# Chapter 10: Inventory - Part 2 Types of Inventory and Demand Availability Cost vs. Service Tradeoff Pull vs. Push Reorder Point System Periodic Review.

## Presentation on theme: "Chapter 10: Inventory - Part 2 Types of Inventory and Demand Availability Cost vs. Service Tradeoff Pull vs. Push Reorder Point System Periodic Review."— Presentation transcript:

Chapter 10: Inventory - Part 2 Types of Inventory and Demand Availability Cost vs. Service Tradeoff Pull vs. Push Reorder Point System Periodic Review System Joint Ordering Number of Stocking Points Investment Limit Just-In-Time

Optimal Inventory Control For perpetual (continual) demand. Treat each stocking point independently. Consider 1 product art 1 location. Periodic Determine: Review System How much to order: M-q i When to (re)order: T Find optimal values for: M & T.

Periodic Review System Receive 1st order Place 1st order Receive 2nd order Place 2nd order Receive 3rd order Place 3rd order LT1 LT2 LT3 Order M - minus amount on hand every T time units. T = 20 days and M=90 in this example.

Periodic Review Useful when: –Inventory is reviewed on a fixed schedule (e.g., every week). –Multiple items are ordered from one supplier. –A common order interval allows transportation economies of scale. Each order must last for time T + LT -> Must protect against stockout during T + LT We will consider only constant lead time: s LT = 0 T + LT s’ d = s d

D = demand (usually annual)d = demand rate S = order cost (\$/order)LT = (constant) lead time I = carrying cost k = stockout cost (% of value/unit time)P = probability of being in C = item value (\$/item) stock during lead time s d = std. deviation of demand s LT = std. deviation of lead time = 0 s’ d = std. deviation of demand during lead time M = maximum level T = time between orders TC = total cost (usually annual) N = number of orders/year Q = average order quantity Inventory Variables

Periodic Review - Optimal Ordering Optimal order interval: Optimal number of orders/year: Maximum level : M = d(T+LT) + z s’ d Optimal cost: 2S T = IDC T 1 N = TC = T S + IC 2 dT + IC z s’ d + T 1 k s’ d E(z) T + LT s’ d = s d Std. deviation of demand during lead time: Average inventory: AIL =(dT)/2 + z s’ d

3 Cases 1. Stockout cost k is known; P is not known. -> Calculate optimal P (similar to reorder point) We will not consider this case. 2. Stock cost k is not known; P is known. -> Can not use last term in TC. 3. Stockout cost k is known; P is known. -> Could use k to calculate optimal P.

Periodic Review Example (same as Reorder Point) D = 5000 units/year d = 96.15 units/week S = \$10/orders d = 10 units/week C = \$5/unit I = 20% per year LT = 2 weeks (constant) s LT = 0

Periodic Review Example - Case 2 D = 5000 units/year d = 96.15 units/week S = \$10/orders d = 10 units/week C = \$5/unit I = 20% per year LT = 2 weeks (constant) s LT = 0 k is not known; P =90% -> z = 1.28 T = 2x10 0.2x5000x5 = 0.0632 years = 3.29 weeks Solution: TC = 158.05 + 157.92 + 29.44 = \$345.41/year M = 96.15(3.29+2) + 1.28x23.0 = 538.07 3.29 + 2 s’ d = 10 = 23.0

Periodic Review Example - Case 3 D = 5000 units/year d = 96.15 units/week S = \$10/orders d = 10 units/week C = \$5/unit I = 20% per year LT = 2 weeks (constant) s LT = 0 k = \$2/unit; P =90% -> z = 1.28 T = 3.29 weeks as in Case 2 Solution: TC = 158.05 + 157.92 + 29.44 + 34.53 = \$379.94/year s’ d = 23.0 as in Case 2 M = 538.07 as in Case 2

Reorder Point Example - Case 3 k =\$2/unit; P =90% T = 3.29 weeks Solution: TC = \$379.94/year M = 538.07 units Could use k=\$2/unit to find optimal P Cost would decrease (as with Reorder Point).

Expected number of units out-of-stock/year Service Level - Periodic Review Annual demand SL= 1 - % of items out-of-stock = 1 - (1/T) x s’ d x E(z) D = 1 - s’ d E(z) DT = 1 - time units must agree

Service Levels for Cases 2 & 3 SL = 1 - 23.0(.0475) 5000(0.0632) = 0.9965 units/year years

Comparison Reorder Point Casek P Q ROP TC(\$/yr) SL 12.9678 322 218 347.97.9994 2-.90 316 210 334.33.9979 32.90 316 210 355.58.9979 Periodic Review Casek P T(wks) M TC(\$/yr) SL 2-.90 3.29 538 345.41.9965 32.90 3.29 538 379.94.9965

Comparison Reorder Point System has: –Lower cost. –Lower average inventory. –Higher service level. –Less certainty about timing of future orders.

Change T to a Convenient Value: 3 weeks T = 3.29 weeks = 0.0632 years s’ d = 23.0 units M = 538.1 units TC = \$379.94/year Ordering every 3.29 weeks is not very convenient! Suppose you order every 3 weeks: T = 3 weeks TC = 173.33 + 144.23 + 28.62 + 36.82 = \$383.00/year M = 96.15(3+2) + 1.28x22.36 = 509.37 3 + 2 s’ d = 10 = 22.36

Change T to a Convenient Value: 4 weeks T = 3.29 weeks = 0.0632 years s’ d = 23.0 units M = 538.1 units TC = \$379.94/year Consider T = 4 weeks T = 4 weeks TC = 130.00 + 192.30 + 31.35 + 30.25 = \$383.90/year M = 96.15(4+2) + 1.28x24.49 = 607.35 4 + 2 s’ d = 10 = 24.49 Cost for T = 3 weeks, T=4 weeks, T = 3.29 weeks are about the same!

Joint Ordering Suppose several items are ordered from the same supplier. Each items would have an optimal order interval T. This would require frequent orders to the same supplier. Alternative: Find a common order interval T and order all items from the same supplier together. - This is a variation of the periodic review system.

D i = demand for item i (usually annual) d i = demand rate for item i s di = std. deviation of demand for item i S i = order cost for item i (\$/order) O = common order cost (\$/order) (pay once per order) C i = value for item i(\$/item) P i = probability of being in stock for item i k i = stockout cost for item i LT = (constant) lead time I = carrying cost (% of value/unit time) s’ di = std. deviation of demand during lead time M i = maximum level T = time between orders TC = total cost (usually annual) Joint Ordering Variables

Joint Ordering - Optimal Ordering Optimal order interval: Maximum level : M i = d i (T+LT) + z i s’ di Optimal cost: 2(O+  S i ) T = I  D i C i TC = T O+  S i TI  D i C i 2 1 + I  C i z i s’ d i + T 1  k i s’ di E(z i ) T + LT s’ di = s di Std. deviation of demand during lead time: +

Joint Ordering - continued Service Level: Optimal number of orders/year: Average inventory: AIL i =(d i T)/2 + z i s’ di s’ di E(z i ) SL i = 1 - TDiTDi T 1 N =

X Y D i = Annual demand 26003900 d i = demand rate 50/week75/week s di = std. deviation10/week15/week S i = order cost \$20\$16 O = common order cost \$50 C i = value\$60\$44 P i = in-stock probability 80%90% k i = stockout cost \$30/item\$10/item LT = (constant) lead time 2 weeks I = carrying cost20%/year Consider ordering X and Y jointly. Joint Ordering Example - 2 items: X & Y

X Y D i = Annual demand 26003900 d i = demand rate 50/week75/week s di = std. deviation10/week15/week S i = order cost \$20\$16 O = common order cost \$50 C i = value\$60\$44 P i = in-stock probability 80%90% k i = stockout cost \$30/item\$10/item LT = (constant) lead time 2 weeks I = carrying cost20%/year Joint Ordering Example - 2 items: X & Y 2(50 + 20 + 16) T = (0.2)(2600x60 + 3900x44) = 0.0512 years = 2.66 weeks

Joint Ordering Example: Item X Optimal order interval: P X = 0.8 -> z X = 0.84 -> E(z X ) = 0.1120 M X = 50(2.66+2) + 0.84(21.59) = 251.14 AIL X = (50x2.66)/2 + 0.84(21.59) = 84.64 SL X = 1 - (21.59x0.1120)/(2.66x50) = 0.9818 T = 0.0512 years = 2.66 weeks 2.66 + 2 s’ dX = 10 = 21.59

Joint Ordering Example: Item Y Optimal order interval: P Y = 0.9 -> z Y = 1.28 -> E(z Y ) = 0.0475 M Y = 75(2.66+2) + 1.28(32.38) = 390.95 AIL Y = (75x2.66)/2 + 1.28(32.38) = 141.20 SL Y = 1 - (32.38x0.0475)/(2.66x75) = 0.9923 T = 0.0512 years = 2.66 weeks 2.66 + 2 s’ dY = 15 = 32.38

Joint Ordering Example - Total Cost T = 0.0512 years = 2.66 weeks TC = 0.0512 50+20+16 0.0512(0.2)(2600x60 + 3900x44) 2 1 + (0.2) (60x0.84x21.59 + 44x1.28x32.38) 0.0512 1 (30x21.59x0.1120 + 10x32.38x0.0475) + TC = T O+  S i TI  D i C i 2 1 + I  C i z i s’ d i + T 1  k i s’ di E(z i ) + + TC = 1681.20 + 1675.80 + 582.44 + 1718.79 = \$5658.23/year

Joint Ordering Summary For item X: Every 2.66 weeks order 251 - amount on hand. For item Y: Every 2.66 weeks order 391 - amount on hand. Total cost = \$5658/year If X was ordered separately (with S = 50+20=70) T X = 3.5 weeks, M X = 295 and TC X = \$3503.71/year If Y was ordered separately (with S = 50+16=66) T Y = 3.22 weeks, M Y = 435 and TC Y = \$2776.80/year Total cost for X and Y = \$6280.51/year

Joint Ordering - Adjust T For item X: Every 2.66 weeks order 251 - amount on hand. For item Y: Every 2.66 weeks order 391 - amount on hand. Total cost = \$5658/year Could adjust T to 3 weeks: s’ dX = 22.36 M X = 268.8 TC = 1490.67 + 1890.00 + 603.18 + 1578.39 = \$5562/year s’ dY = 33.54 M Y = 417.9

Min-Max System Approximate, but easy. Hybrid Reorder Point/Periodic Review system. Useful when inventory decreases in large steps (lumpy demand). Order M - amount on hand, when inventory falls below ROP. M = ROP + Q*

Min-Max System Order M - amount on hand, when inventory falls below ROP. ROP = dxLT + zs’ d M = ROP + Q Q = IC 2DS ROP M

Min-Max System Example Q = 350; ROP = 100; LT = 2 days M = 450 DaySales Inventory 0 112 1 8104 2 64 40order 410 = 450 - 40 3 10 30 4 25415= 30 - 25 + 410 5 35380 …... 32 …121 33 23 98order 352 = 450-98 34 3 95 35 40 407= 95 - 40 + 352

Other Issues Stock to demand. –Estimate safety stock in terms of time. –Example: Order quantity = forecast of demand during time between orders + lead time + 1 week safety stock. Multi-item, Multi-location inventories. –Inventories at plants, regional distribution centers, field warehouses, retail outlets, etc. –Complex, computer models (integer programming). Pipeline inventories. –Reducing transit time can reduce inventory (regular and safety stock), but increase transportation cost.

Aggregate Control of Inventories Turnover Ratio. –Annual sales/Average inventory. –Can be for one items or all items. –Assumes sales and inventory increase proportionally. ABC Classification –Based on (annual) sales amount. Example on page 353 is wrong. –Give most attention to A items. (differentiation)

Average Inventory and Throughput Square Root Rule –Amount of inventory in a facility is proportional to the square root of the throughput (sales) at the facility. Annual Throughput (cwt) Average Inventory (cwt) Average inventory = k Throughput

Square Root Rule An organization has total throughput (sales) of X units per year. If there were n stocking locations, then each would expect throughput of X/n per year. I T = total inventory if there was a single stocking point. I i = average inventory at one of n identical stocking points. I T = k X I i = k X/n Square root rule: I T = I i n Total inventory in system = nI i

Square Root Rule Example 1 Suppose the current system has 10 warehouses and each one has \$60,000 of inventory on average. Q1: How much will the inventory investment change if there are to be 5 (not 10) warehouses? Investment with 10 warehouses = 10x60,000 = \$600,000 I T = 60000 10 = \$189,737 With 5 warehouses: 189,737 = I i 5 Investment with 5 warehouses = 5x84853 = \$424,265 So I i = \$84,853 Inventory change = 600,000-424,265 = \$175,735 (-29%)

Square Root Rule Example 2 Suppose the current system has 10 warehouses and each one has \$60,000 of inventory on average. Q2: How many warehouses are needed if the average inventory in each is to be \$30,000? I T = 60000 10 = \$189,737 With \$30,000 in each warehouse: n So n = (189,737/30,000) 2 = 40 warehouses I T = 189,737 = 30,000

Square Root Rule Example 3 Suppose the current system has 10 warehouses and each one has \$60,000 of inventory on average. Q2: How many warehouses are needed if the total inventory in the system is to be cut in half? Investment with 10 warehouses = 10x60,000 = \$600,000 I T = 60000 10 = \$189,737 With \$300,000 in all warehouses: I i = 300,000/n n So n = (300,000/189,737) 2 = 2.5 warehouses I T = 189,737 = (300000/n) n 300000 =

Total Investment Limit Suppose several products are stored in the same warehouse. A maximum total investment limit (L) for the warehouse is specified. Order sizes (Q) may need to be decreased to reduce the total inventory investment. L = Maximum amount invested in inventory. TI = Total inventory. TI =  C i (average inventory for item i) =  C i (Q i /2) <= L

Total Investment Limit Calculate order size Q i for each product. If TI <= L, then the limit is not violated. If TI > L, then the limit is violated, so reduce the order sizes by a fraction R: Qi=Qi= IC i 2D i S i R = L  C i (Q i /2) New order size (Q) = R x old order size

Total Investment Limit Example Product S i C i D i Q i C i Q i /2 1 50 20 12000 2 50 10 25000 3 50 15 8000 Three products are stored in 1 warehouse. The total inventory investment can not exceed \$10,000. I = 30% per year.

Total Investment Limit Example Product S i C i D i Q i C i Q i /2 1 50 20 12000 447.21 \$4,472.10 2 50 10 25000 912.87 \$4,564.35 3 50 15 8000 421.64 \$3,162.30 \$12,198.75 Investment exceeds \$10,000 so R = 10,000 12,198.75 Three products are stored in 1 warehouse. The total inventory investment can not exceed \$10,000. I = 30% per year. = 0.8198

Total Investment Limit Example Product S i C i D i Q i new Q i 1 50 20 12000 447.21 366.6 2 50 10 25000 912.87 748.3 3 50 15 8000 421.64 345.6 Now investment equals \$10,000. Three products are stored in 1 warehouse. The total inventory investment can not exceed \$10,000. I = 30% per year. R = 0.8198

Just-In-Time (JIT) Production system that originated in Japan. Toyota is best known example. Now used by many U.S. manufacturers and suppliers. Producer: Produce small lots of high quality products with low inventory. Supplier: Deliver small lots of high quality components on time.

Just-In-Time Goals Philosophy: –Synchronize the supply chain to respond to customers. –Eliminate all waste (inventory and scrap). Goals: –Zero inventory. –Lot size of 1. –100% quality. Continuously strive to reduce inventory and lot size - and improve quality.

JIT vs. “Traditional” Operations Just-In-Time –Inventory = Liability –Setups = Liability –Defects/scrap = Liability Eliminate/reduce: inventory, setups, defects, scrap. “Traditional” Operations –Inventory = Protection against stockouts –Setups = necessary evil. –Defects/Scrap = Expected. Optimize tradeoffs.

JIT and “Traditional” Operations Just-In-Time Reducing setup cost S: –Reduces Q. –Reduces lot sizes and reduces inventory. No inventory requires high quality components, high quality production and high quality transportation: –No defects. –No late deliveries. Q= IC 2DS p-d p

JIT Requires Frequent, Small, On-time Deliveries Encourages suppliers to locate near customers. –Locate auto part plants near assembly plants. –Locate auto parts warehouse near assembly plants. Requires high quality transportation carriers. –Deliveries can not be late, without inventory. Favors truck and air. Small shipments favor truck and air. –Example: Electronics manufacturers may fly parts from Asia several times per week.

Benefits of Just-In-Time Lower inventory. –Frees up \$ for other uses. –Less space required. –Especially important when interest rates are high. Higher quality. –Leads to higher prices, better sales, etc. –More efficient use of space. Less scrap/defects to take up space.

Concerns with Just-In-Time Low inventory requires secure, reliable supply chains. –Reliable suppliers and transportation. –Reliable infrastructure. Difficult to implement. –Requires major changes in business processes. –Hard to do “partial just-in-time”. –Difficult to phase in. –Works best if suppliers use just-in-time.

Japan and Just-In-Time How does Japan differ from the USA? –Area. –Population. –Land use. –Resources. What are implications for success of just-in-time?

Download ppt "Chapter 10: Inventory - Part 2 Types of Inventory and Demand Availability Cost vs. Service Tradeoff Pull vs. Push Reorder Point System Periodic Review."

Similar presentations