Systems Engineering and Analysis Chapter 9 Optimization in Design and Operations 2/16/2019 rd

Gaussian Reduction Solve 2x + 3y = 1 5x + 2y = 8 3 1 1 3/2 1/2 1 0 2 3 1 1 3/2 1/2 1 0 2 5 2 8 0 -11/2 11/2 0 1 -1 2/16/2019 rd

AFTER STUDYING THIS CHAPTER, YOU SHOULD BE ABLE TO: Use calculus to optimize functions; Understand the origins of optimization; Do decision theory with optimization techniques; Structure unconstrained and constrained optimizations; Evaluate decision making opportunities; Do graphical and simplex method of linear programming. L E A R N I N G O B J E C T I V E S 2/16/2019 rd

The "Best" System The word optimal should not appear in the statement of the problem, because there is no single optimal solution to complex systems problems. "the best is the enemy of the good" "systems engineering is the art of the good enough" Balanced View Satisficing Desired Minimum acceptable 2/16/2019 rd

Derivative f '(y), y', dy/dx, Df Where was the farmer when he jumped off the roof? Q P x x + h 2/16/2019 rd

Optimization Given a length of L units to create a square and a circle, find the dimensions to maximize the area. Of all planar areas the circle has minimum perimeter. For L = 10; 6.25,3.5;7.96 x L - x x/4 C = 2r = L - x For L = 10 and x = 40/(pi + 4); f = 3.5; f(L= 10) = 6.25; f(L= 0) = 7.96 f(x) = 0 x L f '(x) = f(0) = => all for the circle; absolute maximum f(L) = => all for the square f '(x) = x/8 + (2*x - 20)/(4*pi) for L = 10 f() = ; absolute minimum f(x) = x2/16 + (L – x)2/4, 0  x  L; f(0) = L2/4; L2 /16 f'(x) = x/8 - (L – x)/2 = 0 at x = 4L/( + 4) f(L) = L2/16; f[4L/( + 4)] = L2/4(4 + ) absolute minimum 2/16/2019 rd

Partial Differentiation y = f(x, y, z, t, …) is differentiated with respect to one of these variables, the other variables are held constant. Let w = f(x, z) = x2 + xz2. Then Wx = Wz = Consider right circular cylinder V(d, h) = d2h/4; Vd = dh/2 and Vh = d2/4; the area 2/16/2019 rd

Unconstrained Optimization Find the area bounded by the curve y = f(x) = x3 - 6x2 + 9x + 1 and the x-axis between the curve's x-extrema. y'(x) = 3x2 –12x + 9 = 0 when x = 1 or x = 3 pt of inflection y''(x) = 6x – 12; x = 2 is Pt-Inflection y''(1) < 0 => relative max at(1,5); y''(3) > 0 => relative min at (3,1) 1 2 3 = 6 (U-poly-eval '(1 9 -6 1) 1) syms x; f = x^3 - 6*x^2 + 9*x + 1; fx = diff(f); roots([3 -12 9]);ans = 3.00 1.00; fxx1 = -6.00 fxx2 = subs(diff(fx),3) =fxx2 = 6.00; subs(f,1) = 5.00; subs(f,3) = 1.00 2/16/2019 rd

Second Derivative Test + + " - - " Second Derivative positive => minimum Second Derivative negative => maximum 2/16/2019 rd

Cost to Construct A box has base length 3 times the width. The cost to construct is $10/ ft2 for the top and bottom and $6/ft2 for the sides. The volume must be 50 ft3. Find the dimensions to minimize the cost to construct. C = 2(10)(3w2) + 2(6)(hw) + 2(6)(3hw) = 60w2 + 48hw; 50 = 3w2h => h = 50/3w2 C = 60w2 + 800/w C' = 120w - 800/w2 = 0 when 12w3 = 80 => w = 1.88 C" = 120 + 1600/w > 0 => relative minimum. h l=3w w 2/16/2019 rd

Minimize the Difference The difference between two numbers is 20. Make their product as small as possible. x – y = 20; A(x, y) = xy = x(x – 20) = x2 - 20x A'(x) = 2x - 20 = 0 when x = 10 A"(x) = 2 implying a relative minimum x = 10; y = -10 2/16/2019 rd

Calculations with Functions f(x) = x2; g(x) = x – 1 f(x) + g(x) = (f + g)(x) = x2 + x – 1 f(x) – g(x) = (f – g)(x) = x2 – x + 1 f(x)g(x) = x2 (x – 1) = x3 – x2 … 2/16/2019 rd

Inverses If y = f(x) is equivalent to x = g(y), the g and f are inverses and 1-1. Find the inverse of g(x) = x2 – 2 y = g(x) = x2 – 2; x2 = y – 2; x =  (y + 2)1/2 Not 1-1; hence no inverse f(x) = (x + 1)/5 f- (x) = 5x – 1 f(x) = x3 f-1(x) = x1/3 Find the inverse of y = 2x – 3. Solve for x: x = (y + 3)/2; then switch; y = (x + 3)/2 2/16/2019 rd

Two Variable Optimization Given Z = f(x, y) Zx(a, b) = 0; Zy(a, b) = 0 => existence of an extremum or a saddle point at point (a, b). Regard the determinant D = Zxx Zyy – Zxy Zyx evaluated at (a, b). If D(a, b) > 0 and Zxx < 0; Zyy < 0 => rel maximum and Zxx > 0; Zyy > 0 => rel minimum If D(a, b) < 0, critical point is a saddle point D(a, b) = 0, critical point may not be one of the above types, i.e., test fails. 2/16/2019 rd

Z = f(x, y) = x3 + y3 + 3xy Zx = 3x2 + 3y; Zy = 3y2 + 3x Solve '3x2 + 3y = 0' to get (a, b) = (-1, -1); (0, 0) '3y2 + 3x = 0' Zxx = 6x; Zxy = 3; Zyy = 6y D(0, 0) = -9 < 0 => saddle point Z(0, 0) = 0 Z(-1, -1) = 1 D = D(-1, -1) = 36 – 9 = 27 > 0 Zxx(-1, -1) = -6 < 0 and Zyy (-1, -1) = -6 < 0 Point (-1, -1, 1) is a relative maximum. 2/16/2019 rd

Z = f(x, y) = x3 + y3 + 3xy clear; clc; format compact syms x y f = x^3 + y ^3 + 3 * x * y % equation to optimize fx = diff(f, x); fy = diff(f,y); fxy = diff(fx, y); fyx = diff(fy, x) fxx = diff(f, x, 2); fyy = diff(f, y, 2) hess = [fxx fxy; fyx fyy] dval = det(subs(hess, {x, y},{-1, -1})) z = subs(f, {x, y},{-1,-1}) solve('3*x^2+3*y', '3*y^2 + 3*x')  x: [4x1] y: [4x1] 2/16/2019 rd

Z = f(x, y) = x3 + y3 + 3xy Zx = 3x2 + 3y = 0; 3y2 + 3x = 0; Candidates: (0, 0); (-1 -1) x2 + y2 + x + y = 0; or (x + ½)2 + (y + ½)2 = ½ Zxx = 6x D = At (-1 -1), D = 36 and Zxx < 0 => rel max; Z(-1, -1) = 1 At (0, 0), test fails 2/16/2019 rd

Two Variable Optimization f(x, y) = x2 – 8y + 2y2 – 6x + 30 fx = 2x – 6 = 0 => x = 3 fy = -8 + 4y = 0 => y = 2 f(3, 2) = 32 – 8*2 + 2*22 – 6* 3 + 30 = 13 Check to see if max or min. fxx = 2 fxy = 0 fyx = 0 fyy = 4 = 8 The value of the determinant is 8 and fxx > 0 => relative minimum. f(3, 2) = 5 2/16/2019 rd

Nonlinear Systems Solve y = x2 - 2x + 1 y = x3 + x2 - 2x + 3 x2 - 2x + 1 = x3 + x2 - 2x + 3 x3 = -2 (cubic 1 0 0 2)  -1.259921 x = -1.259921 y = 5.107243 2/16/2019 rd

Nonlinear Optimization Find the max of f(x) = 12x – 3x4 - 2x6 f'(x) = 12(1 – 1x3 - 1x5 ); f(-) = -; f() = - (optimize '(0 12 0 0 -3 0 -2))  (0.83762 7.883946) (root '(0 12 0 0 -3 0 -2) 1.5)  1.250963 f"(x) = -36x2 - 60x4 < 0 for x, x  0 => absolute maximum x f(x) 0 0 1 7 1.250963 2 -152 6 roots  0 -0.98 -0.98 0.35 0.35 1.25 2/16/2019 rd

Problem 9-4 220x + 15,000 for 0 < x  1000 f(x) = 120x + 115,000 for 1000  x  2500 205x – 97,500 for x > 2500 Use the above cost function to determine the production rate in order to maximize profit if the selling price is $200/unit. Profit = (200 – 220)x – 15K = -20x -15K (can't break even) = (200 – 120)x – 115K = 80x -115K => BE = 1437.5 ** = (200 – 205)x + 97,500 = -5x + 97,500 => BE = 19500. dP/dx = -20; 80; -5 Level of production is (1000, 2500) Best is 2500 items 2/16/2019 rd

Problem 9-4 Code ;;; Systems Engineering ;;; Profit = Revenue - Cost (defun p9-4 (x) "Args: x" (cond ((<= x 1000) (- (* -20 x) 15E3)) ((< 1000 x 2501) (- (* 80 x) 115E3)) (t (+ (* -5 x) 97.5E3)))) 2/16/2019 rd

Linear Programming A company makes 2 auto cleaners, X and Y returning profits of $10 and $30 respectively. X needs 4 hours of Process A and 8 hours of Process B; Y needs 6 of A and 4 of B. There are 12 hours of capacity A and 16 hours of capacity B. Optimize profit. Max 10X + 30Y 4 Y Subject to 4X + 6Y <= 12 8X + 4Y <= 16 2 (3/2, 1) X, Y >= 0 0 2 3 X (LP '((-10 -30 0)(4 6 12)(8 4 16)))  #2A((10 0 5 0 60)(2/3 1 1/6 0 2)(16/3 0 -2/3 1 8)) (solve '((4 6 12)(8 4 16)))  (3/2 1) 10 0 5 0 60 2/3 1 1/6 0 2 16/3 0 -2/3 1 8 2/16/2019 rd

Linear Programming (Lp '((-0.6 -0.7 -0.5 0) (2.4 3 2 1200) (0 2.5 1.5 600) (5 0 2.5 1500))) Max 0.6X1 + 0.7X2 + 0.5x3 s.t. 2.4X1 + 3.0X2 + 2.0X3 <= 1200 0.0X1 + 2.5X2 + 1.5X3 <= 600 5.0X1 + 0.0X2 + 2.5X3 <= 1500 Initial Tableau -0.6 -0.7 -0.5 0.0 0.00 0.0 0.00 2.4 3.2 2.0 1.0 0.0 0.0 1200 0.0 2.5 1.5 0.0 1.0 0.0 600 5.0 0 2.5 0.0 0.0 1.0 1500 continued  … -0.6 0 -0.08 0.0 0.28 0.0 168.0 2.4 0 0.2 1.0 -1.2 0.0 480.0 0.0 1 0.6 0.0 0.4 0.0 240.0 5.0 0 2.5 0.0 0.0 1.0 1500.0 0 0.0 -0.03 0.25 -0.02 0.0 288.0 1 0.0 0.083333 0.416667 -0.5 0.0 199.999984 0 1.0 0.6 0.0 0.4 0.0 240.0 0 0.0 2.083334 -2.083333 2.5 1.0 500.000064 Tableau?(Y/N) y 0.0 0.0 0 0.22 0.016 0.0144 295.2 1.0 0.0 0 0.5 -0.6 -0.04 180.0 0.0 1.0 0 0.6 -0.32 -0.288 95.999984 0.0 0.0 1 -1.0 1.2 0.48 240.000016 3 steps LP Optimum 295.2 #2A((0.0 0.0 0 0.22 0.016 0.0144 295.2)(1.0 0.0 0 0.5 -0.6 -0.04 180.0)(0.0 1.0 0 0.6 -0.32 -0.288 95.999984)(0.0 0.0 1 -1.0 1.2 0.48 240.000016)) 2/16/2019 rd

Final Tableau LP Optimum = 295.2 0.0 0.0 0 0.22 0.016 0.0144 295.2 1.0 0.0 0 0.5 -0.6 -0.04 180 0.0 1.0 0 0.6 -0.32 -0.288 96 0.0 0.0 1 -1.0 1.2 0.48 240 (M* (inverse #2A((2.4 3 2.0)(0.0 2.5 1.5)(5 0 2.5))) #2A((1200)(600)(1500)))  #2A((180.0)(96.0)(240.0)) [f x flag] = linprog([-0.6 -0.7 -0.5],[2.4 3 2; 0 2.5 1.5; 5 0 2.5], [1200 600 1500]')  180 96 240 x = -295.2 (mmult (inverse #2A((2.4 3 2.0)(0.0 2.5 1.5)(5 0 2.5))) #2A((1200)(600)(1500))) 5-29-2009 rd

Linear Programming P9-27 (LP '((-7.8 -9.4 -2.6 0)(4.2 11.7 3.5 1800)(0.8 4.3 1.9 2700) (12.7 3.8 2.5 950))) 0 0.0 0.74 0.67 0.0 0.39 1588.8 0 1 0.26 0.096 0.0 -0.032 142.28 0 0.0 0.70 -0.39 1.0 0.066 2062.43 1 0.0 0.12 -0.03 0.0 0.09 32.23 0 0.0 0.744191 0.676619 0.0 0.390409 1588.803456 0 1.0 0.255975 0.095755 0.0 -0.031667 142.27552 0 0.0 0.703099 -0.388826 1.0 0.065596 2062.429184 1 0.0 0.120259 -0.028651 0.0 0.088215 32.232524 x2 = 142.276 2/16/2019 rd

LP Example (LP '((-2 -3 0)(3 5 15)(6 2 12)))  Max 2x1 + 3x2 -2 -3 0 0 0 subject to 3x1 + 5x2 <= 15 3 5 1 0 15 x2 6x1 + 2x2 <= 12 6 2 0 1 12 x1, x2 >= 0 3 -1/5 0 3/5 0 9 3/5 1 1/5 0 3 24/5 0 -2/5 1 6 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 Z(0, 0) = 0; Z(0, 3) = 9; Z(2, 0) = 6; Z(5/4, 9/4) = 37/4 (Mmult (inverse #2A((3 5)(6 2))) #2A((15)(12)))  #2A((5/4)(9/4)) 6 (5/4, 9/4) 0 2 5 x1 Feasible Region. Evaluate all corner points y 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 2 steps LP Optimum 37/4 #2A((0 0 7/12 1/24 37/4)(0 1 1/4 -1/8 9/4)(1 0 -1/12 5/24 5/4)) 2/16/2019 rd

subject to 3x1 + 5x2 <= 15  3x1 + 5x2 + x3 = 15 Max 2x1 + 3x2 subject to 3x1 + 5x2 <= 15  3x1 + 5x2 + x3 = 15 6x1 + 2x2 <= 12  6x1 + 2x2 + x4 = 12 Basis X1 X2 X3 X4 Q Z-row -2 -3 0 0 0 X3 3 5 1 0 15 X4 6 2 0 1 12 -1/5 0 3/5 0 9 3/5 1 1/5 0 3 24/5 0 -2/5 1 6 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 2/16/2019 rd

LP Example Max 2X1 + 3X2 s.t. 3X1 + 5X2 <= 15 6X1 + 2X2 <= 12 x1 x2 x3 x4 -2 -3 0 0 0 X3 3 5 1 0 15 X4 6 2 0 1 12 -1/5 0 3/5 0 9 X2 3/5 1 1/5 0 3 X4 24/5 0 -2/5 1 6 2/16/2019 rd

s.t. 7/10X1 + 1X2 <= 630 Cutting and dyeing Max 10 X1 + 9X2 s.t. 7/10X1 + 1X2 <= 630 Cutting and dyeing 1/2 X1 + 5/6X2 <= 600 Sewing 1 X1 + 2/3X2 <= 708 Finishing 1/10X1 + 1/4X2 <= 135 Inspection and packaging 0 -7/3 0 0 10 0 7080 0 8/15 1 0 -7/10 0 672/5 0 1/2 0 1 -1/2 0 246 1 2/3 0 0 1 0 708 0 11/60 0 0 -1/10 1 321/5 0 0 35/8 0 111/16 0 7668 0 1 15/8 0 -21/16 0 252 0 0 -15/16 1 5/32 0 120 1 0 -5/4 0 15/8 0 540 0 0 -11/32 0 9/64 1 18 (mmult (inverse #2A((7/10 1)(1 2/3))) #2A((630)(708)))  #2A((540)(252)) 2/16/2019 rd

Minimization Problem Min 2Y1 + 3Y2 s.t. 1Y1 >= 125 Demand for product 1Y1 + 1Y2 >= 350 Total production 2Y1 + 1Y2 >= 200 Processing time Z(350, 0) = 700 Y1,Y2 >= 0 350 Dual: Max 125X1 + 350X2 + 200X3 200 s.t 1X1 + 1X2 + 2X3 <= 2 125 0X1 + 1X2 + 1X3 <= 3 X1, X2 >= 0 100 350 Final Tableau 225 0 500 350 0 700 1 1 2 1 0 2 -1 0 -1 -1 1 1 2/16/2019 rd

A Blending Problem The B. A. Nutt Company sells mixed nuts of two quality levels. The expensive mix should not contain more than 25% peanuts nor less than 40% cashews. The cheap mix should not have more than 60% peanuts and no less than 20% cashews. Cashews cost 50 cents a pound and peanuts cost 20 cents a pound. The expensive mix sells for 80 cents a pound and the cheap mix for 40 cents a pound. What should the blend of each mix be in order to maximize profit. The company has $100 a day with which to purchase nuts.

The Model Let x1 = pounds of cashews in expensive mix x2 = pounds of peanuts in expensive mix y1 = pounds of cashews in cheap mix y2 = pounds of peanuts in cheap mix Max z = 0.8 (x1 + x2) + 0.4 (y1 + y2) - 0.5(x1+ y1) - 0.2(x2 + y2) subject to: 0.5 (x1 + y1) +0.2(x2 + y2 ) <= 100 x2 / (x1+x2) <= 0.25 y2/(y1 + y2) <= 0.6 x1 / (x1+x2) >= 0.4 y1/(y1 + y2) >= 0.20

Inventory Two Conflicting Costs Too frequent orders increase ordering costs. Less frequent orders increases inventory. How frequently should materials be ordered? Stock is simply cash in disguise. Continuous or Perpetual ~ checkbook, laser scanner fixed-order quantity (EOQ) Periodic or fixed time period ~ textbooks, variable amount 2/16/2019 rd

Inventory Examples Blood in blood bank Unused telephone numbers phone company is holding Standby crew members Cash in bank Unused space in warehouse Empty seats in classrooms. Raw materials, purchased parts, components, sub-assemblies, work-in-progress, finished goods and supplies 2/16/2019 rd

Purpose of Inventory Protection against fluctuating demand Protection against delayed supply Decouple successive stages to ease breakdown of system Protection against inflation Benefits of large quantities Primary basis for business Savings on ordering cost Other reasons such as bargaining power, labor 2/16/2019 rd

Dependent and Independent Demands Dependent demands ~ raw materials, components, subassemblies used in end items. keyboards for computers. Largely predictable Cars => 4 tires Independent demands ~ Finished products; Highly uncertain demand. Need forecasting. 2/16/2019 rd

Inventory Model TC = total system cost per period L = procurement level or replenishing level Q = procurement quantity D = demand rate in units per period T = lead time in periods N = number of periods per cycle; Q/D R = replenishment rate in units per period Ci = item cost per unit Cp= procurement cost per procurement Ch = holding or carrying cost per unit or % of Q/2 per period CS = shortage cost per unit short per period continued  On hand 2/16/2019 rd

Inventory Costs Holding Order/Setup Satisfaction Cost of capital Labor Goodwill Rent Communication Loss in future sales Utilities Transportation Labor Insurance Communication Labor Taxes Shrinkages Spoilages Obsolescence TC = IC + PC + HC + SC 2/16/2019 rd

Inventory Classifications By Process By Importance By Life Raw materials A, B, C Perishable Work in progress Nonperishable Finished goods Usually 5 to 10% of the stock keeping units (SKUs) account for A, 40 to 50% for B, and the remaining 50% for C. Stock keeping unit 2/16/2019 rd

ABC Classification In stock are 6400 items, 400 are in class A, 1000 in B and 5000 in C. The company operates 250 days per year and wants to count A, B, C items with a relative frequency of 5, 2, and 1 times a year. How many items should he company count per day on average? Item Frequency Total Type Number A 400 5 2000 B 1000 2 C 5000 1 5000 9000/250 36 items/day 2/16/2019 rd

EOQ Assumptions: Certain known and constant demand, constant lead time, shortages are not allowed, instantaneous receipt of order. Max Inventory Q Average inventory Q/2 2t t Time 2/16/2019 rd

EOQ Cost Model Annual Cost ($) Q Optimal order Q* order size Slope = 0 Total Cost Carrying Cost = ChD/Q Ordering Cost = CoD/Q Q Optimal order Q* order size 2/16/2019 rd

Constant demand Week 1 2 3 4 5 6 7 Demand 2005 2005 2000 2000 2025 1950 2010 About 2000 cases of beverages a week; 104,000/year Holding cost – to buy and maintain stock, cost of capital expressed as a percentage of amount invested; 25% insurance, taxes, breakage, pilferage, warehouse overhead Ordering cost – preparation of voucher, processing the order, postage, salary, etc. Assumed fixed regardless of quantity ordered Suppose $32 Case of beer costs $8 => .25 * 8 = $2 holding cost (EOQ (* 2000 52) 32 2)  $3648.56 (Total Cost) = $1824.28 (Ordering Cost) + $1824.28 (Holding Cost) Q-star = 1824.28 Cycle time = 0.02 (EOQ 2000 32 0.0384) weekly 2/16/2019 rd

Inventory Pattern Inventory consumed at constant demand rate Q Average ½ Q Inventory T ~ time to deplete inventory Q 2/16/2019 rd

Holding and ordering Costs Annual Carrying or Holding Cost = Average inventory * holding cost of 1 unit = ½ QCh Annual Ordering Cost = number of orders/year * cost/order = (D/Q)Co Total Annual Cost = Annual Holding Cost + Annual Ordering Cost TC = ½ QCh + (DCo/Q) (EOQ model) 2/16/2019 rd

Inventory Model TC = CiD + CpD/Q + ChQ(1 – D/R)/2 – Ch(DT – L) + Ch(DT – L)2/[(2Q(1 – D/R)] + Cs(DT – L)2/2Q(1 – D/R) d is daily demand rate; p is daily production rate D and R for annual demand and production Infinite rate of replenishment makes D/R  0 2/16/2019 rd

Total Annual (EOQ Model) Cost TC = ½ QCh + (D/Q)Co TC' = Ch/2 – DCo/Q2 = 0 when Q* = (2DCo/Ch)1/2 TC'' = 2DCoQ-3  0 => relative minimum 2/16/2019 rd

Single Source Alternative - (Inventory D R time Ci Cp Ch Cs) (INVENTORY 10 20 8 4.82 100 0.20 0.10)  Q* = 244.95 L* = -1.65 TC* = 56.36 (INVENTORY 10 800000 8 4.82 100 0.20 10000) (EOD Demand Cost-Ordering Cost-Holding) 2/16/2019 rd

Economic Order Quantity Annual Demand 104,000 Ordering Costs $32.00 Annual Inventory Holding Rate % 25 Cost per Unit $8.00 Working Days per Year 250 Lead Time (Days) 2 Optimal Inventory Policy 1824.28068 Annual Inventory Holding Cost $1,824.28 Annual Ordering Cost Total Ordering Cost $3,648.56 Maximum Inventory Level 1824.28 Average Inventory Level 912.14 Reorder Point 832.00 Number of Orders per Year 57.01 Cycle Time (Days) 4.39 2/16/2019 rd

EOQ Formula Q* = (2 * 50 * 200,000/0.2)1/2 = 10,000 Annual demand is 200,000 items. Ordering cost is $50 and holding cost is 20% of the value of the item. Find the EOQ, optimal number of orders and the total inventory costs. TC = T0 + TH = DC0/Q + CHQ/2 T' = -DCO /Q2 + CH/2 = 0 when Q* = Q* = (2 * 50 * 200,000/0.2)1/2 = 10,000 Optimal number = 200,000/10,000 = 20 orders at Q = 10,000 each Inventory costs = TC = (200,000 * 50)/10,000 + 10,000 * 0.2/2 = $2,000. (EOQ 2E5 50 0.2)  $2000 (Total Cost) = $1000 (Ordering Cost) + $1000 (Holding Cost) Q-star = 10000.00 units (EOQ (/ 2E5 12) 50 (/ 0.2 12)) $ 166.67 (Total Cost) = $ 83.33 (Ordering Cost) + $ 83.33 (Holding Cost) 2/16/2019 rd

Sensitivity Q* = Note that if the ordering cost CO or the demand quadruples, Q* doubles, and if the holding cost CH quadruples, Q* halves. Note also that TC is relatively insensitive to changes in the order quantity and that sensitivity for decreasing the EOQ is larger than for increasing. 2/16/2019 rd

Sensitivity (EOQ (* 2000 52) 32 (* 0.25 8))  Manipulate Ch and Co $3648.56 (Total Cost) = $1824.28 (Ordering Cost) + $1824.28 (Holding Cost) Q-star = 1824.28 units Cycle time = 0.02 (EOQ (* 2000 52) 34 (* 0.25 8))  $3760.85 (Total Cost) = $1880.43 (Ordering Cost) + $1880.43 (Holding Cost) Q-star = 1880.43 units Cycle time = 0.02 2/16/2019 rd

Inventory Demand is 60 per week and holding costs are 2 cents each with an annual interest rate of 25%. Ordering cost is $12. Find the EOQ, total cost, and cycle time. (EOQ (* 60 52) 12 (* 0.02 0.25))  annual Q* = (2*12*60*52/0.02*0.25)1/2 (EOQ 60 12 (* 0.02 (/ 0.25 52)))  weekly $19.35 (Total Cost) = $9.67 (Ordering Cost) + $9.67 (Holding Cost) Q-star = 3869.88; Cycle time = 1.24 years = Q* /D (EOQ 60 12 (* 0.02 (/ 0.25 52))) (EOQ (* 52 60) 12 (* 0.02 0.25)) 2/16/2019 rd

Production Costs (1) Q (2) TFC (3) TVC (4) TC (5) AFC (6) AVC (7) AC (8) MC 1 60 30 90 10 2 40 100 20 50 5 3 45 105 15 35 4 55 115 13.75 28.75 75 135 12 27 6 120 180 TFC Total Fixed Cost TVC Total Variable Cost 2/16/2019 rd

Optimize syms x y f = -pi*0.6*0.12*x*y + (x - 0.5)^2 + (y - 0.3)^2 fx = diff(f,x); fy = diff(f,y); fxx = diff(fx,x); fyy = diff(fy,y) fxy = diff(fx,y); fyx = diff(fy, x) hess = [fxx, fxy' fyx, fyy] solve('2*x - (9*pi*y)/125 -1 = 0','2*y - (9*pi*x)/125 - 3/5 = 0') ans = x: [1x1 sym] y: [1x1 sym] 2/16/2019 rd

f = x2yz; A = 3x2yi +yz2j - xzk syms x y z ; f = x^2 * y * z; f = x^2*y*z; A = [3*x^2 * y, y*z^2, -x* z]; A = [ 3*x^2*y, y*z^2, -x*z]; f * A  [ 3*x^4*y^2*z, x^2*y^2*z^3, -x^3*y*z^2] diff(ans, z)  [ 3*x^4*y^2, 3*x^2*y^2*z^2, (-2)*x^3*y*z] diff(ans, y)  [ 6*x^4*y, 6*x^2*y*z^2, (-2)*x^3*z] subs(ans, {x, y, z}, {1, -2 ,-1})  -12 -12 2 2/16/2019 rd

F(x, y ,z)= sin(xy + 2z) Find Fxzy = F132 Fx = y cos(xy + 2z) Fxz = -2y sin(xy + 2z) Fxzy = -2xy cos(xy + 2z) -2 sin(xy + 2z) 2/16/2019 rd

F(x, y) = 2x4 + y2 – x2 - 2y Determine extrema Fx = 8x3 – 2x = 0 when x = 0,  ½ Fy = 2y – 2 = 0 when y = 1 Candidates are (-½, 1); (½ , 1); (0, 1) Fxx = 24x2 – 2; Fxy = 0; Fyy = 2; Fyx = 0 D = = -4 when x = 0; 8 when x = ½ Conclude relative minimums at (-½, 1); (½ , 1) with F = -9/8 D = -4 > Saddle Point at (0, 1) where F = -1. 2/16/2019 rd

Lagrange's Method Of Multipliers To find extrema of f(x, y) with the constraint g(x, y), form L(x, y, ) = f(x, y) + g(x, y);  is Lagrange multiplier. Example: Find extrema of f(x, y) = 7(1 - x/4 – y/6) with constraint g(x, y) = (x – 2)2 + (y – 3)2 – 1 = 0; L(x, y, ) = 7(1 – x/4 – y/6) + [(x – 2)2 + (y – 3)2 – 1] Lx = -7/4 + 2 (x – 2) = 0 =>  = 7/8(x - 2) Ly = -7/6 + 2 (y – 3) = 0 =>  = 7/12(y - 3) L = (x – 2)2 + (y – 3)2 – 1; and 2x – 3y + 5 = 0 x = (3y – 5)/2; (1.5y – 4.5)2 + (y – 3)2 – 1 = 0 (quadratic 3.25 -22.5 28.25)  (5.275358 1.647719) y (5.413037 -0.02842) x 2/16/2019 rd

Lagrange Multiplier Suppose we have a problem: Maximize f(x, y) = 5 – (x – 2)2 – 2(y - 1)2 subject to g(x, y) = x + 4y = 3 Ignore the constraint to get fy and fx = 0 => x = 2 and y = 1 L(x, y, ) = 5 – (x – 2)2 – 2(y - 1)2 + (3 - x - 4y) Lx = -2(x – 2) -  = 0; Ly = -4(y – 1) - 4 = 0  = - 2x + 4 = - y + 1 or 2x – 1y = 3 1x + 4y = 3 x = 5/3; y = 1/3; f(5/3, 1/3) = 4, maximum;  = 2/3. x = 3 - 4y => 2(3 – 4y - 2) =  = 2 – 4y and  = 1 - y 2/16/2019 rd

Lagrange Multiplier Use a Lagrange multiplier to find the maximum rectangular area fenced in by 100 feet of fencing. L(x, y, ) = xy + (2x + 2y -100) Lx = y + 2; Ly = x + 2; and L = 2x + 2y – 100 = 0 (solve '((0 1 2 0)(1 0 2 0)(2 2 0 100)))  (25 25 -25/2) y + 2 = x + 2 => x = y => 4x = 100 or x = y = 25. x y 2/16/2019 rd

Lagrange Multiplier Max Z = -2x2 – y2 + xy + 8x + 3y s.t. 3x + y = 10 Observe y = 10 - 3x can be substituted. L(x, y, ) = -2x2 – y2 + xy + 8x + 3y + (3x + y – 10) Lx = -4x + y + 8 + 3 = 0 => y = -3 - 8 + 4x (1) Ly = -2y + x + 3 +  = 0 => 2y =  + 3 + 1x (2) L = 3x + y – 10 = 0 => y = 10 – 3x (3) (solve '((-4 1 3 -8)(1 -2 1 -3)(3 1 0 10)))  (69/28 73/28 -1/4) Z = 15.0179 Without constraint, the max occurs at x = 19/7, y = 20/7, max = 15.142857 2/16/2019 rd

Lagrange Multiplier Find critical points for f(x, y) = 25 - x2 - y2 with constraint x2 + y2 - 4y = 0 D = -2 0 L(x, y, ) = 25 – x2 - y2 + (x2 + y2 – 4y) 0 -2 Lx = -2x + 2x = 0 =>  = 1; x  0 Ly = -2y + 2y - 4 = 0 =>  = 2y/(2y – 4) L = x2 + y2 - 4y = 0 Conclude x = 0 => y(4 – y) = 0 with critical points (0, 0) and (0, 4) f(0, 0) = 25 which is absolute maximum 2/16/2019 rd

Lagrange Multipliers Find minimum of f(x, y, z) = 3x2 + 2y2 + 4z2 subject to 2x + 4y - 6z + 5 = 0 L(x, y, z, ) = 3x2 + 2y2 + 4z2 + (2x + 4y - 6z + 5) Lx = 6x + 2 Ly = 4y + 4 Lz = 8z - 6 L = 2x + 4y – 6z + 5 (solve '((6 0 0 2 0)(0 4 0 4 0)(0 0 8 -6 0)(2 4 -6 0 -5)))  ( x y z ) = (-2/11 -6/11 9/22 6/11) f(-2/11, -6/11, 9/22) = 1.289256 and |D| = 192 > 0 => minimum

Summary Use of optimization in systems engineering and analysis Differential Calculus Engineering Economy Mutually Exclusive Alternatives Life-cycle Cost Estimation Inventory Models Linear Programming Constrained Optimization 2/16/2019 rd

P9-10 Buy a car for $5500 with O&M $1200 increasing 15% per year, first year maintenance $350 increasing $100 per year. Salvage loss is 15%/yr &16% APR. Find economic life without considering time value of money. n MV loss O&M TCost CumTC EAC $825 $1550 $2375.00 $2375.00 $2375.00 701.25 1650 2351.25 4726.25 2363.13 596.06 1750 2346.06 7072.31 2357.44 506.65 1850 2356.65 9428.96 2357.24 ** 430.65 1950 2380.65 11,809.61 2361.92 6 366.00 2050 2416.00 14,225.61 2370.94 2944.28 3149 3369.67 2/16/2019 rd

P9-18 Demand 12/period Purchase Manufacture Ch = 0.02 Lead time 18 periods 13 periods Item Cost $11 $9.60 Procure Cost $20 $90 Replenish  25 units/period TC* = 11*12 + (2 * 20 * 0.02 * 12)1/2 = $135.10 Buy = 9.6*12 + [(2 * 90(1 – 12/25) * 0.02 * 12)]1/2 = $119.94 Make Make vs. Buy => $15.16 cheaper to make. Q* = sqrt(2 * 90 * 12)/[0.02(1 -12/25)] = 456 units L* = 12 * 13 = 156 units Page 260 for formulas 2/16/2019 rd

Or select fx to see math formula MINVERSE Excel inverse of a matrix. Highlight an empty array for the inverse, enter =MINVERSE(A1:C3) Where A1:C3 are the NW and SE corners of the matrix and press Control SHIFT Enter Or select fx to see math formula MINVERSE 2 -1 3 1 -5 -7 2/16/2019 rd

Power Plant A power plant (PP) is 1000 feet down stream on opposite bank of a 200 feet wide river. Cable costs $50/ft on land and $80/ft under water. Find best cost of installation. 1000 - x x d C(x) = 50(1000 – x) + 80(x2 + 2002)1/2 C'(x) = -50 + 80/ (x2 + 2002)1/2 = 0 when x = 160.1 => d = 256.2 => 1000 - x = 840 Cost = 840 * 50 + 256 * 80 = $62,488. 200 PPlant x 2/16/2019 rd

Unconstrained Extrema Given f(x, y) = 2x4 + y2 - x2 -2y, determine the relative extreme of f, if any. fx = 8x3 - 2x = 0; fy = 2y – 2 = 0 x = 0; ½, -½ ; y = 1 3 candidates (0, 1), (½ , 1) ( -½, 1) fxx = 24x2 - 2 > 0 for x = ½ & -½ and negative for x = 0. D = = -4 for (0, 1) saddle, 8 for (½, 1) f( ½ , 1) = -9/8; f(0, 1) = -1 relative minimum at ( ½, 1) and saddle point at (0, 1) 2/16/2019 rd

Constrained Minimum Find minimum value of W(x, y, z) = x2 + y2 + z2 subject to x + y + z =1. Let L(x, y, z, ) = x2 + y2 + z2 + (x + y + z – 1) Then Lx = 2x +  = 0, Ly = 2y +  = 0, Lz = 2z +  = 0, L = x + y + z – 1 = 0 (solve '((2 0 0 1 0)(0 2 0 1 0)(0 0 2 1 0)(1 1 1 0 1))) (1/3 1/3 1/3 -2/3) => W = 1/3. 2 0 0 0 2 0 = 8 > 0 => minimum 0 0 2 2/16/2019 rd

Constrained Maxima Find the maximum value of w = xy + z subject to x2 + y2 + z2 = 1. L(x, y, z, ) = xy + z = (x2 + y2 + z2 - 1) Lx = y + 2x = 0 y2z + 2  xyz = 0 Ly = x + 2y = 0 x2z + 2  xyz = 0 Lz = 1 + 2z = 0 xy + 2  xyz = 0 L = x2 + y2 + z2 -1 = 0 y2z = x2z = xy => z = 0 or y2 = x2 z = 0 can't be; thus y2 = x2; if x = y then x2z = xy = x2 => x = 0 or z = 1. Max at (0, 0, 1); w(0, 0, 1) = z = 1. 2/16/2019 rd

Critical Points Find and label the critical points of F(x, y) = 2x2 + y2 – 2xy + 5x - 3y + 1 Fx = 4x - 2y + 5 = 0 Fxx = 4 Fxy = -2 Fy = 2y - 2x - 3 = 0 Fyx = -2 Fyy = 2 \ (solve '((4 -2 -5)(-2 2 3)))  (-1, ½ ) D = 12 => (-1, ½ ) is relative minimum F(-1, ½) = 2 + ¼ + 1 - 5 - 3/2 + 1 = -9/4. 2/16/2019 rd

Find the inverse of the following matrices by forming the adjacency matrix with the identity matrix attached and converting the initial matrix to the identity matrix.   A = B = A-1 = B-1 = 1 4 1 0 1 4 1 0 1 0 -3/5 4/5 2 3 0 1 0 -5 -2 1 0 1 2/5 -1/5  2/16/2019 rd

Chapter 9 Problems Maximize the rectangular area with perimeter p. Symmetry => p/4 per side with p2/16 area. 2. f(x) = 2x2 –10x + 50 for x  0. 5/2 f'(x) = 4x -10 = 0 when x = 5/2, f(5/2) = 37.5 f"(x) = 4 => minimum, f(0) = 50; f() =  25 41.6 f(x) = x3 – 100x2 + 3125x => f(0) = 0 f'(x) = 3x2 -200x + 3125 = 0 and f"(x) = 6x – 200 (quadratic 3 -200 3125)  (41.667 25) f(25) = 31,250 is relative maximum and f(41.67) = 28,935.18 is relative minimum (poly-eval '(3 -200 3125) '(41 42)) (-32 17) 2/16/2019 rd

Practice S-1 Write the equation of the tangent to the curve y = x3 - 6x2 + 9x + 1 where its slope is a minimum. S-2 The demand is 20,000, ordering cost $50 and the holding cost is 0.1. Find Q*. S-3 Find the inverse of matrix 1 2 3 4 S-4 The cost for a machine is $1000 the first 5 years and $2000 the next 5 years. At i = 10% per year, the annual worth is closest to a) $1225 b) $1302 c) $1383 d) $1426 S-5 Max 3x1 + 4x2 subject to 2x1 + 3x2 <= 12 and 5x1 + 3x2 <= 15 S-6 Min 12x1+ 15x2 subject to 2x1 + 5x2 >= 3 and 3x1 + 3x2 >= 4 2/16/2019 rd

Example Write the equation of the tangent to the curve y = x3 - 6x2 + 9x + 1 where its slope is a minimum. Slope function is S(x) = 3x2 –12x + 9 = 0 S'(x) = 6x – 12 = 0 when x = 2 S(2) = -3 is the minimum slope at x = 2, y = 3 (y – 3) = -3(x – 2) or slope fn y = -3x + 9 (cubic 1 -6 9 1)  -0.1038 (quadratic 3 -12 9)  1, 3 2/16/2019 rd

Example The demand is 20,000, ordering cost $50 and the holding cost is 0.1. Find Q*. TC = CoD/Q + CHQ/2 TC' = -C0D/Q2 + CH/2 = 0 when Q* = $447.42 = (Total Cost) = $ 223.61 (Ordering Cost) + $ 223.61 (Holding Cost) Q* = 4472.14 2/16/2019 rd

Example Find the inverse of matrix 1 2 3 4 Inverse matrix Method I a + 3b = 1; 2a + 4b = 0 => a = -2b, => b = 1 c + 3d = 0 => c = -3d; 2c + 4d = 1 => -2d = 1 or d = -1/2 c = 3/2, b = 1, a = -2 Method II 1 2 1 0 1 2 1 0 1 0 -2 1 3 4 0 1 0 -2 -3 1 0 1 3/2 -1/2 (solve '((1 3 0 0 1)(2 4 0 0 0)(0 0 1 3 0)(0 0 2 4 1)))  (-2 1 3/2 -1/2) 2/16/2019 rd

Example The cost for a machine is $1000 for each of the first 5 years and $2000 for each of the next 5 years. At i = 10% per year, the annual worth is closest to a) $1225 b) $1302 c) $1383 d) $1426 0 5 10 P = 2000(P/A, 10%, 5)(P/F, 10%, 5) + 1000(P/A, 10%, 5) = 7581.57 * 0.6209 + 3790.79 = $8498.35 A = $8498.35(A/P, 10%, 10) = $1383.07. (AGP (+ (PGA 1000 10 10) (PGF (PGA 1000 10 5) 10 5)) 10 10)  $1383.07 2/16/2019 rd

Example Max 3x1 + 4x2 s.t. 2x1 + 3x2 <= 12 5x1 + 3x2 <= 15 Maximum See graph Z(0, 0) = 0; Z(0, 4) = 16; Z(3, 0) = 9; Z(1, 10/3) = 49/3 ** Minimum formulation Min 12u1 + 15u2 s.t. 2u1 + 5u2 >= 3 3u1 + 3u2 >= 4 Z**(11/9, 1/9) =147/9 = 49/3 ** 5 x2 4 (1, 10/3) 3 6 x1 0 0 11/9 1/9 49/3 0 1 5/9 -2/9 10/3 1 0 -1/3 1/3 1 Objective Function = (0 0 11/9 1/9 49/3) x1 = 1 x2 = 10/3 (11/9, 1/9) 2/16/2019 rd

Tableaus Max 3x1 + 4x2 s.t. 2x1 + 3x2 <= 12 AX = B 5x1 + 3x2 <= 15 Initial Tableau at corner point (0, 0) -3 -4 0 0 0 2 3 1 0 12 5 3 0 1 15 Tableau at corner point (0, 4) -1/3 0 4/3 0 16 2/3 1 1/3 0 4 3 0 -1 1 3 Final Tableau Tableau at corner point (3, 0) 0 0 11/9 1/9 49/3 0 -7/5 0 ¾ 9 0 1 5/9 -2/9 10/3 0 7/5 1 -2/5 6 1 0 -1/3 1/3 1 1 3/5 0 1/5 3 X2 5 4 X1 0 3 6 2/16/2019 rd

Final Tableau 0 0 11/9 1/9 49/3 Max 3x1 + 4x2 0 1 5/9 -2/9 10/3 s.t. 2x1 + 3x2 <= 12 1 0 -1/3 1/3 1 5x1 + 3x2 <= 15 AX = B A-1AX = A-1B = X = A-1B Objective Function row = (0 0 11/9 1/9 49/3) x1 = 1 x2 = 10/3 2/16/2019 rd

(LP '((-5 -3 -2 0)(2 3 2 24)(4 2 1 8)(1 4 6 36))) Max 5x1 + 3x2 + 2x3 How much for another yard of wool? s.t. 2x1 + 3x2 + 2x3 <= 24 (wool) cotton? 4x1 + 2x2 + 1x3 <= 8 (cotton) silk? 1x1 + 4x2 + 6x3 <= 36 (silk) Look at tableau; solve problem mentally without tableau. Specify the values for x1 to x6 using the final tableau below. 1/8 0 0 0 5/4 1/8 29/2 -25/8 0 0 1 -5/4 -1/8 19/2 23/8 1 0 0 3/4 -1/8 3/2 -7/4 0 1 0 -1/2 1/4 5 (MMULT (inverse #2A((2 1)(4 6))) #2A ((8)(36))) 2/16/2019 rd

Final Tableau Basis X1 X2 X3 X4 X5 X6 Q Z 1/8 0 0 0 5/4 1/8 29/2 X4 -25/8 0 0 1 -5/4 -1/8 19/2 X2 23/8 1 0 0 3/4 -1/8 3/2 X3 -7/4 0 1 0 -1/2 1/4 5 How much more another unit of cotton ? wool? silk? 2/16/2019 rd

Problem 9-6 BTUs needed for t thick: 2200(3000)(75 - 45) 10% S= $850 (1/0.13 + t/0.27) (0.5) = 39.6E7/ (1/0.13 + t/0.27) BTUs per year Annual cost = 39.6E7/ (1/0.13 + t/0.27) * (4.40/1000)(1/2000) = 875.6 / (1/0.13 + t/0.27) Thickness First cost CR Cost of Heat Total cost 0 $0 0 $113.83 113.83 2 2200*0.18 = 396 -19.24 57.98 38.74 4 2200*0.30 = 660 41.37 38.90 80.27 6 2200*0.44 = 968 112.09 29.27 141.36 2/16/2019 rd

Lagrange Multipliers You need 100 units made at 2 plants with the total cost function C = 0.1x2 + 7x + 15y + 1000. How should you divide the units between plants x and y to minimize the cost? L(x,y,) = 0.1x2 + 7x + 15y +1000 + (x + y – 100) Lx = 0.2x + 7 +  = 0 Ly = 15 +  = 0 =>  = -15 and 0.2x + 7 = 15 or x = 40; y = 60. (solve '((.2 0 1 -7)(0 0 1 -15)(1 1 0 100)))  (40 60 -15) C(40,60) = $2349 min See that Lxx and Lyy are > 0; D > 0 2/16/2019 rd

Lagrange Multipliers Find critical points for f(x,y,z) = x2 + 2y – z2 ; subject to 2x – y = 0; y + z = 0 L(x,y,z,1 2) = x2 + 2y –z2 + 1 (2x – y) + 2(y + z) Lx = 2x + 21 = 0 => 1 = -x Ly = 2 - 1 + 2 = 0 => 1 - 2 = 2 Lz = -2z + 2 = 0 => 2 = 2z L 1 = 2x – y = 0 => y = 2x L 2 = y + z = 0 => y = -z (solve '((2 0 0 2 0 0)(0 0 0 -1 1 -2)(0 0 -2 0 1 0) (2 -1 0 0 0 0)(0 1 1 0 0 0)))  (2/3 4/3 -4/3 -2/3 -8/3) 2/16/2019 rd

Lagrange Multipliers Find minimum of f(x, y, z) = 3x2 + 2y2 + 4z2 subject to 2x + 4y - 6z + 5 = 0 L(x, y, z, ) = 3x2 + 2y2 + 4z2 + (2x + 4y - 6z + 5) Lx = 6x + 2 Ly = 4y + 4 Lz = 8z - 6 L = 2x + 4y – 6z + 5) (solve '((6 0 0 2 0)(0 4 0 4 0)(0 0 8 -6 0)(2 4 -6 0 -5)))  (-2/11 -6/11 9/22 6/11) x = - /3, y = -, z = 3/4 -2/3 - 4 - 9/2 + 5 = 0  = -6/11, x = 2/11, y = 6/11, z = -9/22 f(1/11, 6/11, -9/22) = 1.289256 and |D| = 192 > 0 => minimum

Extrema for functions of 2 Variables Find the critical points of f(x, y)= x2 - 4x + 2y2 + 4y + 7 fx = 2x – 4 = 0 => x = 2 fy = 4y + 4 => y = -1 Critical point at (2, -1) fxx = 2; fyy = 4 => D = 8 > 0 and fxx > 0 => relative minima f(2, -1) = 4 - 8 + 2 – 4 + 7 = 1 2 0 Where D = = 8 0 4 (poly-eval '(1 -4 2/16/2019 rd

Problem 9-19 Demand = 250, R = 600, C0 = 400, Ch = $0.15, Cs = $3.25 Q* = [(1/(1 – 250/600)]1/2 [(2 *400*250)(0.15 + 1/3.25)]1/2 = 1546 L* = (250 * 12) = (1 – 250/600)1/2 [(2 * 400 * 250)/(3.25(1 + 3.25 /0.15)]1/2 = $2960 TC* = (90 * 250) + (1 – 250/600)1/2 * [(2 * 400 *0.15 *3.25 *250)/(0.15 + 3.25)]1/2 = $22,629 2/16/2019 rd

Inventory Analysis Over Time Annual Basis: D = 10,000, Co = $150, Ch = $0.75 Monthly Basis: D = 833.3, Co = $150, Ch = $0.0625 (EOQ 10E3 150 0.75)  Total Cost = $1500 = $ 750 (Ordering Cost) + $ 750 (Holding Cost) Q-star = 2000 units Cycle time = 73 (365/5) working days (EOQ 833.3 150 0.0625)  Total Cost = $125 = $ 62.50 (Ordering Cost) + $ 62.50 (Holding Cost) Q-star = 2000 units Cycle time = 876.02 ?? working days 2/16/2019 rd

Inventory Example Annual demand is 4050 cleaners, 6 ft2 for each, 900 ft2 available, holding cost is 16% of the $50 cleaner cost. The ordering cost is $32. It cost $0.50 per square foot for additional space. Is more space needed if 900 ft2 exist? (EOQ 4050 32 8)  $1440.00 Total Cost = $ 720.00 (Order Cost) + $ 720.00 (Hold Cost) Q* = 180 units AI 900/6 = 150 units costing 4050 * 32 / 150 + 8 * 75 = $1464 Expanding 30 units requires (6 * 30) 180 ft2 costing 0.5 * 180 = $90 making total cost $1440 saving $24 but costing $90 to expand or $1464 + 90 = $1554 2/16/2019 rd

Polar Coordinates Definition of Polar Coordinates (r, )  x x = r cos  y = r sin  2/16/2019 rd

Converting Convert polar coordinates (r, ) (6, 5/6) to rectangular coordinates (x, y). x = 6 cos 5/6 = 6 * y = 6 sin 5/6 = 6 * ½ = 3 2/16/2019 rd

F(x, y, z) Find the equation of the plane tangent to the surface x3 + 2xy2 - 7z3 + 3y + 1 = 0 at (1, 1, 1) Let 7z3 = F(x, y) = x3 + 2xy2 + 3y + 1 21z2Zx = 3x2 +2y2 => Zx = (3 + 2)/21 = 5/21 21z2Zy = 4xy + 3 => Zy = 7/21 = 1/3 (z – 1) = (5/21)(x – 1) + (1/3) (y – 1) (5/21)x + (1/3) y – z – 9/21 = 0 or 5x + 7y – 21z + 9 = 0 2/16/2019 rd

Point of plane nearest to Given Point Find the point of the plane 2x – 3y – 4z = 25 which is nearest to the point (3, 2, 1) D2 = (x – 3)2 + (y – 2)2 + (z – 1)2 where z = ¼ (2x – 3y – 25) D2 = (x – 3)2 + (y – 2)2 + [½ x – ¾y – (29/4)]2 D2x = 2(x – 3) + 2[½ x – ¾y – (29/4)]½ = 0 D2y = 2(y – 2) = 2[½ x – ¾y – (29/4)]-¾ Simplify to 10x – 3y = 53 -6x + 25 y = -55 (solve '((10 -3 53)(-6 25 -55)))  (5, -1) => z = -3 2/16/2019 rd

Circle in Polar Coordinates Graph r = 5 2/16/2019 rd

Partials Given z = 3x3y2 – 4x2y4, find zxx b) zyx c) zyy d) zxxx zx = 9x2y2 – 8xy3 => zxx = 18xy2 – 8y3 zy = 6x3y – 16x2y3 => zyx = 18x2y – 32xy3 zyy = 6x3 – 48x2y2 zxxx = 18y2 2/16/2019 rd

Minimum Distance Find the minimum distance s between the origin and the plane 2x + 2y + z = 6. S2 = f(x, y) = x2 + y2 + z2 = x2 + y2 + (6 – 2x – 2y)2 fx = 2x + 2(6 – 2x – 2y)(-2) = 0 = 2x - 4(6 – 2x – 2y) 10x + 8y = 24 fy = 2y + 2(6 – 2x – 2y)(-2) = 0 = 2y – 4(6 – 2x – 2y) 8x + 10y = 24 (Solve '((10 8 24)(8 10 24)))  (4/3, 4/3) Z = 18/3 – 16/3 = 2/3 S = (16/9 + 16/9 + 4/9)1/2 = 2 2/16/2019 rd

Diophantine Equation Find positive integral solutions to 12x + 7y = 220. x = (220 – 7y)/12 = 18 – (7y - 4)/12 for x > 0 (7y - 4)/12 <= 17 or 0 < y  29 x = 2; y = 28 2/16/2019 rd

Average Speed 60 You travel from a to b averaging 30mph. How fast must you travel back from b to a to average 60 mph for the round trip? 2/16/2019 rd