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### Inventory Control Models

### Properties of the Optimal Solutions

Ch 4 (Known Demands)

R. R. Lindeke

IE 3265, Production And Operations Management

Reasons for Holding Inventories

- Economies of Scale
- Uncertainty in delivery lead-times
- Speculation. Changing Costs Over Time
- Smoothing: to account for seasonality and/or Bottlenecks
- Demand Uncertainty
- Costs of Maintaining Control System

Characteristics of Inventory Systems

- Demand
- May Be Known or Uncertain
- May be Changing or Unchanging in Time
- Lead Times - time that elapses from placement of order until it’s arrival. Can assume known or unknown.
- Review Time. Is system reviewed periodically or is system state known at all times?

Characteristics of Inventory Systems

- Treatment of Excess Demand.
- Backorder all Excess Demand
- Lose all excess demand
- Backorder some and lose some
- Inventory who’s quality changes over time
- perishability
- obsolescence

Real Inventory Systems: ABC ideas

- This was the true basis of Pareto’s Economic Analysis!
- In a typical Inventory System most companies find that their inventory items can be generally classified as:
- A Items (the 10 - 20% of sku’s) that represent up to 80% of the inventory value
- B Items (the 20 – 30%) of the inventory items that represent nearly all the remaining worth
- C Items the remaining 50 – 70% of the inventory items sku’s) stored in small quantities and/or worth very little

Real Inventory Systems: ABC ideas and Control

- A Items must be well studied and controlled to minimize expense
- C Items tend to be overstocked to ensure no runouts but require only occasional review
- See mhia.org – there is an “e-lesson” on the principles of ABC Inventory management – check it out! – do the on-line lesson!

Relevant Inventory Costs

- Holding Costs - Costs proportional to the quantity of inventory held.
- Includes:
- Physical Cost of Space (3%)
- Taxes and Insurance (2 %)
- Breakage Spoilage and Deterioration (1%)
- Opportunity Cost of alternative investment. (18%)

Here these holding issues total: 24%

- Therefore, in inventory systems, the holding cost would be taken as:
- h .24*Cost of product

Lets Try one:

- Problem 4, page 193 – cost of inventory
- Find h first (yearly and monthly)
- Total holding cost for the given period:
- THC = $26666.67
- Average Annual Holding Cost
- assumes an average monthly inventory of trucks based on on hand data
- $3333

Relevant Costs (continued)

- Ordering Cost (or Production Cost).

Includes both fixed and variable components

slope = c

KC(x) = K + cx for x > 0; 0 for x = 0.

Relevant Costs (continued)

- Penalty or Shortage Costs. All costs that accrue when insufficient stock is available to meet demand. These include:
- Loss of revenue due to lost demand
- Costs of book-keeping for backordered demands
- Loss of goodwill for being unable to satisfy demands when they occur.

Relevant Costs (continued)

- When computing Penalty or Shortage Costs inventory managers generally assume cost is proportional to number of units of excess demand that will go unfulfilled.

The Simple EOQ Model – the most fundamental of all!

- Assumptions:

1. Demand is fixed at l units per unit time – typically assumed at an annual rate (use care).

2. Shortages are not allowed.

3. Orders are received instantaneously. (this will be relaxed later).

Simple EOQ Model (cont.)

- Assumptions (cont.):

4. Order quantity is fixed at a value “Q” per cycle. (we will find this as an optimal value)

5. Cost structure:

a) includes fixed and marginal order costs (K + cx)

b) includes holding cost at h per unit held per unit time.

Finding an Optimal Level of ‘Q’ – the so-called EOQ

- Requires us to take derivative of the G(Q) equation with respect to Q
- Then, Set derivative equal to Zero:
- Now, Solve for Q

Properties of the EOQ (optimal) Solution

- Q is increasing with both K and and decreasing with h
- Q changes as the square root of these quantities
- Q is independent of the proportional order cost, c. (except as it relates to the value of h = I*c)

Try ONE!

- A company sells 145 boxes of BlueMountain BobBons/week (a candy)
- Over the past several months, the demand has been steady
- The store uses 25% as a ‘holding factor’
- Candy costs $8/bx and sells for $12.50/bx
- Cost of making an order is $35
- Determine EOQ (Q*) and how often an order should be placed

But, Orders usually take time to arrive!

- This is a realistic relaxation of the EOQ ideas – but it doesn’t change the model
- This requires the user to know the order “Lead Time”
- And then they trigger an order at a point before the delivery is needed to assure no stock outs
- In our example, what if lead time is 1 week?
- We should place an order when we have 145 boxes in stock (the one week draw down)
- Note make sure lead time units match units in T!

But, Orders usually take time to arrive!

- What happens when order lead times exceed T?
- We proceed just as before (but we compute /T)
- is the lead time is units that match T
- Here, lets assume = 6 weeks then:
- /T = 6/3.545 = 1.69 – in the Blue Mount Bon-Bon case
- Place order: 1.69 cycles before we need product!
- Trip Point is then 0.69*Q* = .69*514 = 356 boxes
- This trip point is not for the next stock out but the one after that (1.69 T or 6 weeks from now!)– be very careful!!!

Sensitivity Analysis

Let G(Q) be the average annual holding and set-up cost function given by

and let G* be the optimal average annual cost. Then it can be shown that:

Sensitivity

- We find that this model is quite robust to Q errors if holding costs are relatively low
- We find, for a given Q – a mistake in the ordering quantity
- that Q* +Q has smaller error than Q* - Q (Error means we incur extra inventory maintenance costs – a penalty cost)
- That is, we tend to have a smaller penalty cost if we order too much compared to too little

EOQ With Finite Production Rate

- Suppose that items are produced internally at a rate P (> λ, the consumption rate). Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely:
- Except that h’ is defined as h’= h(1-λ/P)

Lets Try one:

- We work for Sam’s Active Suspensions
- They sell after market kits for car “Tooners”
- They have an annual demand of 650 units
- Production rate is 4/day (working at 250 d/y)
- Setup takes 2 technicians working 45 minutes @$21/hour and requires an expendable tool costing $25

Continuing:

- Each kit costs $275
- Sam’s uses MARR of 18%, tax at 3%, insurance at 2% and space cost of 1%
- Determine h, Q*, H, T and break T down to:
- T1 = production time in a cycle (Q*/P)
- T2 = non producing time in a cycle (T – T1)

Engineering Teams: (You can) Do It

Quantity Discount Models

- All Units Discounts: the discount is applied to ALL of the units in the order. Gives rise to an order cost function such as that pictured in Figure 4-9

Quantity Discount Models

- Incremental Discounts: the discount is applied only to the number of units above the breakpoint. Gives rise to an order cost function such as that pictured in Figure 4-10.

Properties of the Optimal Solutions

- For all units discounts, the optimal will occur at the bottom of one of the cost curves or at a breakpoint. (It is generally at a breakpoint.). One compares the cost at the largest realizable EOQ and all of the breakpoints beyond it. (See Figure 4-11).
- For incremental discounts, the optimal will always occur at a realizable EOQ value. Compare costs at all realizable EOQ’s. (See Figure 4-12).

To Find EOQ in ‘All Units’ discount case:

- Compute Q* for each cost level
- Check for Feasibility (the Q computed is applicable to the range) – we say it is “Realizable”
- Compute G(Q*) for each of the realizable Q*’s and the break points.
- Chose Q* as the one that has lowest G(Q)

Lets Try one:

- Product cost is $6.50 in orders <600, $3.50 above 600.
- Organizational I is 34%
- K is $300 and annual demand is 900

Lets Try one:

- Both of these are Realizable (the value is ‘in range’)
- Compute G(Q) for both and breakpoint (600)
- G(Q) = c + (*K)/Q + (h*Q)/2

Order 674 at a time!

In an Incremental Case:

- Cost is a strictly varying function of Q -- It varies by interval!
- Calculate a C(Q) for the applied schedule
- Divide by Q to convert it to a “unit cost” function
- Build G(Q) equations for each interval
- Find Q* from each G(Q) Equation (derivative!)
- Check if “Realizable”
- Compute G(Q*) for realizable Q*’s

Trying the previous problem (but as Incremental Case):

- Cost Function: Basically states that we pay 6.50 for each unit up to 600 then 3.50 for each unit ordered beyond 601:
- C(Q) = 6.5(Q), Q < 600
- C(Q) = 3.5(Q – 600) + 6.5*600, Q 600
- C(Q)/Q = 6.5, Q < 600(order up to 600)
- C(Q)/Q = 3.5 + ((3900 – 2100)/Q), Q 600 or
- C(Q)/Q = 3.5 + (1800)/Q(orders beyond 601)

Trying the previous but as Incremental Case:

- For the First Interval:
- Q* = [(2*300*900)/(.34*6.50)] = 495 (realizable)
- For order > 600, find Q* by writing a G(Q) equation and then optimizing:
- [G(Q) = c + (*K)/Q + (h*Q)/2]

Differentiating G2(Q)

Realizable!

Now Compute G(Q) for both and “cusp”

- G(495) = 900*6.5 + (300*900)/495 + .34((6.5*495)/2) = $6942.43
- G(600) = 900*6.5 +(300*900/600) + .34((6.5*600)/2) = $6963.00
- G(1763) = 900*(3.5 +(1800/1783)) + (300*900)/1783 + .34*(3.5 +(1800/1783))*(1783/2) = $5590.67

Lowest cost – purchase 1783 about every 2 years!

Lets jump back into our teams and do some!

TRY 22b and 23 on Pg 211

Resource Constrained Multi-Product Systems

- Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following constraint on the system:

Resource Constrained Multi-Product Systems

- When the condition that:
- is met, the solution procedure is straightforward. If the condition is not met, one must use an iterative procedure involving Lagrange Multipliers.

EOQ Models for Production Planning

- Consider n items with known demand rates, production rates, holding costs, and set-up costs. The objective is to produce each item once in a production cycle. For the problem to be feasible the following equation must be true:

Issues:

- We are interested in controlling Family MAKESPAN (we wish to produce all products within our chosen cycle time)
- Underlying Assumptions:
- Setup Cost (times) are not Sequence Dependent (this assumption is not always accurate as we will later see)
- Plants uses a “Rotation” Policy that produces a single ‘batch’ of each product each cycle – a mixed line balance assumption

EOQ Models for Production Planning

- The method of solution is to express the average annual cost function in terms of the cycle time, T. The optimal cycle time has the following mathematical form:
- We must assure that this time allows for all setups and all production times.

Working forward:

- This last statement means:
- (sj+(Qj/Pj)) T
- sj is set up time
- Of course: Qj = j*T
- So – with substitution: (sj+((j*T)/Pj) T
- Or: T((sj/(1- j/Pj)) = Tmin
- Finally, we must Choose T(chosen cycle time) = MAX(T*,Tmin)

Lets Try Problem 30

Given: 20 days/month and 12 month/year; $85/hr for setup

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