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OPERATIONS MANAGEMENT ASSIGNMENT 3 SUBMITTED BY :John Michael franklin SECTION: A ROLL NO:SMBA10080 Chapter 10 Q3) Average Mean = x = (3.06 + 3.15 + 3.11 + 3.13 + 3.06 + 3.09) / 6 = 3.1 Range = R = (0.42 + 0.5 + 0.41 + 0.46 + 0.46 + 0.45 ) / 6 = 0.45 From table 10.3, for n=20, we have , A2 = 0.18, D3 = 0.41, D4 = 1.59 UCLx = x + A2 R = 3.1 + (0.18 x 0.45) = 3.181 LCLx= x – A2 R = 3.1 – (0.18 x 0.45) = 3.019 UCLR = D4 R = 1.59 x 0.45 = 0.7155 LCLR = D3 R = 0.41 x 0.45 = 0.1845 The largest sample mean = 3.15 , smallest = 3.06. Both are well within the control limits. Similarly, the largest sample range is 0.50 and the smallest is 0.41.Both are well within the control limits. Hence, the result suggests that the process is in control. Q4) From the given data, then, using MS Excel, we compute as follows:   | Sample |   | 1 | 2 | 3 | 4 | 5 | 6 |   | 79.2 | 80.5 | 79.6 | 78.9 | 80.5 | 79.7 |   | 78.8 | 78.7 | 79.6 | 79.4 | 79.6 | 80.6 |   | 80 | 81 | 80.4 | 79.7 | 80.4 | 80.5 |   | 78.4 | 80.4 | 80.3 | 79.4 | 80.8 | 80 |   | 81 | 80.1 | 80.8 | 80.6 | 78.8 | 81.1 | Total | 397.4 | 400.7 | 400.7 | 398 | 400.1 | 401.9 | Mean | 79.48 | 80.14 | 80.14 | 79.6 | 80.02 | 80.38 | Largest No. | 81 | 81 | 80.8 | 80.6 | 80.8 | 81.1 | Smallest No. | 78.4 | 78.7 | 79.6 | 78.9 | 78.8 | 79.7 | Range | 2.6 | 2.3 | 1.2 | 1.7 | 2 | 1.4 | Then, using MS Excel, we compute as follow: Average Mean = x | 79.96 | Average Range = R | 1.866667 | From table 10.3, for n=5, we have , A2 = 0.58, D3 = 0, D4 = 2.11 UCLx = x + A2 R = 79.96+ (0.58 x 1.866667) = 81.0446 LCLx= x – A2 R = 79.96- (0.58 x 1.866667) = 79.96 – 1.0846 =78.8754 UCLR = D4 R = 2.11 x 1.866667 = 3.9457 LCLR = D3 R = 0 x 1.866667 = 0 The largest sample mean is 80.38 and the smallest is 79.48. Both are well within the control limits. Similarly, the largest sample range is 2.6 and the smallest is 1.2.Both are well within the control limits. Hence, the result suggests that the process is in control. Chapter 12 Q15) Annual usage = D= 4900 Quantity | Unit Price | Holding Cost (H) | Less than 1000 | $5 | .4x5 = $2 | 1000 – 3999 | $4.95 | .4x4.95 = $1.98 | 4000 – 5999 | $4.90 | .4x4.90 = $1.96 | More than 6000 | $4.85 | .4x4.85 = $1.94 | Ordering cost = S = $50 Holding cost = H = .4PD Minimum point Q5 = 2DSH * 24900502 * 494.97 * 495 Minimum point Q4.95 = 2DSH * 24900501.98 * 497.46 * 498 Minimum point Q4.90 = 2DSH * 24900501.96 * 500 Minimum point Q4.85 = 2DSH * 24900501.94 * 502.57 * 503 Therefore the feasible minimum quantity is 495 units. Now we will find out the total costs TC = Carrying costs + Ordering costs + Purchasing costs * (Q2)H + DQS + PD * 4952*2 + 4900495*50 + (4900)*5 * 495 + 495 + 24500 * TC = $25490 Now finding the total cost for the quantity at which the price is minimum Q = 6000 TC = Carrying costs + Ordering costs + Purchasing costs * (Q2)H + DQS + PD * 60002*1.94 + 49006000*50 + (4900)*4.85 * 5820 + 40.8333 + 23765 * TC = $29625 The optimal quantity to order to minimize cost is thus 495 units. Q 16) Annual Demend = D = 800x12 = 9600 Ordering cost = S= $40 Holding cost = H = .25xUnitPrice Supplier A SUPPLIER A | Quantity | Unit Price | Holding Cost | 1-199 | $14 | .25x14 = $3.5 | 200 – 499 | $13.80 | .25x13.8 = $3.45 | 500 + | $13.60 | .25x13.6 = $3.4 | Minimum Quantity Q14 = 2DSH Q14 = 29600403.5 Q14 = 468.43 Units * 469 Units SUPPLIER B | Quantity | Unit Price | Holding Cost | 1-149 | $14.1 | .25x14.1 = $3.525 | 150 – 349 | $13.90 | .25x13.9 = $3.475 | 350 + | $13.70 | .25x13.7 = $3.425 | Supplier B Minimum Quantity Q14.1 = 29600403.525 Q14.1 = 466.76 * 467 Units Q13.8 = 29600403.45 Q13.8 = 471.81 Units * 472 Units Q13.6 = 29600403.4 Q13.6 = 475.27 * 476 Units Therefore the optimal quantity is 472 Units TC = Carrying costs + Ordering costs + Purchasing costs * (Q2)H + DQS + PD * (4722)3.45 + 960047440 + 13.8x9600 Q13.9 = 29600403.475 Q13.9 = 470.11 Units * 471 Units Q13.7 = 2DSH Q13.7 = 29600403.425 * 473.53 Units * 474 Units Therefore the optimal quantity is 474 Units TC = Carrying costs + Ordering costs + Purchasing costs * (Q2)H + DQS + PD * (4742)3.425 + 960047440 + 13.7x9600 * 811.725+810.12+131520 * 133141.84 Now finding out the quantity at which price is minimum Q = 350 * TC = (Q2)H + DQS + PD * TC = (3502)3.425 + 960035040 + 13.7x9600 * TC = 599.375 + 1097.14 + 131520 * 133216.515 Now finding the quantity at which price is min. For Q = 500 TC is as follows: * TC = (5002)3.4 + 960050040 + 13.6x9600 * TC = 850+ 768+ 130560 * TC = 132178 Therefore we can observe that Supplier A supplies at the lowest rate for a quantity 500 units, so supplier A will be preferred. Q) Difference between different types of EOQ. Economic Order Quantity: The economic order quantity (EOQ) is the fixed order quantity (Q) that minimizes the total annual costs of placing orders and holding inventory (TC). This type of model is used when i) Demand is independent. ii) Compute how much to order. Formula: TCEOQ = (Q2)H + DQS + PD TC= Total Cost Q= Quantity H= Holding Cost D= Demand S= Ordering cost P= Unit Price D= Demand Economic Production Quantity: The economic production quantity (EPQ) is the production quantity (lot size) that minimizes the total annual cost of setups and holding inventory. This type of model is used when i) Demand is dependent. ii) Compute how much to make at one time. Formula: TC = Total Cost D= Demand Q= Quantity S= Order Quantity Imax= Maximum Inventory H= Holding Cost Reorder Point: EOQ tells us about how much to order but not when to order. ROP tells us about when to order. The reorder point occurs when the quantity on hand falls below a threshold value. There are 4 determinants of the reorder point: i) The rate of demand. ii) The lead time. iii) The extent of demand and/or lead variability. iv) The degree of stockout risk acceptable to management.