b The holding cost are 9 off the value with a value off 100 per pound for the

B the holding cost are 9 off the value with a value

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b) The holding cost are 9% off the value, with a value off \$100 per pound for the breakdown and \$125 per pound for the finish . This leads to: 9% × (\$100×526,540+\$125×182,837)/(526,540+182,837) = \$9.58 The fixed cost are 25\$x8x2= \$400 The total demand is 526540 + 182387 = 750377 units per year Q * = 2DK h = 2×400×750377 9,58 = 7515,931 = 7516 Average throughout the process: 570 + 450 = 1020. 7916/1020 = 7,761 Quantity for breakdown: 570*7,761 = 4423,608 Quantity for finish : 450 * 7,761 = 3492,45 526540 /4423,608 *55 minutes = 6546,62 minutes = 109 uur (optimal use Breakdown ) 182837 /3492,45* 120 minutes = 6282,249 minutes = 104,7 hours = 105 hours (109+104)/24 = 8,90 days + 2*8 hours ( changeover shifts ) = 9,566 hours This is exactly two weeks a) The large mill has an uptime of 85%, therefore the total uptime per year is: (52-4)= 48 weeks with a changeover time of 8 hours. The machine operates 5 days a week: 47*5 = 240 days hence, the uptime is: 0,85*240= 204 days The amount of cycles per year are: 709,377/7697= 92,16 cycles = 93 cycles 93 cycles *1,46 days = 134,57 days, which is less than 204 days resulting in the large mill having enough capacity. d) The manufacturing lead-time would thus significantly reduce. When an order arrives, the breakdown phase is a lot faster compared to the two-week cycle. The order can go through the process of breakdown faster and go to the finish phase faster, as the products don’t have to wait anymore. 7
6. Consider material 1003 in Figure 3-31. This product has high sales volumes with a reasonably low variability since its monthly sales has an average of 629.0 units and a standard deviation of 235.2 units. As a result, H.C. Starck decided to keep inventory for this final product. For simplicity, they assume the following: - the sales numbers represent the actual demand, - the demand follows a normal distribution, - the replenishment lead time is given by Figure 3-26 (that is, the average is 2.3 weeks and the standard deviation is 1.8 weeks), - there is a two-week production cycle and - there are 4.3 weeks in one month. a) Which type of inventory replenishment policy would you recommend for this final product? b) What should the value(s) of the policy parameter(s) be to achieve a cycle service level equal to 90%? Include the formula that you used. c) For the value(s) of the policy parameter(s) found in Question 7b, what is safety stock level? d) For the value(s) of the policy parameter(s) found in Question 7b, what is the average fill rate? Include the formula that you used. a) That I would recommend for H.C. Starck is continuous reviews. There is a lot off variation in the demand off Starck as the average demand is 629 units and the standard deviation is 235,2. Therefore it is necessary to pay attention on the stock. This won’t be necessary if the company would have a high safety stock, however the goal is too minimize stock making it no option. b) AVG = 629 STD = 235,2 8
AVGL = 2,3/4,3 = 0,5349 STDL = 1,8/4,3 = 0,4186 Z value = 1.29 R = AVG× AVGL + z × AVGL×STD 2 + AVG 2 ×STDL 2 . R = 629 × 0,5349 + 1,29 × 0,5349 × 235,2 2 + 629 2 × 0,4186 2 . R = 742,17 c) Safety Stock = z × AVGL × STD 2 + AVG 2 × STDL 2 Safety stock = 1,29 × 0,5349 × 235,2 2 + 629 2 × 0,4186 2 . Safety Stock = 405,718 d) Z= 1.29, L(z)= 0.0465 (in Table) STD = 253,2 L = 0,5349 Q = 629 Fill rate= 1- STD* √L* L(z)/Q Fill rate = 1 – 235,2 * √0,5349 * 0,0465/629 Fill rate = 0,987 = 99% 9
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• Fall '16
• B. Beije
• H.C. Starck