Chapter 10 Quality Control Review

1. Chapter 10 Quality Control Review Processing new accounts at a bank is intended to average 10 minutes each. Five samples of four observations each have been taken. The overall mean of the samples is 10.04 and the average range is 0.52. What are the upper and lower control limits for the X-bar and R charts? Solution: X-double-bar = 10.04 R-bar = 0.52 UCLx = X-double-bar + A2(R-bar) Look up A2 in the Table based on a sample size of 4  A2 = 0.729 So, UCLx = 10.04 + (0.729)(0.52) = 10.41 LCLx = 10.04 - (0.729)(0.52) = 9.66 CL = 10.01 UCLR = D4(R-bar) LCLR = D3(R-bar) Look up D4 and D3 in the Table based on a sample size of 4 UCLR = 2.282(0.52) = 1.18 LCLR = 0(0.52) = 0 CLR = 0.52

2. An Inspector counted the number of scratches in the exterior paint of each of the automobiles being prepared for shipment and found an average of 3.9 scratches. What are the three-sigma control chart limits? Solution This is a C-chart because we are interested in the “number of scratches.” C-bar = 3.9 Sigma = square root (3.9) = 1.97 Z = “three-sigma” = 3 UCLC = C-bar + Z(square root (c-bar)) = 3.9 + (3)(1.97) = 9.81 LCLC = C-bar - Z(square root (c-bar)) = 3.9 - (3)(1.97) = -2.01  0 CLC = 3.9

3. Out of 500 cars inspected on the dealer’s lot, 30 had severe hail damage that would require immediate repair. Find the two-sigma limits for the percent of defective cars. Solution This is a P-chart because we are interested in the “percent of defective cars” N = 500 Z = 2 P-bar = 30/500 = 0.06 SigmaP = Square Root((P-bar(1-P-bar))/n) = Square Root((.06(1-.06))/500) = .0106 UCLP = P-bar + Z(sigmaP) = 0.06 + (2)(.0106) = .0812 LCLP = P-bar - Z(sigmaP) = 0.06 - (2)(.0106) = .0388 CLP = 0.06

4. Find the initial solution using the Intuitive Method and the Total Cost X X 100 100 X 200 200 100 100 Total Cost = 3250 250 250 X 50 50 Use the MODI Method for this iteration. D-A: 5 – (0+3) = +2 D-B: 4 – (0 + 4) = 0 E-A: 8 – (0 + 3) = 5 F-C: 5 – (3 + 3) = -1 Make an improvement in cell F-C and transfer the new solution to the next table

5. Use the Stepping Stone method on this iteration. D-A: +5 – 3 + 5 – 6 = +1 D-B: +4 – 3 + 3 – 4 = 0 E-A: +8 – 3 + 5 – 6 = +4 F-B: +7 – 4 + 3 – 5 = +1 100 250 50 50 250 Total Cost = 3200

6. Basic EOQ (Economic Order Quantity) Model • Inventory Management • A toy manufacturer uses approximately 32,000 silicon chips annually. The chips are used at a steady rate during the 240 days a year that the plant operates. Annual holding cost is $3 per chip, and ordering cost is $120. • D = 32,000 H = 3 S = 120 Work 240 Days/yr • Determine • Optimal order quantity. Q0 = SQRT((2DS)/H) = SQRT((2(32000)(120))/3) = 1600 • The number of orders per year = D/Q0 = 32000/1600 = 20 times per year • c. The length of the order cycle (the number of workdays in an order cycle). = Q0/D = 1600/3200 = 1/20 year x 240 days/year = 12 days • d. Total Cost = D/Q(S) + Q/2 (H) = (32000/1600)(120) + (1600/2)(3) = = 2400 + 2400 = 4800

7. Quantity Discount Model Inventory Management A small manufacturing firm uses roughly 3,400 pounds of chemical dye a year. Currently the firm purchases 300 pounds per order and pays $3 per pound. The supplier has just announced that orders of 1,000 pounds or more will be filled at a price of $2 per pound. The manufacturing firm incurs a cost of $100 each time it submits an order and assigns an annual holding cost of 17% of the purchase price per pound. Determine the optimal order size and total cost. D = 3400 S = 100 H = .17(P)Step 1. Calculate Q0 for each discount level using Q0 = SQRT((2DS)/H) Q1 = 1154.7 Q2 = 1414.2Step 2. Round upward if you have to. Since both Q1 is above the first discount range, we don’t round down and leave it alone. Since Q2 is within the second discount range, we don’t do anything to it. Step 3. Calculate Total Cost for each discount level using TC = (D/Q)S + (Q/2)H + PD TC1 = $10,788.90 TC2 = $7,280.83 Step 4. Choose the lowest TC and order… Choose $7,280.83 and order 1414.2 units

8. Safety Stock Model • Inventory Management • The housekeeping department of a motel uses approximately 400 washcloths per day. The actual number tends to vary with the number of guests on any given night. Usage can be approximated by a normal distribution that has a mean of 400 and a standard deviation of 9 washcloths per day. A linen supply company delivers towels and washcloths with a lead time of three days. If the motel policy is to maintain a stockout risk of 2 percent, how much safety stock should be maintained and what is the ROP? • = 9, Lead Time (LT) = 3, d = 400, Stockout risk = 2%, Service level = 1 - .02 = 0.98 • Z = (from table or calculator) = 2.05 • SS = zs = 2.05(4) = 18.45 • Old ROP = dLT = 400(3) = 1200 • New ROP = dLT + SS = 1200 + 18.45 = 1218.45

9. Inventory Management Suppose we have the following information: p=200/day D = u*working days = 50 * 220 = 11,000/yr u=50/day H=$2 S=$70 We operate 220 days/year What model is it? EPQ model Find Q0 = SQRT((SDS)/H) SQRT(p/(p-u)) = SQRT((2(11000)(70))/2) * SQRT(200/(200-50)) = 1013.2 Find the number of production runs = D/Q0 = 11000/1013.2 = 10.86 runs per year What is the maximum inventory level Imax = (Q0/p)(p-u) = (1013.2/200) * (200 – 50) = 760 How much time between production runs (cycle time)? = Q0/u = 1013.2/50 = 20.28 days How many days in production (run time)? = Q0/p = 1013.2/200 = 5.07 days

10. Know these terms for the exam. • Supply Chain • Functions of a Supply Chain • Movement of- Materials (Inventory velocity)- Information (Information velocity) • Value Chain • Value Added • Supply / Demand • Strategic Partnering • Bullwhip Effect • Benefits of Supply Chain • Elements of Supply Chain • Logistics • Traffic Management • DRP • Supplier Certification • EDI • e-Commerce • e-Business • CPFR • Supply Chain Performance Drivers • Inventory Velocity • Information Velocity • Purchasing- Roles of Purchasing- Centralized Purchasing- Decentralized Purchasing • Value Analysis • RFID • SCOR