = 100 pipes 300

10. 10 Contribution per unit CalculationTotal contribution $ Screwdrivers$10$10 x 3503,500 Spanners$14$14 x 2002,800 Pipes$8$8 x 100800 Total maximum possible contribution 7,100 The entire hours have been utilised in the production of screwdrivers and spanners. It is not possible to manufacture pipes as there are no machine hours left. As a result, only these two types of hardware items will be produced. The maximum contribution is calculated as follows: Example

11. 11 Linear programming (LP) Linear programming Used to arrive at optimal production plan Applied when more than one LF Ensures optimal utilisation of resources Mathematical technique Formulate a linear programming problem involving two variables Steps to formulate a linear programming problem Identification of constraints Limiting factors that restrict production are converted into mathematical equations Formulation of the objective function Linear equation indicating relationship between the output and profit 2 1 Maximisation of profit or minimisation of cost

12. 12 30 hours25 hoursTotal time available $82.5 hours5 hoursSmall bag $53 hours4 hoursBig bag Profit per unitSewing departmentCutting department Asmees Plc manufactures two types of shopping bags: big and small. The details of the processing time in the cutting and sewing departments are given below, as is the profit per kilogram. Formulate a linear programming problem to arrive at the optimal product mix Answer Let the units to be produced of the big bags be ‘x’ and the units of the small bags to be produced be ‘y’. The profit per unit of the big bag is $5 and per unit of the small bag is $8. Therefore, the objective function will be expressed as: Maximise 5x + 8y Cutting department constraint The hours required per unit for the big bag are 4 hours, therefore the hours required to produce x units will be = 4x. Example Continued…

13. 13 The hours required per unit for the small bag are 5 hours, therefore the hours required to produce y units will be = 5y. The total hours available are 25. The constraint will be 4x + 5y ≤ 25 hours. Sewing department constraint The hours required per unit for the big bag are 3 hours, therefore the hours required to produce x units will be = number of units x hours required = 3x The hours required per unit for the small bag are 2.5 hours, therefore the hours required to produce y units will be = number of units x hours required = 2.5y The total hours available are 30 The constraint will be 3x + 2.5y ≤ 30 hours The linear programming problem will be Maximise the objective function 5x + 8y Subject to constraints 4x + 5y ≤ 25 hours 3x + 2.5y ≤ 30 hours The non-negativity constraint x, y ≥ 0 Continued…

14. 14 Question In a given linear programming problem, the constraint is that three times the production of the product denoted by x cannot exceed four times the production of the product denoted by y. The constraint will be expressed as: a) 3x > 4y b) 3x ≥ 4y c) 4x ≥ 3y d) 3x ≤ 4y Answer The correct option is (d). Three times the quantity of x produced must not exceed four times the quantity of y produced. Three times x is 3x and four times y is 4y. The phrase “must not exceed” is interpreted as being either equal to or less than. Therefore 3x can be less than or equal to 4y which is expressed as 3x ≤ 4y.

15. 15 Question The graph given below is drawn for the constraint: a) 2x + 3y ≤ 12 b) 5x + 4y ≤ 20 c) Y ≤ 20 d) X ≤ 10

16. 16 Answer a) The line passes through co-ordinates x = 6 on the X - axis and y = 4 on the Y - axis. The co-ordinates at the point x = 6 are (6, 0) and at the point y = 4 are (0, 4). We will have to check which of the given equations satisfies these values. (a) 2x +3y <= 12 Converting inequality into equality: 2x + 3y = 12 At (6, 0) - 2(6) + 3(0) = 12 At (0, 4) – 2(0) + 3(4) = 12 As a result, this equation is satisfied by the co-ordinates on the graph. Since all the other options are bound to be wrong we do not need to solve them. There is no option which says that two of the options are correct. However, in the examination you might not find the first option to be the correct one. Therefore you will have to do this exercise for all the options in order to find the correct one.

17. 17 Question The following graph has been established for a given set of constraints The objective function (OF) for the company has also been plotted on the graph and the feasible region is bounded by the area ABCD. At which point on the graph will profits be maximised? A B C D

18. 18 Answer (d) Profit is maximised when the objective function (OF) line intersects with point D in the feasible area. Since the feasible area is given as the bounded region ABCD, we will have to move the given line OF within the area. The farthest point to which the line OF can move in the feasible area is vertex D and hence the profit will be maximised at point D.

19. 19  Identify a single limiting factor?  Determine the optimal production plan where an organisation is restricted by a single limiting factor?  Formulate a linear programming problem involving two variables?  Determine the optimal solution to a linear programming problem using a graphical approach?  Use simultaneous equations, where appropriate, in the solution of a linear programming problem? RECAP

20. [training@getthroughguides.com]