Inventory Control System for a New Product: 22-Mec-B4 Integrated Manufacturing Systems, May 2018, Q3

Question 3: Inventory Control System for a New Product (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Data as printed
Symbol	Quantity	Value
$Q$	(a) Production economic lot size	1,000 units
$p$	(b) Production (supply) rate	50 units per day
$d$	(c) Usage rate	20 units per day
$L$	(d) Delay from order release to production start	10 ± 5 days
$H$	(e) Annual cost of storing one unit	$5.00 per unit-year
$c$	(f) Production cost of the product	$15.00 per unit
$N$	(g) Production and sales days per year	240

Find. A complete operating policy — annual demand, number of production runs, the length and shape of one inventory cycle, maximum and average inventory, the set-up cost implied by calling 1,000 units “economic”, the reorder point and safety stock, and the resulting annual cost — together with the checks that prove the policy is internally consistent.

Question 3 — inventory profile of the recommended (Q, R) system. Stock builds at 30 units a day for 20 days, peaks at 700 units, then depletes at 20 a day; the order is released when the position reaches 300 units.

Approach. Because stock is supplied gradually while it is being consumed, the governing model is the finite-production-rate (economic production quantity) model rather than plain EOQ. Convert the daily rates to an annual demand, build the inventory profile, invert the EPQ formula to recover the set-up cost the stated lot size implies, and set the reorder point from the lead time.

Check: how the “10 ± 5 days” start-up delay is read. The paper gives no distribution and no service level, so the band is treated here as a worst case: the order is released early enough to survive a 15-day delay, which makes the safety stock $d \times 5 = 100$ units and the reorder point $d \times 15 = 300$ units. The statistical alternative — reading ±5 days as roughly a two-standard-deviation band, so $\sigma_{L} = 2.5$ d and $\sigma_{DDLT} = 20(2.5) = 50$ units, and setting a 95 per cent cycle service level — gives $z\sigma = 1.645(50) = 82$ units of safety stock, only 18 units (under one day of usage) below the worst-case figure. The policy is insensitive to the choice; the worst-case reading is carried through below and stated here per Note 1.

Confirm the model applies, and get annual demand. Stock can only accumulate if production outruns usage, and it does: $p = 50 > d = 20$ units per day, so the build-up rate is $p - d = 30$ units per day. Annual demand is $$D = d\,N = 20 \times 240 = 4{,}800\ \text{units per year}$$ and the number of production runs per year is $m = D/Q = 4{,}800/1{,}000 = 4.8$.
Lay out the inventory cycle. Each lot takes $t_{p}=Q/p=1{,}000/50=20$ working days to build. One full cycle lasts $t_{c}=Q/d=1{,}000/20=50$ working days, of which the remaining $50-20=30$ days are pure depletion with the machine turned over to other work. As a check, $N/t_{c}=240/50=4.8$ cycles a year, matching $m$ exactly.
Compute the peak and average inventory. Stock rises at $p-d$ only for as long as the run lasts, so the peak is not $Q$ but $$I_{\max}=Q\left(1-\frac{d}{p}\right)=1{,}000\left(1-\frac{20}{50}\right) =1{,}000(0.60)=600\ \text{units}$$ which the build-up form confirms, $(p-d)t_{p}=30 \times 20=600$. The cycle stock averages $\bar{I}=I_{\max}/2=300$ units, and its annual carrying cost is $\bar{I}H = 300 \times 5 = $ $1,500.
Recover the set-up cost the question withholds. The paper never states a set-up or ordering charge, but it does describe 1,000 units as the economic lot, and that word is a datum: it means 1,000 satisfies the EPQ expression $Q^{*}=\sqrt{2DS/[H(1-d/p)]}$. Inverting for the set-up cost, $$S=\frac{Q^{2}H\left(1-\dfrac{d}{p}\right)}{2D} =\frac{(1{,}000)^{2}(5)(0.60)}{2(4{,}800)} =\frac{3{,}000{,}000}{9{,}600} =\boxed{S = 312.50\ \text{dollars per set-up}}$$
Check the recovery two ways. Substituting $312.50 back into the EPQ formula reproduces $Q^{*}=\sqrt{2(4{,}800)(312.50)/[5(0.60)]}=\sqrt{1{,}000{,}000}=1{,}000$ units exactly. More usefully, the annual set-up cost is $mS = 4.8 \times 312.50 = $ $1,500, which equals the annual carrying cost of $1,500 computed in step 3. Set-up cost equalling carrying cost is the balance that defines an economic lot size, so if those two figures had not matched, a datum would have been misread. Both checks pass.
Set the reorder point and the safety stock. Production begins $10 \pm 5$ days after the order is released, so the trigger must cover the longest credible delay of $L_{\max}=15$ days. Usage during that delay is $$R = d\,L_{\max} = 20 \times 15 = \boxed{\text{reorder point } R = 300\ \text{units}}$$ made up of $d\,L_{\text{nom}} = 20 \times 10 = 200$ units of expected lead-time demand plus $SS = d \times 5 = 100$ units of safety stock against the five-day uncertainty. Carrying that safety stock costs $100 \times 5 = $ $500 a year, and it lifts the peak on the shelf to $I_{\max}+SS = 600 + 100 = 700$ units.
Assemble the annual cost of the policy. Material dominates and does not depend on the lot size: $Dc = 4{,}800 \times 15 = $ $72,000 a year. The inventory-related costs are set-up $1,500, cycle-stock carrying $1,500 and safety-stock carrying $500, a total of $3,500 a year, so the whole system costs $75,500 a year. Inventory investment peaks at $700 \times 15 = $ $10,500 at production cost, and the stock turns $D/(\bar{I}+SS) = 4{,}800/400 = 12$ times a year.
State the system as an operating rule. The result is a fixed-order- quantity (Q, R) system: hold a perpetual record of on-hand plus on-order; when the position falls to 300 units, release an order for 1,000 units; production starts about ten days later, runs for twenty working days at 50 units a day, and the shelf peaks near 700 units before depleting at 20 a day. Expect 4.8 runs a year, one about every ten weeks. Review the parameters once the new product's demand history is long enough to estimate variability, since every figure above rests on a deterministic usage rate of exactly 20 units a day.

Final results — Question 3
Quantity	Symbol	Value
Annual demand	$D$	4,800 units per year
Production runs per year	$m$	4.8
Run length / cycle length	$t_p$ / $t_c$	20 days / 50 days
Maximum cycle inventory	$I_{\max}$	600 units
Average cycle inventory	$\bar{I}$	300 units
Implied set-up cost	$S$	$312.50 per set-up
Safety stock	$SS$	100 units
Reorder point	$R$	300 units
Peak stock on the shelf	$I_{\max}+SS$	700 units
Annual set-up cost	$mS$	$1,500
Annual carrying cost (cycle + safety)		$1,500 + $500 = $2,000
Annual cost excluding material		$3,500
Total annual cost including material		$75,500
Inventory turns		12 per year

Topic. Finite-production-rate lot sizing (the economic production quantity) and the reorder-point policy built on it.

Key relations. $$Q^{*}=\sqrt{\frac{2DS}{H\left(1-\dfrac{d}{p}\right)}},\qquad I_{\max}=Q\left(1-\frac{d}{p}\right),\qquad R=d\,L+SS$$ with annual set-up cost $\left(D/Q\right)S$ equal to annual carrying cost $\tfrac{1}{2}Q\!\left(1-d/p\right)H$ at the optimum.

Why this works. Plain EOQ assumes a lot arrives instantaneously, so the shelf jumps to $Q$; here the lot dribbles in at 50 a day while 20 a day are being taken out, so the shelf never rises above $Q(1-d/p)$. That single factor, 0.60 in this problem, is the whole difference between the two models, and it reduces both the peak inventory and the carrying cost by 40 per cent, which in turn justifies a larger economic lot than plain EOQ would give. Recovering $S$ by inverting the formula is legitimate because “economic lot size” is a statement that the first-order condition already holds, and the resulting number is confirmed independently by the set-up-equals-carrying balance.

Common pitfall. Using $I_{\max}=Q=1{,}000$ instead of $600$ overstates the carrying cost by two thirds and, through the inverted formula, overstates the set-up cost to $520.83. The second pitfall is confusing the run length with the cycle length: production occupies the machine for 20 days, but the lot must last 50 days, which is what tells the planner the machine is free for other work for 30 days in every cycle.

Source. Nahmias and Olsen, Production and Operations Analysis, 7th ed., ch. 4 (Inventory Control Subject to Known Demand), the finite production rate model and the reorder point; Buffa and Sarin, Modern Production / Operations Management, 8th ed., ch. 6 (Inventory Control Systems).

Question 3 of 5: Inventory Control System for a New Product

Question 3: Inventory Control System for a New Product (20 marks)