Nonlinear Programming and Inventory Control (Multiple Items)

1. Nonlinear Programmingand Inventory Control (Multiple Items)

2. Multiple Items • Materials management involves many items and transactions • Uneconomical to apply detailed inventory control analysis to all items • Because a small percentage of inventory items accounts for most of the inventory value • Must focus on important items • Isolate those items requiring precise control • ABC analysis indicates where managers should concentrate

3. ABC Analysis • Divides inventory into three classes according to dollar volume (dollar volume=Annual demand *unit purchase cost) • A class is high value items whose dollar volume accounts for 75-80% of the total inventory value, while representing 15-20% of the inventory items • B class is lesser value items whose dollar volume accounts for 10-15% of the total inventory value, while representing 20-25% of the inventory items • C class is low value items whose volume accounts for 5-10% of the total inventory value, while representing 60-65% of the inventory items • Thus, the same degree of control is not justified for all items

4. Example

5. Example-Cont. $A:75-80 %A:15-20 $B:10-15 %B:20-25 $C:5-10 %C:60-65

6. EOQ and Multiple Items • Most inventory systems stock many items • May treat each item individually and then add them up • Restrictions may be imposed (limited warehouse capacity, upper limit on the maximum dollar investment, number of orders per year, and so on) • Will be covered after covering NLP

7. Overview of NLP • Many realistic problems have nonlinear functions • When LP problems contain nonlinear functions, they are referred NLP • Have a separate name, because they are solved differently • In LP, solutions are found at the intersections • In NLP, there may no corner point • Solution space can be undulating line or surface • Like a mountain range with many peaks and valleys • Optimal point at the top of any peak or at the bottom of any valley • In NLP, we have local and global points

8. Local and Global Optimal Point • Solution techniques generally search for high points or low points • Difficulty of NLP is to determine whether the identified is a local or global optimal point • Global optimal point can be found by the very complex mathematical techniques Global optimal point Local optimal points

9. Constrained and UnconstrainedOptimization • Constrained optimization • Profit function: Z= vp-cf-vcv, v=1500-24.6p, cf=$10,000 and cv=$8 • vp creates a curvilinear relationship • Results in quadratic function: Z= 1696.8p-24.6p2-22,000 • δZ/δp=0, 1696.8-49.2p=0, p=34.49 • Called classical unconstrained optimization profit price

10. Constrained Optimization Max Z= 1696.8p-24.6p2-22,000 s.t. p<=20 • Referred NLP • Solution is on the boundary formed by constraint • Change RHS from 20 to 40 • Solution is no longer on the boundary • Makes process of finding optimal solution difficult, particularly, with more variables and constraints Feasible space P=20 price Feasible space price P=40

11. Single Facility Location Problem • Locate a centralized facility that serves several customers • Minimize the total miles traveled between the facility and all customers • Locations of the cities and the number of trips are: (20,20,75), (10,35,105), (25,9,135), (32,15, 60), (18,8,90) • d=[(xi-x)2+(yi-y)2]1/2 • Min Σditi city1 city3 city5 city4 city2

12. NLP with Multiple Constraints • Consider a problem with two constraints • A company produces two products and the production is subject to resource constraints • Demand of each product is dependent on the price (x1=1500-24.6p1 and x2=2700-63.8p2) • Cost of producing x1 and x2 are 12 and 9 • Production resources are (2x1+2.7 x2<=6000, 3.6x1+2.9x2<=8500, 7.2x1+8.5 x2<=15,000) • Model: Max z=(p1-12)x1+(p2-9)x2 s.t. 2x1+2.7 x2<=6000 3.6x1+2.9x2<=8500 7.2x1+8.5 x2<=15,000 • Where x1=1500-24.6p1 and x2=2700-63.8p2 • Decision variables are?

13. Solution Techniques • Very complex • Two method • Least or Substitution method • Transferring a constraint optimization to an unconstraint optimization • Lagrange multiplier

14. Substitution Method • Restricted to models containing only equality constraints • Involves solving the constraint for one variable for another • New expression will be substituted into the objective function to eliminate it completely

15. Example • Max Z= vp-cf-vcv s.t. v=15,00 -24.6 p • cf=$10,000 and cv=$8 • vp creates a curvilinear relationship • Constraint has been solved v for p • Substitute it in objective function Z= 1500p-24.6p2-cf-1500cv+24.6 pcv Z= 1696.8p-24.6p2-22,000 • Differentiating and setting it equal to zero • 0=1696.8-49.2p • p=34.49

16. Example • Consider the Furniture Company • Assume contribution of each declines as the quantity increases • Relationship for x1: $4-0.1x1 • Relation for x2: $5-0.2 x2 • Profit earned form each:($4-0.1x1)x1 and ($5-0.2 x2)x2 • Total profit: z=4x1+5x2-0.1x12-0.2x22 • Consider just one constraint: x1+2x2=40 or x1=40-2x2 • z=4(40-2x2)+5x2-0.1 (40-2x2)-0.2x22 • z=13x2-0.6x22 • x1=18.4, z=$70.42

17. Limitation of Substitution • Highest order of decision variable was a power of two • Dealt only with two decision variables and single constraint

18. Lagrange Multiplier • Used for constraint optimization consisting of nonlinear objective function and constraints • Transform objective function into a Lagrangian function • Constraints as multiples of Lagrange multipliers are subtracted from the objective function • Consider an example Max z=4x1+5x2-0.1x12-0.2x22 s.t. x1+2x2=40 • Forming Lagrangian function L=4x1+5x2-0.1x12-0.2x22 - (x1+2x2-40)

19. Lagrange Multiplier-Cont. • Partial derivatives of L with respect to each of variables • δL/δx1=0, δL/δx2=0, δL/δ=0 • 4-0.2x1=0, 5-0.4-2=0, x1-2x2+40=0 • X1=18.3, x2=10.8, =0.33, z=70.42

20. Sensitivity Analysis • Lagrangian multiplier, , is analogous to dual variables in LP • Shows changes in objective function value by changing the RHS • Positive value of  shows the increase in objective function • Example Max z=4x1+5x2-0.1x12-0.2x22 s.t. x1+2x2=41 • x1=18.8, x2=11.2, =0.27, z=70.75

21. Mathematical Notations • Di=Annual demand for item i in units • Ci=Unit purchase cost of item i in dollar • Ai= ordering cost of item i in dollar • fi=Required storage space for item i in square foot • F=Maximum total storage space available • TC=Total average annual costs in dollar • Hi= Annual holding cost per unit i per year in dollar

22. Application of NLP and Inventory Modeling

23. Example-Limited Working Capital Total=17120 Available budget=15,000 Inventory carrying charge=0.2

24. Solution • Solved by the method of Lagrange‑multiplier • Before applying the method, we should solve the total cost function by ignoring the constraint • Otherwise, we form the Lagrangian expression

25. Lagrangian Expression • Total budget=17120> available budget=15000 • Form the Lagrangian Expression • Derivative with respect to Qi and lambda

26. Final Result • From the first equation • From the second equation

27. Problem Solution

28. HW Assignment #1 • Assume that the annual inventory carrying charge is 10% • and that 15000 sq.ft of floor space are available. • What is the optimal inventory policy for these items. • Determine the cost of having only 15,000 sq.ft of floor space?

29. HW#2 Classify these items into A, B, and C.

30. HW#3 • The Furniture Company has developed the following NLP model to determine the optimal number of chairs and tables to produce daily. Max z=7x1-0.3x12 +8x2-0.4x22 S.t. 4x1+5x2=100 hr • Determine the optimal solution to this NLP using the substitution method. • Determine the optimal solution to this NLP using the Lagrange multipliers.

31. HW#4 • Consider the following NLP model to determine solution. Max z=30x1-2x12 +25x2-0.5x22 S.t. 3x1+6x2=300 • Determine the optimal solution to this NLP using the substitution method. • Determine the optimal solution to this NLP using the Lagrange multipliers.

32. HW#5 • Section 11.2c, Problems 1 and 4