# 6.3 - MATH 1081  Monday, March 14   ...

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Unformatted text preview: MATH 1081  Monday, March 14    Chapter 6 – Section 3    FURTHER BUSINESS APPLICATIONS    Homework #9 (due 3/28):    Section 6.3 #10, 18, 22                    Section 9.1 #28, 46, 48    Clicker Check­in:  Choose any letter to check in now.    Suppose that you are a manufacturer.  Each time you run a  batch through production, there are expenses of set up in the  factory.  So, to minimize those costs, you will want a few  production runs as possible.    On the other hand, a large production run introduces the  expense of storing the product until the consumer’s purchase  it.  So, to minimize storage costs, you want as small a batch of  product at a time as possible.    Where is the point that minimizes total cost overall?    This is called the economic lot size. To make this more concrete, suppose a company manufactures  12,000 units of a product each year (for which the demand is  uniform throughout the year) and that the product can be  manufactured in several small batches (of equal size)  throughout the year.    The company could, for example, schedule one production run  at the start of the year making all 12,000 units; or they could  schedule 4 production runs (once every 3 months) making  3000 in each batch; or they could schedule 12 production runs  (one at the start of each month) making just 1000 in each  batch.            Assuming uniform demand  through the year, the  inventory in the warehouse  would look like one of the  following:    So, the average in storage is  half the batch size.   Let’s say the number in each batch is  q .  This is the variable in  our problem, as we can decide how many per batch (and  therefore how many batches in the year) we make.    Then in our current scenario,   q we have   units in storage throughout the year and,  2 12000 we will have a total of   production runs.  q   Suppose it costs \$2 to store a unit of the product for a year,  \$500 to set up the factory for each production run, and \$5 per  unit to produce.    This means that the total cost for the year will be given by the  function    ⎛ 12000 ⎞ ⎛ q⎞         C (q ) = 2 ⎜ ⎟ + 500 ⎜ ⎟ + 5 (12000 ) .  ⎝ 2⎠ ⎝ q ⎠ product storage set up   This is the function that we must minimize.  (What makes  sense for the domain of this function?)  What is the value of  q   that does the trick?  If we go back and look at this problem in general, we can  derive the general formula (and avoid the calculus every time).    Namely, if we let    q = number of units in each batch     k = cost of storing one unit for a year   f = fixed setup costs for a production run       g = cost of production of each unit     M = total annual demand for the product     we get the following cost function  ⎛M⎞ ⎛ q⎞     C (q ) = k ⎜ ⎟ + f ⎜ ⎟ + g ( M )   ⎝ 2⎠ ⎝ q⎠ product storage set up Keeping in mind that the only actual variable in that problem is  q , we find that there is one critical number found at    2 fM             q= .  k   Why isn’t the cost per unit,  g , part of this formula?      As we did in our specific example, this can be shown to be a  minimum, and since it is the only critical number, that is where  the absolute min occurs.    A very similar problem is one regarding the economic order  quantity.  In this case, we are ordering and storing product to  sell.  Each time we order, there are fees for ordering (in  addition to the cost per unit ordered) and there are fees to  store any product currently on inventory.    How many/how often should we order to minimize costs,  2 fM assuming that we know the annual demand?   q =   k    q = number of units in each batch k = cost of storing one unit for a year   f = fixed setup costs to place an order M = total annual demand for the product In this section, we also consider another application that is  actually not one of optimization.  This application, called  elasticity of demand, is a measure of how much a change in the  price of the product affects the demand.      If the price goes up, generally the demand will drop.  However,  depending on what the item is, the sensitivity of demand  varies.    To quantify this notion of elasticity, we can compare the  percent change in price to the percent change in demand.    If a “small” percent change (increase) in price results in a  “large” percent change (drop) in demand, then raising the price  will reduce revenue.  This is probably something we want to  avoid.        If  E < 1, then the relative change in demand is less than the  relative change in price.  So, total revenue increases as price  increases.    If  E > 1, then the relative change in demand is greater than the  relative change in price and total revenue decreases as price  increases.    What happens when  E = 1?      This is called unit elasticity and indicates that the relative  change in demand is equal to the relative change in price at  that point.   This is the point at which total revenue is  maximized.  (See page 393 for further detail.)    Example:  Suppose for a certain commodity,  q = 48, 000 − 10 p 2 .    Find  E  and determine at which price (if any)  revenue is maximized.    Clicker Check­out:  Choose any letter to check out now.      Next time – Section 9.1    ...
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## This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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