13.6 Practice Problems
Problem 13.6.1. A shop sells \(2400\) units a year; ordering costs Rs. 100 per order and holding Rs. 3 per unit per year. Find the EOQ, the number of orders, the cycle time and the annual cost, and the reorder point if delivery takes \(5\) working days out of \(300\).
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Solution. \(Q^* = \sqrt {2 \times 100 \times 2400/3} = 400\); \(D/Q^* = 6\) orders a year, one every \(2\) months (\(50\) working days); cost \(\sqrt {2 \times 100 \times 2400 \times 3} =\) Rs. 1200. Daily demand is \(2400/300 = 8\), so reorder when the stock falls to \(r = 5 \times 8 = 40\).
Problem 13.6.2. In the previous problem, a supplier offers a \(2\%\) discount on orders of \(1000\) or more, the unit price being Rs. 20 and holding \(15\%\) of the price a year. Should the shop accept?
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Solution. Without the discount: \(h = 3\), cost \(1200 + 20 \times 2400 = 49\,200\). With it, price \(19.6\), \(h = 2.94\), and at \(Q = 1000\) (the EOQ \(\sqrt {2 \times 100 \times 2400/2.94} = 404\) is below the minimum), the cost is \(100 \times 2.4 + 2.94 \times 500 + 19.6 \times 2400 = 240 + 1470 + 47\,040 = 48\,750\). Accept: the saving of Rs. 960 on purchases outweighs the extra Rs. 510 of ordering and holding.
Problem 13.6.3. Derive the EOQ with backorders: show that with order size \(Q\) and maximum backlog \(B\) the annual cost is \(KD/Q + h(Q - B)^2/2Q + pB^2/2Q\), and minimise it.
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Solution. In each cycle of length \(Q/D\) the stock is positive for a fraction \((Q - B)/Q\) of the time with average \((Q - B)/2\), and negative for a fraction \(B/Q\) with average backlog \(B/2\), giving the holding and backorder terms. Setting \(\partial /\partial B = 0\): \(-h(Q - B) + pB = 0\), \(B = hQ/(h + p)\). Substituting, the cost becomes \(KD/Q + \frac {hp}{h + p}\cdot \frac Q2\), an EOQ cost with holding rate \(hp/(h + p)\), minimised at \(Q^* = \sqrt {2KD(h + p)/hp}\).
Problem 13.6.4. A baker sells cakes at Rs. 50, costing Rs. 20 to make; unsold cakes are given away. Daily demand is uniform on \(\{20, 21, \ldots , 39\}\). How many should be baked?
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Solution. \(c_u = 30\), \(c_o = 20\), critical fractile \(\frac {30}{50} = 0.6\). \(F(Q) = (Q - 19)/20\) for \(Q\) in the range, and \(F(Q) \geq 0.6\) first at \(Q - 19 = 12\), \(Q = 31\). Bake \(31\) cakes, more than the mean demand \(29.5\), because a lost sale costs more than a wasted cake.
Problem 13.6.5. Lead-time demand is normal with mean \(200\) and standard deviation \(30\). Find the reorder point for a \(95\%\) probability of no stock-out per cycle, and the safety stock.
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Solution. \(z = 1.645\), so \(r = 200 + 1.645 \times 30 = 249.4\), say \(250\); the safety stock is about \(50\) units. Raising the service level to \(99\%\) (\(z = 2.326\)) needs \(r = 270\): the last few percentage points of service are expensive.
Problem 13.6.6. Show that in the newsvendor model the optimal expected profit equals \((p - c)\mu \) minus a term proportional to \(\sigma \) when demand is normal, and interpret.
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Solution. Write \(Q = \mu + z\sigma \). The expected leftover is \(\mathbb E(Q - X)^+ = \sigma \,\mathbb E(z - Z)^+\) and the expected shortage \(\sigma \,\mathbb E(Z - z)^+\) with \(Z\) standard normal, so the expected profit is \((p - c)\mu - \sigma \bigl (c_o\mathbb E(z - Z)^+ + c_u\mathbb E(Z - z)^+\bigr )\). At the critical fractile the bracket equals \((c_u + c_o)\varphi (z)\), so the loss from uncertainty is \((c_u + c_o)\varphi (z)\sigma \), proportional to \(\sigma \): halving the forecast error halves the cost of uncertainty, which measures what better forecasting is worth.
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