Inventory

Chapter 20 Notes Page 1

Please send comments and corrections to me at [email protected]

XX. Inventory Management: EOQ, JIT and Constraints

Managing your inventory properly is an important means of controlling your costs and, thereby, improving the profitability of your firm. We will discuss the Economic Order Quantity (EOQ) model, which helps determine the optimal amount of inventory to produce or purchase at a given time. We will also briefly discuss the Just In Time Inventory Method (JIT). With JIT, you strive to manage your production or acquisition of inventory so that the inventory arrives or is produced “just in time” to be sold; thereby reducing your

inventory carrying costs. Finally, we will discuss how to make decisions, such the amount and identity of goods to manufacture, when dealing with limited resources (constraints).

Economic Order Quantity When deciding how much inventory to purchase or manufacture, you are dealing with two types of inventory costs: (i) Order/Setup Costs, and (ii) Carrying Costs. When purchasing inventory, Order/Setup costs represent all of the costs involved in placing that order. When manufacturing inventory, Order/Setup Costs represent the cost of setting up your production line in order to produce the inventory. Carrying Costs represent, all of the costs involved in holding inventory (e.g., interest costs on funds invested in inventory, insurance, breakage and storage costs) until the inventory is sold:

Inventory Costs = Order or Setup Costs + Carrying Costs

We will use the following variables in this discussion: TC Total Inventory Costs P Order/Setup Costs Q When ordering inventory, Q is the quantity of units ordered in a single order.

When manufacturing inventory, Q is the lot size for a single production run of the inventory

D Annual number of Inventory needed C Annual Carrying Costs for one unit of inventory. You can calculate the number of orders or production runs that a firm will have:

Number of Orders/Setups = Annual Demand For Units

= _D_

Size of Order/Production Run Q

Inventory Management

Chapter 20 Notes

Please send comments and c

You can calculate the cost to place runs in a given year:

Annual Ordering/Setup Cost =

If we were to store all of the units production run), the Annual C

Annual Carrying Cost of One Run/

We do not, however, retain all of these unitinventory, and the number of unitsassume that the inventory is Usage), then the average amount of of the amount of the units in a single

Using this assumption, the carrying cost of our inventory for the entire year amount that we calculated above

Using this information, we can calculate the T

Total Annual Inventory Costs =

Total Annual Inventory Costs = P(

You can calculate the quantity to generate the minimum annual inventory costs by taking the first derivative of this equation with respect to Q; setting it to zero

corrections to me at [email protected]

to place all of the orders or to set up all of the

Setup/Order Cost x Number of Orders/Setups

of the units ordered in a single order (or produced Annual Carrying Costs for these units would be:

/ Order = Annual Carrying Cost Per Unit x No. of Units in Run/Order

We do not, however, retain all of these units in our inventory. We s, and the number of units in inventory drops until they are all

inventory is sold or used at a constant or uniform rate (the Rate of amount of units in inventory at any given time

a single order or production run:

he carrying cost of our inventory for the entire year amount that we calculated above:

_CQ_ 2

n calculate the Total Inventory Costs of our firm for a year:

Total Annual Inventory Costs = Annual Setup/Order Cost + Annual Carrying Cost

Total Annual Inventory Costs = P( _D_

) + ( _CQ_

Q 2

You can calculate the quantity of inventory that should be ordered or produced in order minimum annual inventory costs by taking the first derivative of this

; setting it to zero; and solving for Q.

Page 2

all of the production

= P( _D_

) Q

produced in a single

of Units in Run/Order = CQ

s in our inventory. We sell or use this all gone. If you rate (the Rate of

ime is equal to half

he carrying cost of our inventory for the entire year is half of the

osts of our firm for a year:

Annual Setup/Order Cost + Annual Carrying Cost

)

ld be ordered or produced in order minimum annual inventory costs by taking the first derivative of this



First, let us rewrite the above equation to make it easier for us to use the Power Rule for taking derivatives:

TC = PDQ-1 + ½CQ ∂/∂Q TC = -PDQ-2 + ½C

0 = -PDQ-2 + ½C -½C = -PDQ-2

-½CQ2 = -PD

Q2 = _2PD_ C

Q

=

_______ / _2PD_ (“EOQ Formula”) √ C

The EOQ Formula gives you the optimal amount of units that should be ordered or manufactured at a given time. The Lead Time is the time it takes to obtain the new inventory once the production is started or the order is placed. For example, it takes two days for the inventory to arrive once an order is placed, or it takes two days for us to manufacture the inventory. This two-day period is the Lead Time. The Reorder Point is the point in time where the existing inventory is just sufficient to cover the demand for your inventory until the new inventory arrives. The Reorder Point can be calculated as follows:

Reorder Point = Rate of Usage x Lead Time

For example, we sell 10 units of inventory a day, and it takes two days to receive new inventory once we place an order for the new inventory (or begin production of inventory). The Reorder Point is 20 units. In other words, the firm should place an order for new inventory once our current inventory level drops to 20 units. Since we assume that the inventory is sold or used at a constant or uniform rate, the Rate of Usage can be calculated as follows:

Rate of Usage = ___Annual Demand For Units___ Number of Working Days in Year

Safety Stock is extra inventory carried as a buffer against fluctuation in demand. The Reorder Point formula can be modified to reflect your need for Safety Stock:

Reorder Point = (Rate of Usage x Lead Time) + Safety Stock



EOQ Example

Since Smith has 250 business days, that means that it sells 20 units a day (5,000/250). If it takes a day to produce the units then, Smith should make it a practice to make a production run once its inventory level drops to 20 units. If Smith wished to have Safety Stock, it would add those units to the 20-unit level. Smith would save the following amount if would employ the optimal run of 1,000 units rather than the 1250-unit run (5,000/4) currently planned:

Total Annual Inventory Costs = P( _D_

) + ( _CQ_

) Q 2

Inventory Cost With Planned Run = [$1000 (5000/1250) + ½ (1250 x 10)] $10,250 Less: Inventory Cost With EOQ Run = [$1000 (5000/1000) + ½ (1000 x 10)] -10,000

Inventory Cost Savings = $250

Just In Time Inventory The Just In Time Inventory Method (JIT) seeks to produce inventory so that its production is completed just as the inventory is needed (e.g., being sold or used). For inventory being purchased, inventory should arrive just as it is being sold or used. A perfect JIT model reduces carrying costs to zero. When purchasing inventory, firms implementing JIT try to develop relationships with their suppliers so that the inventory arrives when needed. The supplier makes more deliveries of smaller orders. The supplier is willing to make such an arrangement in

Smith, Inc. is a Mom & Pop manufacturer of explosives. The annual demand for the explosives is estimated to be 5,000 units. The annual cost to hold one unit in inventory is $10 per year, and the cost to initiate a production run is $1,000. There are no explosives on hand, and Smith has scheduled four equal production runs of explosives for the upcoming year, the first of which is to be run immediately. Smith has 250 business days per year, sales occur uniformly throughout the year, and production is completed within one day.

EOQ = ________________

= 1,000 units √2($1,000)(5,000)/10

Using the EOQ formula, we see that that the optimal run is 1,000 units:



exchange for preferential treatment (e.g., a long-term commitment from the purchaser, better shelf location for products). In order to accomplish this goal when manufacturing goods, firms try to minimize its setup cost & time, and thereby reduce their lead times and setup costs.

Constrained Optimization We will now consider how to maximize Sales Revenue & Contribution Margin; or minimize costs when dealing with limited resources (constraints). We will first discuss using the graphical approach to solving these constrained optimization problems. Then we will discuss how to use Excel to solve these problems.

The first thing you need to do is to state your Objective Function. This is your goal (the “object” of your analysis). The Objective Function should state your goal (e.g., maximize or minimize something) along with a mathematical formula for the item that you are maximizing or minimizing (e.g., Revenue, Contribution Margin or Cost). The Objective Function Variables should be your alternative courses of action (e.g., the number of each type of product that you will produce). For example, assume that Karloff Bolt Company makes two types of bolts, Bolt A and Bolt B. Bolt A has a Unit Contribution Margin of 10¢, and the Contribution Margin per Unit for Bolt B is 12¢. You want to know how many units of each bolt Karloff should produce in order to maximize its total Contribution Margin.

The Objective Function is:

Maximize .10A and .12B where “A” is the number of A Bolts that you will produce and “B” is the number of B Bolts that you should produce. “A” and “B” are the Objective Function Variables. Without more information, Karloff would produce an infinite number of each type of bolt, and thereby make an infinite total Contribution Margin. In reality, there is a limit to the number of bolts that can be produced.



Let us assume that each bolt must pass through the following three machines, and the time required on each machine differs, as shown in the table below:

Machine I Machine II Machine III

A Bolt .1 min .1 min .1 min B Bolt .1 min .4 min .5 min

In a day, there are 240, 720, and 160 minutes available, on Machine I, Machine II and Machine III, respectively. Karloff wants to know how many of each type of bolts it should produce in a day. We now have to add these constraints to our Objective Function:

Description Problem

This is the Objective Function Max .10(A) + .12(B) These are the constraints: subject to: You only have 240 minutes available on Machine I. .1A + .1B ≤ 240 You only have 720 minutes available on Machine II. .1A + .4B ≤ 720 You only have 160 minutes available on Machine III. .1A + .5B ≤ 160 You can't manufacture a negative number of bolts. A , B ≥ 0 Next, you have to graph the area of production that meets all of these constraints (the Feasible Region or Area). The graph should consist of a Cartesian axis with the Objection Function Variables as the x-axis and the y-axis. For example, the last constraint alone means that you are dealing with the top right quarter of the Cartesian axis:

The wiggly line represents that the area goes out to infinity. You now want to graph the other constraints, which are inequalities. They each include the equal sign combined with either a “greater than” (≥) or “lesser than” (≤) symbol. The area covered by such an inequality consists of a line that divides Cartesian plane and



one side of that line. The formula of the line that divides the Cartesian plane is the inequality formula with an equal sign (=) substituted for the inequality (≤, ≥). The way to graph such an inequality is to first graph the line that divides the Cartesian plane and then chose the side of the line that satisfies the inequality. To graph the first constraint (.1A+.1B ≤ 240), first graph the line, .1A+.1B=240. The easiest way to graph the line is identify the points where it crosses each axis. We know that B is zero on any point on the A-axis. Therefore, the point where the line crosses the A-axis has zero as the B coordinate. So, we replace B with zero in the equation and solve for A:

.1A + .1B = 240 .1A + .1(0) = 240

.1A = 240 A = 240/.1 A = 2400

So, we know that the line crosses the A-axis at (2400, 0). We now want to see where the line crosses the B-axis. We know that A is zero at any point on the B-axis. Therefore, the point where the line crosses the B-axis has zero as the A coordinate. So, we replace A with zero in the equation and solve for B:

.1A + .1B = 240 .1(0) + .1B = 240

.1B = 240 B = 240/.1 B = 2400

So, we know that the line crosses the B-axis at (0, 2400). With these two points, you can now graph the line.



Now, we need to pick the side of the line that satisfies this inequality. The easiest way to do this is to test one point on one side of the line. You test a point by substituting the coordinates into the inequality formula, and check to see if coordinates produce a value that satisfies the inequality. If the point that is tested satisfies the inequality, then every point on the same side of the line as the point tested will satisfy the inequality. It is easiest to use the origin (0,0) to test whether the inequality is true because you are dealing with zeros as the variables. So, plug the origin into the inequality and see if the inequality is true for that point:

.1A + .1B ≤ 240 Inequality .1(0) + .1(0) ≤ 240 Test the Origin

0 ≤ 240 True Statement So, we know that the Feasible Region for the first constraint includes the side of the line that contains the origin. The Feasible Region that satisfies the first and last constraint consists of the following:

To graph the second constraint (.1A+.4B ≤ 720), we want to graph the line, .1A+.4B=720. We know that this line crosses the A-axis at (0, 7200):

.1A + .4B = 720 .1A + .4(0) = 720

.1A = 720 A = 720/.1 A = 7200



We know that this line crosses the B-axis at (1800, 0):

.1A + .4B = 720 .1(0) + .4B = 720

.4B = 720 B = 720/.4 B = 1800

By testing the origin, we see that the side that contains the origin satisfies the inequality:

.1A + .4B ≤ 720 Inequality .1(0) + .4(0) ≤ 720 Test the Origin

0 ≤ 720 True Statement So, we know that the Feasible Region for the second constraint includes the side of the line that contains the origin. The Feasible Region that satisfies the first, second and last constraints consists of the following:

To graph the third constraint (.1A+.5B ≤ 160), we want to graph the line, .1A+.5B=160. We know that this line crosses the A-axis at (0, 1600):

.1A + .5B = 160 .1A + .5(0) = 160

.1A = 160 A = 160/.1 A = 1600



We know that this line crosses the B-axis at (320,0):

.1A + .5B = 160 .1(0) + .5B = 160

.5B = 160 B = 160/.5 B = 320

By testing the origin, we see that the side that contains the origin satisfies the inequality:

1A + .5B ≤ 160 Inequality 1(0) + .5(0) ≤ 160 Test the Origin

0 ≤ 160 True Statement So, we know that the Feasible Region for the third constraint includes the side of the line that contains the origin. The Feasible Region that satisfies all of the constraints is the following:

The fact that the Feasible Region no longer touches the lines for the first and second constraints tells you that the first and second constraints are not “binding”. The time on Machines I and II could be unlimited and it would not affect our production possibilities. Once, you have identified the area that satisfies all of the constraints, you have to decide which points within the Feasible Region maximize your Contribution Margin. The corners points of the Feasible Region are the most extreme points, and therefore, they represent the production levels that produce the most extreme Contribution Margins (e.g., the highest and lowest Contribution Margins).



In order to find the highest Contribution Margin, you plug each of these corner points into the original Objective Function [.10(A)+.12(B)] and check to see which corner point produces the highest Contribution Margin:

Corner .10(A)+.12(B) Contribution Margin

(1600 , 0) .1 (1600) + .12 (0) = 160 + 0 = $160.00 (0 , 320) .1 (0) + .12 (320) = 0 + 38.4 = 38.40 (0 , 0) .1 (0) + .12 (0) = 0 + 0 = 0

The firm will generate the highest Contribution Margin by producing 1600 A Bolts and no B Bolts.

Using Excel To Solve Constrained Optimization Problems We can solve these Constrained Optimization Problems using Excel. You need to have one cell on your Excel spreadsheet represent the value of each of your two Objective Function Variables (number of A Bolts and number of B Bolts). Excel will place the optimal value in these cells after it solves the problem. For now, give A Bolts a value of “10” and give B Bolts a value of “20” so that you can tell if your other formulas are correct:

Cell Label Cell Contents

A Bolts 10 B Bolts 20

You also need to have another cell represent the value of the Objection Function formula using the values in your Objective Function Variable cells:


Objective Function =.1*[A Bolt Cell]+.12*[B Bolt Cell] For your constraints, you need to have a separate cell for each constraint. The cell will contain the value of the side of the constraint inequality formula that uses the Objective Function Variables (e.g., the time on each machine uses the number of each type of bolt produced). The side of the constraint inequality formula containing the fixed value will be supplied later:


Time on Machine I =.1*[A Bolt Cell]+.1*[B Bolt Cell] Time on Machine II =.1*[A Bolt Cell]+.4*[B Bolt Cell] Time on Machine III =.1*[A Bolt Cell]+.5*[B Bolt Cell]



Your spreadsheet should look something like the following:

Now, select “Tools” on the Menu Bar, and then click on “Solver”. (If you do not see “Solver”, select “Add-Ins” and then select “Solver Add-In”.) Once you select “Solver”, the following dialogue box will open:

For “Set Target Cell”, click on the cell that contains the Objective Function. For “Equal To”, click whether you want to maximize the Objective Function or minimize it. For “By Changing Cells”, highlight the two cells that contain the Objective Function Variables. For “Subject to the Constraints:”, click on “Add” and the following dialogue box will open:

For each constraint: (i) the left box should name the cell that contains the portion of the constraint inequality that contains the Objective Function Variables; (ii) the middle box



should give the proper inequality symbol; and (iii) the right box should contain the fixed value portion of the constraint inequality. For each non-negative constraint (e.g., A≥O): (i) the left box should name the cell that contains the Objective Function Variable; (ii) the middle box should have “>=” as the inequality symbol; and (iii) the right box should contain “0” as the fixed amount. In our example, we have the following constraints:

Once you have entered all of this information, click on “Solve”. Excel will change the values in the Objective Function Variable cells to reflect their optimal value. You can see that Excel gives the same answer that we got using the graphical method (produce 1600 A Bolts and zero B Bolts):

A “Solver Results” dialogue box will open. Leave “Keep Solver Solution” checked. Also, under “Reports” click on “Answer” and “Sensitivity”. When you click “OK”, two new worksheets will be created on your spreadsheet, the “Answer Report 1” sheet and the “Sensitivity Report 1” sheet. The Answer Report 1 sheet contains a description of your constraints, and indicates which ones were “binding”:

Answer Report 1 indicates that the constraints involving Machines I and II are not binding. We noted the same thing when we used the graphical approach. Notice that



Excel says that restricting the B Bolts to a non-negative answer was binding. This indicates that without this constraint, Excel would have given a negative number for the number of B Bolts to be produced. A negative B Bolt value would allow you to produce more A Bolts. For example, if B could be (-1), then you could obtain an extra 30 seconds on Machine III. This would allow you to produce 5 more A Bolts and stay within the constraint:

.1A+.5B ≤ 160 Constraint

.1(1600) + .5(0) = 160 Optimal Value .1(1605) + .5(-1) = 160 Allowing B to be (-1)

The negative one B Bolt would cost you 12¢ (a negative Contribution Margin) but the five A Bolts would produce an additional Contribution Margin of 50¢. Thus, if you could go negative, the next negative B Bolt would produce a marginal profit of 38¢. This potential is reported as the “Reduced Gradient” on Sensitivity Report 1:

The Lagrange Multiplier is the “Shadow Price” of a minute on Machine III. It reports that you would be willing to pay up to $1 for an extra minute on Machine III. You can calculate this yourself. We know that each A Bolt takes .1 minute on Machine III. You can, therefore, produce 10 A Bolts with an extra minute on Machine III. Each A Bolt produces a Contribution Margin of 10¢. Thus, an extra minute on Machine III, which allows us to produce 10 A Bolts is worth $1 (10¢ x 10) to us.

PROBLEMS E-1. Opportunity costs are: A) not used for decision making. B) the same as variable costs. C) the same as historical costs. D) relevant to decision making.



E-2. Freestone Company is considering renting Machine Y to replace Machine X. It is expected that Y will waste less direct materials than does X. If Y is rented, X will be sold on the open market. For this decision, which of the following factors is (are) relevant?

I. Cost of direct materials used II. Resale value of Machine X

A) Only I B) Only II C) Both I and II D) Neither I nor II E-3. In a sell or process further decision, which of the following costs are relevant? I. A variable production cost incurred prior to the split-off point. II. An avoidable fixed production cost incurred after the split-off point.

A) Only I. B) Only II. C) Both I and II. D) Neither I nor II. Use the following to answer questions E-4 through E-6: The Tingey Company has 500 obsolete microcomputers that are carried in inventory at a total cost of $720,000. If these microcomputers are upgraded at a total cost of $100,000, they can be sold for a total of $160,000. As an alternative, the microcomputers can be sold in their present condition for $50,000. E-4. The sunk cost in this situation is: A) $720,000. B) $160,000. C) $ 50,000. D) $100,000.



E-5. What is the net advantage or disadvantage to the company from upgrading the computers rather than selling them in their present condition?

A) $110,000 advantage. B) $660,000 disadvantage. C) $ 10,000 advantage. D) $ 60,000 advantage. E-6. Suppose the selling price of the upgraded computers has not been set. At what

selling price per unit would the company be as well off upgrading the computers as if it just sold the computers in their present condition?

A) $100. B) $770. C) $300. D) $210. Use the following to answer questions E-7 and E-8: Meacham Company has traditionally made a subcomponent of its major product. Its annual production of 20,000 subcomponents results in the following costs:

Meacham has received an offer from an outside supplier who is willing to provide 20,000 units of this subcomponent each year at a price of $28 per subcomponent. Meacham knows that the facilities now being used to make the subcomponent would be rented to another company for $75,000 per year if the subcomponent were purchased from the outside supplier. Otherwise, the fixed overhead would be unaffected. E-7. If Meacham decides to purchase the subcomponent from the outside supplier,

how much higher or lower will operating income be than if Meacham continued to make the subcomponent?

A) $45,000 higher. B) $70,000 higher. C) $30,000 lower. D) $70,000 lower. E-8. Suppose the price for the subcomponent has not been set. At what price per unit

charged by the outside supplier would Meacham be economically indifferent between making the subcomponent or buying it from the outside?

A) $30.25. B) $29.25. C) $26.50. D) $31.50.

Direct materials............. $200,000Direct labor ................... $180,000Variable overhead ......... $150,000Fixed overhead.............. $100,000



Use the following to answer questions E-9 and E-10: The Melville Company produces a single product called a Pong. Melville has the capacity to produce 60,000 Pongs each year. If Melville produces at capacity, the per unit costs to produce and sell one Pong are as follows:

The regular selling price for one Pong is $80. A special order has been received by Melville from Mowen Company to purchase 6,000 Pongs next year. If this special order is accepted, the variable selling expense will be reduced by 75%. However, Melville will have to purchase a specialized machine to engrave the Mowen name on each Pong in the special order. This machine will cost $9,000 and it will have no use after the special order is filled. The total fixed manufacturing overhead and selling expenses would be unaffected by this special order. E-9. Assume Melville anticipates selling only 50,000 units of Pong to regular

customers next year. If Mowen Company offers to buy the special order units at $65 per unit, the effect of accepting the special order on Melville's operating income for next year should be a:

A) $60,000 increase. B) $90,000 decrease. C) $159,000 increase. D) $36,000 increase. E-10. Assume Melville anticipates selling only 50,000 units of Pong to regular

customers next year. At what selling price for the 6,000 special order units would Melville be economically indifferent between accepting or rejecting the special order from Mowen?

A) $51.50. B) $49.00. C) $37.00. D) $38.50.

Direct materials.................................... $15Direct labor .......................................... 12Variable manufacturing overhead ........ 8Fixed manufacturing overhead ............ 9Variable selling expense ...................... 8Fixed selling expense........................... 3



E-11. Chickenfeet Company manufactures rubber chicken feet. Setup costs are $2.00 and Chickenfeet produces 4,000 feet evenly throughout the year. Using the economic order quantity (EOQ) approach, the optimal production run would be 200 when the cost of carrying one foot in inventory for one year is:

A) $.05 B) $.10 C) $.20 D) $.40 E-12. The Sundust Company manufactures 10,000 brooms evenly throughout the year.

The setup cost is $80, and the cost to carry a broom in inventory for six months is $.20. Sundust's objective is to produce the brooms at the lowest cost possible. Assuming each production run will be for the same number of brooms, how many production runs should Sundust make?

A) 3 B) 4 C) 5 D) 6 E-13. The following information pertains to the Platt Manufacturing Company:

Units required per year............... 60,000 Cost of placing an order.............. $ 800 Unit carrying cost per year........... $1,200 What is the economic order quantity (EOQ) assuming that the units are used evenly throughout the year?

A) 200 B) 283 C) 393 D) 400 P-1 Stereo Goods is a distributor of video tapes. Video Mart is a local retail outlet,

which sells blank and recorded videos. Video Mart purchases tapes from Stereo Goods at $5.00 per tape; tapes are shipped in packages of 25. Stereo Goods pays all incoming freight, and Video Mart does not inspect the tapes due to Stereo Goods’ reputation for high quality. Annual demand is 104,000 tapes at a rate of 2,000 tapes per week. Video Mart earns 15% on its cash investments. The purchase-order lead time is one week. The following cost data are available:

Relevant Ordering Costs Per Purchase Order $94.50 Carrying Costs Per Package Per Year (e.g., Relevant Insurance,

Materials Handling, Breakage, etc. per Year)

$3.50

What is the economic order quantity?



P-2 The Miers Company produces small engines for several manufacturers. The company receives orders from two assembly plants for their Topflight engine. Plant I needs at least 50 engines, and plant II needs at least 27 engines. The company must meet these needs. The company can send at most 85 engines to these two assembly plants. It costs $20 per engine to ship to plant I and $35 per engine to ship to plant II. The company wishes to minimize shipping costs. How many engines should it ship to each plant?

P-3 A company is considering two insurance plans with the types of coverage (in

thousands of dollars) and premiums per thousand dollars as shown below.

Coverage

Fire/Theft Liability Premium

Policy A $10 $80 $50

Policy B $15 $120 $40

The company wants at least $100,000 fire/theft insurance and at least $1,000,000 liability insurance from these plans at a minimum premium cost. How many units of each policy should the firm purchase in order to obtain the desired coverage at the minimum cost?

P-4 The Mostpure Milk Company gets milk from two dairies and then extracts butter

fat from the milk to use in its deluxe ice cream. Dairy A can supply at most 50 gallons of milk averaging 3.7% butterfat, and Dairy B can supply at most 80 gallons of milk averaging 3.2% butterfat. Mostpure can only afford to buy 100 gallons of milk. Most pure wants to find out how much milk to purchase from each supplier in order to get the most raw butterfat.

SOLUTIONS E-1. The answer is d. E-2. The answer is c. The cost savings is relevant. The money you get from selling the old machine

makes your investment in the new machine less. (It is an opportunity cost.) E-3. The answer is b.

The costs up to the split off point are the same regardless of whether you sell or process further. Because they are not different, those costs are not relevant. Only the direct fixed costs (avoidable) after split off are relevant because they will go away if you sell the product rather than processing it further.



E-4.The answer is a. The original cost is the sunk cost. It is not relevant to this decision.

E-5. The answer is c.

Sell Process Further Revenue $50,000 $160,000 Upgrade Cost -0 -100,000 $50,000 $60,000


You would be indifferent if the Upgrade Option produced a net benefit of $50,000. This means that the sales price must be $150,000. Divide it by 500 computers, and you get a sales price per unit of $300.

E-7. The answer is a. The rental income from the sublease would be an opportunity cost of the Make

option. Make Buy Direct Materials $200,000 $ 0 Direct Labor 180,000 0 Variable Overhead 150,000 0 Opportunity Cost 75,000 Purchase Cost _______ 560,000 $28 x 20,000 $605,000 $560,000

E-8. The answer is a.

You would be indifferent if the Upgrade Option cost $605,000. This means that the sales price must be $605,000. Divide it by 20,000 units, and you get a sales price per unit of $30.25.


Take Don’t Take Purchase Price $390,000 $ 0 ($65 x 6,000) Direct Materials -90,000 0 ($15 x 6,000) Direct Labor -72,000 0 ($12 x 6,000) Variable Overhead -48,000 0 ($8 x 6,000) Machine -9,000 0 Variable Selling __-12,000 ___0 $2 x 6,000 $159,000 $0



E-10. The answer is d. Melville would be indifferent if it had no profit from the sale. To do this you would reduce the sales price by $159,000. This would mean sales revenue of $231,000, which means a sales price of $38.50 ($231,000/6,000)

E-11. The answer is d.

√C = 126.49/200 √C = .6324 C = .40


Q = √1600000/.4 Q = √4,000,000 Q = .2,000

production runs = 10,000/ 2,000 = 5 E-13. The answer is b.

Q = √96,000,000/1200 Q = √80,000 Q = 282.84

____ Q = /2PD √ C

_____________ √C = √2 x 2.00 x 4,000 200

____ Q = /2PD √ C

_____________ Q = / 2 x 80 x 10,000 √ .4

____ Q = /2PD √ C

_____________ Q = / 2 x 800 x 60,000 √ 1,200



P-1

Q

=

_______ / _2PD_ √ C

P $94.50 P is the Order/Setup costs, which are given. D 4160 D is the annual demand for units. Here a unit is a package of 25

tapes. So, to get the number of units you have to divide the annual demand for tapes (104,000) by 25 tapes per package to get the number of packages needed.

C $22.25 C is the Carrying Cost per package for a year. We are given that it cost $3.50 to store a package for a year, but the tied-up capital is also a Carry Cost. We can earn 15% on our excess cash, so having the money tied up in inventory costs us 15%. Each package costs $125 ($5 x 25 tapes) and 15% of that amount is $18.75. The total Carrying Costs are $3.50 + $18.75 = $22.25.

Using the EOQ formula, we see that that the optimal run is 1,000 units:

EOQ = __________________

= 188 units √2(4160)(94.50)/(22.25) P-2 In this problem, let I represent the number of engines shipped to plant I and II

represent the number of engines shipped to plant II.

The Objective Function here is to Minimize shipping costs: Min 20(I) + 35(II)

The constraints here are:

You can't ship more than 85 engines: I + II ≤ 85

You have to ship at least 50 engines to plant I I ≥ 50

You have to ship at least 27 engines to plant II II ≥ 27

You can't ship a negative number of engines I , II ≥ 0



Next, you graph the each of these constraints:

Now, you test each corner solution:

Points Objective Function: (20 I + 35 II)

(50,27) 20 (50) + 35 (27) = 1000 + 945 = 1945

(50,35) 20 (50) + 35 (35) = 1000 +1225 = 2225

(58,27) 20 (58) + 35 (27) = 1160 + 945 = 2105

The minimum cost is $1945.

P-3 In this problem, let A represent the number of units of Policy A purchased and B

represent the number of units of Policy B purchased.

The Objective Function: Minimize insurance costs: Min 50(A) + 40(B)


You want at least $100 coverage for fire/theft 10A + 15B ≥100

You want at least $1000 coverage for liability 80A +120B ≥1000

You can't purchase a negative number of units of insurance

A , B ≥ 0





Points Objective Function: (50 A + 40 B)

(25/2,0) 50 (25/2) + 40 (0) = 625 + 0 = 625

(0,25/3) 50 (0) + 40 (25/3) = 0 + 333 = 333

The minimum cost is $333.

P-4 In this problem, let A represent the number of gallons of milk purchased from

Dairy A and B represent the number of gallons of milk purchased from Dairy B.

The Objective Function here is to Maximize butterfat: Max .037(A) + .032(B)


a You can only get 50 gallons from Dairy A A ≤ 50

b You can only get 80 gallons from Dairy B B ≤ 80

c You can't buy more than 100 gallons of milk A + B ≤ 100

d You can't buy a negative number of gallons A , B ≥ 0





Points Objective Function: (.037A + .032B)

(0,80) .037 (0) + .032 (80) = 0 + 2.56 = 2.56

(20,80) .037(20) + .032 (80) = .74 + 2.56 = 3.3

(50,50) .037(50) + .032 (50) = 1.85 + 1.6 = 3.45

(50,0) .037(50) + .032 (0) = 1.85 + 0 = 1.85

The maximum butterfat is 3.45 gallons.