Question

a) FID Corporation produces silver widgets that it uses in joints and relays. FID needs to determine the order quantity, Q, to meet the annual demand at the lowest cost. The price of silver depends on the quantity ordered. Here are the price-breaks (for all-units discounts) and other data for the problem: Price of silver Annual demand Holding cost rate Ordering cost $0.85 per pound (lb) up to 2500 pounds $0.80 per lb for orders more than 2500 lbs and less than 5000 lbs $0.75 per lb for orders 5000 lbs or more 40000 lbs 20% per dollar per year $30 b) Re-solve now with the exception that the discount is incremental, that is for any order quantity, Q. per pound (lb) cost, c(Q), is as follows: $0.85, if 0 < < 2500, $0.85 for first 2500lbs, $0.80 for the next Q - 2500 lbs, if 2500< < 5000 $0.85 for first 2500lbs, $0.80 for the next 2500 lbs, and $0.75 per lb for the next Q - 5000 lbs. if 5000 < 0

Answer 1

Hence, Net heat of combustion - Gross heat of Combustion=881.6 kJ/mol D - Assuming P = 40000 lbs.; ordering cost - $30 g cost, i- 20% per dollar or 0.2 & , Q< 2500 pound (= 0.85 Q- ) 2. DE. J2x40000 x 30 » Q = 3757.34 as it is 72500 Dur assumption is wrong. Q = v ci V 0.85 x 0.2 MULTI Now, if 2500< Q < 5000 , C=0.8 → Q= 2X40000 x 30 v 0.2 x 0.80 => Q - 3872.98 or 1Q=3873 as; i t i cosse it as ber assumption.

be if $0.85 for OC Q5 2500 - $0.85 for first 0-2500 $0.8 for Q-2500 and if a> 5000 - $0.85 for first 2500, for 0.8 for 5000 - 25 and $0.75 for Q-5000. R = 2D [S+V; -9;ci] V - b ci qo=es q ,=2500 q2 = 5000 Vo-0. Viz (2500-0)x0.85 = 2125 V2 = (25000-0)0.85+ (5000 - 2500)x0.8 - 4125 0.85X02 Now, Quus 2x 40 000X 30 3757. > 2500 . not possible Now, if Q >2500 ; C = 0.8 and Q25000 - Qi - 2x40000 (30+ 2125- 2500X0.85 - NOW Now - 0.28 0.8 >> Qiz 8803 7 5000 . Not possible if Q > 5000 > Sooo 2x 40000 (30+4125-50Oox 0.75) Q2 = J 2x - 0.2 x 0.75 Q2 = 14696.94 ~ 14697 obtimal order quanto 14697/ Hence,