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Quick catch up of Linear programing

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- 1. <ul><li>WELCOME </li></ul><ul><li>TO </li></ul><ul><li>CIMA Revision Course </li></ul><ul><li>P2 Performance Management </li></ul><ul><li>Prepared and Compiled By Syed Shehzad Haider Naqvi </li></ul>
- 2. Course Outline <ul><li>Linear Programming </li></ul><ul><li>Pricing </li></ul><ul><li>Product Life Cycle </li></ul><ul><li>Budgeting </li></ul><ul><li>TQM/JIT </li></ul><ul><li>Ratio Analysis </li></ul><ul><li>Miscellaneous </li></ul>
- 3. Why Talk About Linear Programming? <ul><li>Usefull where more than one variable are scarce </li></ul><ul><li>LP is simpler than NLP, hence, good for a foundation </li></ul><ul><li>Linearity has some unique features for optimization </li></ul><ul><li>A lot of problems are or can be converted to a LP formulation </li></ul><ul><li>Some NLP algorithms are based upon LP simplex method </li></ul>
- 4. EXAMPLE <ul><li>Robert Miles operates a small machine shop. Next month he plans to manufacture two new products ( A and B) upon which the unit contribution is estimated to be $50 and $70, Respectively </li></ul><ul><li>For thier manufacture both products require inputs of machine processing time. Raw materials and labour. Each unit of product A require 3 hours of machine processing time, 16 units of raw materials and 6 units of labour. The corresponding per unit requirements for product B are 10,4 and 6. Robert forecats that next month he cab naje available 330 hours of machine processing time. 400units of raw materials abd 240 units of labour. The technology of the manufacturing process is such that at least 12units of product B must be in any given month. </li></ul><ul><li>Robert Miles wishes to formuate a linear programming model so as to determine the number of units of products A and B that he should manufacture next month to maximise contribution. </li></ul>
- 5. 0 12 20 25 33 40 50 60 80 100 110 100 80 70 60 40 35 25 20 Linear Programming (Graphical method) Constraints Equation Coordinates 1 3x 1 + 10x 2 = 330 (33,110) 2 16x 1 + 4x 2 = 400 (25,100) 3 6x 1 + 6x 2 = 240 (40,40) 4 X 2 = 12 (0,12)
- 6. 0 12 20 25 33 40 50 60 80 100 110 100 80 70 60 40 35 25 20 Drawing ISO Contribution line and Feasible Region End select two convenient value of Z substitute into The objective function and plot into the resulting equation ; Here 1750 & 3500 appear to be sensible. 1750 = 50X 1 + 70X 2 x 1 = 35; x 2 = 0 3500 = 50X 1 + 70X 2 x 1 = 70; x 2 = 0
- 7. 0 12 20 25 33 40 50 60 80 100 110 100 80 70 60 40 35 25 20 Feasible Region E D C B A Points X1 x2 Z A 22 12 1940 B 20 20 2400 C 10 30 2600 D 0 33 2310 E 0 12 840
- 8. <ul><li>CIMA`s Official Terminology </li></ul><ul><li>“ an increase in value which would be created by having available one additional unit of limiting factor at its original cost. This represents the opportunity cost of not having the use of the one extra unit” </li></ul>Shadow Pricing In our example following is the equation of labour. 6x 1 + 6x 2 = 240 If one more labour hour will be available then, 6x 1 + 6x 2 = 241 know the value of Z will be Z=2607.1 The contribution has increased by 7.1 by increase 1hour of scarce resource. This is the shadow price a unit of labour
- 9. The Simplex Method <ul><li>“ Method for more than two decision variables” </li></ul>Woodhurst is a furniture company that specialises in high-quality products. The company can manufacture four different types of coffee table ( small, meduim, large and ornate) Each type of table requires time for the cutting of the component parts, for assembly and for finishing. The data in the table below has been collected for the year now bieng planned. Owing to other commitments, no more than a total of 1,800 coffee tables can be made in any given year. Also, market analysis reveals that the annual demand for the company`s small coffee table is at least 800. The Company wishes to determine how many of each type of coffee table it should produce in the coming year to maximise contribution hours required per table Contribution on each table Table Cutting Assembly Finishing small 2 5 1 60 Medium 2 4 4 123 Large 1 3 5 135 Ornate 6 2 3 90 Capacity in hours 3,000 9,000 4,950
- 10. Formulating into the equations <ul><li>Maxmise to : </li></ul><ul><li>Z=60x 1 + 123x 2 + 135x 3 +90x 4 </li></ul>Subject to : 2x 1 + 2x 2 + 1x 3 + 6x 4 + ≤ 3,000 5x 1 + 4x 2 + 3x 3 + 2x 4 + ≤ 9,000 1x 1 + 4x 2 + 5x 3 + 3x 4 + ≤ 4,950 x 1 + x 2 + x 3 + x 4 + ≤ 1,800 x 1 ≥ 800 x 1 ,x 2 ,x 3 ≥ 0 Objective funtion variable(z) 168,750 Variable Value relative loss X1 950 0 X2 250 0 X3 600 0 X4 0 48 Constraint slack/surplus worth 1 0 9 2 1450 0 3 0 21 4 0 21 5 150 0

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