University:
California State University, Northridge
Course:
MATH 460 | Abstract Algebra II
Academic year:

2024
Views:

356

Pages:

26
Author:

hello3xc

Nonlinear Programming and Inventory Control (Multiple Items) 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 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 510% of the total inventory value, while representing 6065% of the inventory items • Thus, the same degree of control is not justified for all items Example Item Annual usage cost Annual dollar usage Rank G-1 40,000 0.07 2,800 5 G-2 195,000 0.11 21,450 1 G-3 4000 0.10 400 9 M-1 100,000 0.05 5000 3 M-2 2,000 0.14 280 10 M-3 240,000 0.07 16,800 2 M-4 16,000 0.08 1280 6 P-1 80,000 0.06 4800 4 P-2 10,000 0.07 700 7 P-3 5,000 0.09 450 8 53,960 $A:75-80 %A:15-20 $C:5-10 Example-Cont. $B:10-15 %B:20-25 %C:60-65 Item Annual dollar usage Cumulative dollar usage % Cumulative % class G-2 21,450 21,450 39.8 39.8 A M-3 16,800 38,250 31.1 70.9 A M-1 5000 43250 9.3% 80.2 B P-1 4800 48050 8.8% 89.0 B G-1 2800 50850 5.2% 94.2 B M-4 1280 52130 2.4% 96.6 C P-2 700 52,830 1.3% 97.9 C P-3 450 53,280 0.8% 98.7 C G-3 400 … … C M-2 280 … 100.0 C 53,960 53,960 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 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 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 Constrained and Unconstrained Optimization • 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: • profit Z= 1696.8p-24.6p2-22,000 δZ/δp=0, 1696.8-49.2p=0, p=34.49 • Called classical unconstrained optimization price 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 P=40 price 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 city2 city4 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? Solution Techniques • Very complex • Two method – Least or Substitution method • Transferring a constraint optimization to an unconstraint optimization – Lagrange multiplier 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 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 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 Limitation of Substitution • Highest order of decision variable was a power of two • Dealt only with two decision variables and single constraint 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 -λ (x1+2x2-40) Lagrange Multiplier-Cont. • Partial derivatives of L with respect to each of variables • δL/δx1=0, δL/δx2=0, δL/δλ=0 • 4-0.2x1=0, 5-0.4-2λ=0, x1-2x2+40=0 • X1=18.3, x2=10.8, λ=0.33, z=70.42 Sensitivity Analysis • Lagrangian multiplier, λ, is analogous to dual variables in LP • Shows changes in objective function value by changing the RHS • Positive value of λ 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, λ=0.27, z=70.75 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 Application of NLP and Inventory Modeling Example-Limited Working Capital Demand Item cost Order cost Item 1 Item 2 Item 3 EOQ QC 1000 50 20 1000 20 50 2000 80 50 100 158 112 Available budget=15,000 Inventory carrying charge=0.2 5000 3160 8960 Total=17120 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 Lagrangian Expression • Total budget=17120> available budget=15000 • Form the Lagrangian Expression L= n n n ∑D C + ∑ A D /Q + ∑ i =1 i i i i =1 i i i =1 n H i Qi + γ ( Qi Ci − B ) 2 i =1 ∑ • Derivative with respect to Qi and lambda ∂L = 0 ⇒ − Ai Di / Qi2 + H i / 2 + λCi ∂Qi ∂L =0⇒ ∂γ n ∑C Q − B = 0 i =1 i i Final Result • From the first equation 2 Ai Di Ci (i + 2λ ) Qi* = • From the second equation n ∑C Q i =1 n ∑C i =1 i n ∑ i =1 i i =B 2 Ai Di Ci =B Ci (i + 2λ ) 2 Ai Di Ci =B (i + 2λ )

Nonlinear Programming and Inventory Control (Multiple Items)

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock