IEOR150F10_hw02_sol

# IEOR150F10_hw02_sol - IEOR 150 Fall 2010 Suggested Solution...

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Unformatted text preview: IEOR 150, Fall 2010 Suggested Solution to Homework 2 1. We have the production rate P = 10000 pounds per day or 2.5M pounds per year, demand rate λ = 0 . 6M pounds per year, setup cost K = \$1500 dollars per order, purchasing cost c = \$3 . 5 per pound, annual interest rate I = 0 . 22+0 . 12 = 0 . 34, holding cost h = cI = 3 . 5 × . 34 = \$1 . 19 per pound per year, and the effective holding cost h ′ = h (1- λ P ) = \$0 . 9044 per pound per year. (a) The optimal production lot size is Q ∗ = √ 2 Kλ h ′ = 44612 . 43 pounds . (b) • (Method 1) The optimal cycle time is T ∗ = Q * λ = 0 . 0744 years. The uptime in a cycle is T 1 = Q * P = 0 . 0178 years. The downtime in a cycle is T 2 = H λ = ( P − λ ) T 1 λ = 0 . 0564 years. Therefore, we have the proportion of uptime T 1 T = 24% and the proportion of downtime T 2 T = 76% . • (Method 2) Note that P = 2 . 5M, λ = 0 . 6M, so T 1 T = λ P = 24% and T 2 T = P − λ P = 76%. (c) The total holding and setup cost is G ∗ = √ 2 Kλh ′ = \$40347 . 49 . Therefore, the annual holding cost and the annual setup cost are both G * 2 = \$20173 . 75 . The annual purchasing cost is λc = \$2100000 . The annual total cost is thus 2100000 + 40347 . 49 = \$2140347 . 49 . The annual profit is (3 . 9- 3 . 5) λ- G * = 240000- 40347 . 49 = \$199652 . 51 . 2. We have the annual interest rate I = 0 . 2, setup cost K = \$100 per order, annual demand λ = 20000 units, unit purchasing cost from the three suppliers c A = \$2 . 5, c B = \$2 . 4, and c C = \$2 . 3. It follows that the annual holding costs per unit are h A = \$0 . 5, h B = \$0 . 48, and h C = \$0 . 46. Let L A = 0 unit, L B = 3000 units, and L C = 4000 units be the lower bounds of order quantity of the three suppliers.= 4000 units be the lower bounds of order quantity of the three suppliers....
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IEOR150F10_hw02_sol - IEOR 150 Fall 2010 Suggested Solution...

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