Ever since the introduction of the first home video systems for television, Steve Goodman has dreamed about manufacturing his own video system for professionals. During the early years of home video, Steve watched a lot of his favorite old movies on his home video and planned the eventual development of his own video system. He intended it to be used primarily by television stations, advertising agencies, and other individuals and groups that wanted the best in video systems. The overall configuration of this system is shown in the illustration. The basic system includes a comprehensive control box, two separate videotape systems, a video disk, and a professional-quality television set. All these devices are fully integrated. In addition, the basic system comes with an elaborate remote control device. This device can operate both video systems, the video disk, and the TV system with ease. The remote control device works by sending infrared signals to the control box, which in turn controls the other devices in the system.
Steve’s unique contribution to the video systems is the control box. The control box is an advanced microprocessor with the ability to coordinate the use and function of the other devices attached to it.
Steve’s professional video system has numerous advantages over similar systems. To begin with, special effects can be introduced easily. Images from the video disk, one of the video systems, and the television system can easily be placed on the other video system. In addition, it is possible to connect the control box to several popular microcomputers including the Macintosh, the IBM PC, the Radio Shack Model 3000, and advanced Zenith and Coleco computer systems. This makes it possible to develop attractive graphics on the microcomputer and to transfer them directly to the video system. It is also possible to hook a stereo system to the control box to integrate the highest quality stereo sound into the system and record it on one of the video systems. The two video systems also offer remarkable flexibility in editing. Several special editing buttons were placed on the remote control station. It is possible to first record a program on one video system and then edit it by using the other videotape system to add and delete sections. One of the best features of Steve’s professional video system is the price. The basic system, including the control box, both video systems, the video disk, and the television system, has a retail price of $1,995.
Steve found manufacturers for the television system, the control box, and the video disk system in the United States. Because videotape systems are more popular, Steve had more choices. After extensive research, he was able to eliminate all of the potential suppliers but two. Both of these suppliers are Japanese companies. Toshiki is a new company located outside of Tokyo, Japan. Like other suppliers, Toshiki offers quantity discounts. For quantities ranging from 0 to 7,000 units, Toshiki would charge Steve a price of $250 per video system. For quantities that ranged between 7,000 and 8,000 units, the per unit cost would be $230. For quantities ranging from 8,000 to 20,000 units, the unit price drops to $220. For more than 20,000 units, the per unit price of the video systems would be only $210. The other Japanese supplier is Kony. Although Kony originally started in Japan, also outside of Tokyo, it now has offices and manufacturing facilities around the world. One of these manufacturing facilities is located less than 100 miles north of Atlanta, Georgia. Like Toshiki, Kony offers quantity discounts for its videotape systems. For quantities ranging from 0 to 1,000, Kony’s per unit cost is $230. For quantities that range from 1,000 to 5,000 units, the unit cost is $240; and for over 5,000 units, the unit cost drops to $220.
Because Kony has manufacturing facilities located in the United States, the cost to place an order and the delivery time are much more favorable than they are with Toshiki. The estimated per order cost from Kony is $40, and the expected delivery time is two weeks. On the other hand, the ordering cost is higher and delivery time is longer for Toshiki. The additional paperwork and problems associated with ordering directly from Japan would increase Steve’s cost to $90 per order. Furthermore, the delivery time for Toshiki is three months. Steve estimates that his carrying cost would be 30 percent. This is due primarily to storage and handling cost as well as the potential for technological obsolescence. For the first year or so of operations, Steve decided to sell only the basic unit: the control box, the television set, the video disk, and the two videotape systems. The demand for the complete system was fairly constant during the past six months. For example, June sales were 7,970; July sales were 8,070; August sales were 7,950; and September sales were 8,010. This demand pattern is expected to continue for the next several months.
Quantitative Analysis for Management
Thirteenth Edition
Chapter 6
Inventory Control Models
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Learning Objectives (1 of 2)
After completing this chapter, students will be able to:
6.1 Understand the importance of inventory control.
6.2 Understand the various types of inventory related
decisions.
6.3 Use the economic order quantity (EOQ) to determine
how much to order.
6.4 Compute the reorder point (ROP) in determining when
to order more inventory.
6.5 Handle inventory problems that allow noninstantaneous
receipt.
6.6 Handle inventory problems that allow quantity
discounts.
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Learning Objectives (2 of 2)
After completing this chapter, students will be able to:
6.7 Understand the use of safety stock.
6.8 Compute single period inventory quantities using
marginal analysis.
6.9 Understand the importance of ABC analysis.
6.10 Describe the use of material requirements planning in
solving dependent-demand inventory problems.
6.11 Discuss just-in-time inventory concepts to reduce
inventory levels and costs.
6.12 Discuss enterprise resource planning systems.
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Chapter Outline (1 of 2)
6.1 Importance of Inventory Control
6.2 Inventory Decisions
6.3 Economic Order Quantity: Determining How Much to
Order
6.4 Reorder Point: Determining When to Order
6.5 EOQ Without the Instantaneous Receipt Assumption
6.6 Quantity Discount Models
6.7 Use of Safety Stock
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Chapter Outline (2 of 2)
6.8 Single-Period Inventory Models
6.9 ABC Analysis
6.10 Dependent Demand: The Case for Material
Requirements Planning
6.11 Just-in-Time Inventory Control
6.12 Enterprise Resource Planning
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Introduction (1 of 3)
• Inventory is an expensive and important asset
• Any stored resource used to satisfy a current or future
need
– Raw materials
– Work-in-process
– Finished goods
• Balance high and low inventory levels to minimize costs
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Introduction (2 of 3)
• Lower inventory levels
– Can reduce costs
– May result in stockouts and dissatisfied customers
• All organizations have some type of inventory planning
and control system
• Determine what goods/services are produced or
purchased
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Introduction (3 of 3)
Figure 6.1 Inventory Planning and Control
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Importance of Inventory Control (1 of 3)
• Five uses of inventory
1. The decoupling function
2. Storing resources
3. Irregular supply and demand
4. Quantity discounts
5. Avoiding stockouts and shortages
• Decoupling Function
– Reduces delays and improves efficiency
– A buffer between stages
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Importance of Inventory Control (2 of 3)
• Storing resources
– Seasonal products stored to satisfy off-season
demand
– Materials stored as raw materials, work-in-process, or
finished goods
– Labor can be stored as a component of partially
completed subassemblies
• Irregular supply and demand
– Not constant over time
– Inventory used to buffer the variability
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Importance of Inventory Control (3 of 3)
• Quantity discounts
– Lower prices may be available for larger orders
– Higher storage and holding costs
– More cash invested
• Avoiding stockouts and shortages
– Stockouts may result in lost sales
– Dissatisfied customers may choose to buy from
another supplier
– Loss of goodwill
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Inventory Decisions
• Two fundamental decisions
1. How much to order
2. When to order
• Major objective is to minimize total inventory costs
1. Cost of the items (purchase cost or material cost)
2. Cost of ordering
3. Cost of carrying, or holding, inventory
4. Cost of stockouts
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Inventory Cost Factors (1 of 2)
Table 6.1 Inventory Cost Factors
ORDERING COST FACTORS
CARRYING COST FACTORS
Developing and sending purchase orders
Cost of capital
Processing and inspecting incoming
inventory
Taxes
Bill paying
Insurance
Inventory inquiries
Spoilage
Utilities, phone bills, and so on for the
purchasing department
Theft
Salaries and wages for purchasing
department employees
Obsolescence
Supplies such as forms and paper for the
purchasing department
Salaries and wages for warehouse employees
Blank
Blank
Utilities and building costs for the warehouse
Supplies such as forms and paper for the warehouse
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Inventory Cost Factors (2 of 2)
• Ordering costs are generally independent of order quantity
– Many involve personnel time
– The amount of work is the same no matter the size of
the order
• Holding costs generally vary with the amount of inventory
or order size
– Labor, space, and other costs increase with order size
– Cost of items purchased can vary with quantity
discounts
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Economic Order Quantity (1 of 2)
• Economic order quantity (EOQ) model
– One of the oldest and most commonly known
inventory control techniques
– Easy to use
– A number of important assumptions
• Objective is to minimize total cost of inventory
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Economic Order Quantity (2 of 2)
• Assumptions:
1. Demand is known and constant over time
2. Lead time is known and constant
3. Receipt of inventory is instantaneous
4. Purchase cost per unit is constant
5. The only variable costs are ordering cost and holding
or carrying cost, and these are constant throughout
the year
6. Orders are placed so that stockouts or shortages are
avoided completely
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Inventory Usage Over Time
Figure 6.2 Inventory Usage over Time
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Inventory Costs in the EOQ Situation
Average inventory level =
Q
2
•
Annual ordering cost is number of orders per year times cost of placing each order
•
Annual carrying cost is the average inventory times carrying cost per unit per year
Table 6.2 Computing Average Inventory
INVENTORY LEVEL
DAY
BEGINNING
ENDING
AVERAGE
April 1 (order received)
10
8
9
April 2
8
6
7
April 3
6
4
5
April 4
4
2
3
April 5
2
0
1
Maximum level April 1 = 10 units
Total of daily averages = 9 + 7 + 5 + 3 + 1 = 25
Number of days = 5
Average inventory level = 25 5 = 5 units
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Inventory Costs in the EOQ Situation
(1 of 3)
Q = number of pieces to order
EOQ = Q * = optimal number of pieces to order
D = annual demand in units for the inventory item
Co = ordering cost of each order
Ch = holding or carrying cost per unit per year
Number of
Ordering cost
Ordering = orders placed
per order
cost
per year
An nu a l
Annual demand Ordering cost D
=
= Co
Number of units per order Q
in each order
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Inventory Costs in the EOQ Situation
(2 of 3)
Q = number of pieces to order
EOQ = Q * = optimal number of pieces to order
D = annual demand in units for the inventory item
Co = ordering cost of each order
Ch = holding or carrying cost per unit per year
Carrying cost
Average
holding =
per unit
inventory
cost
per year
Order quantity
=
(Carrying cost per unit per year)
2
Q
= Ch
2
Annual
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Inventory Costs in the EOQ Situation
(3 of 3)
Figure 6.3 Total Cost as a Function of Order Quantity
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Finding the EOQ (1 of 2)
• When the EOQ assumptions are met, total cost is
minimized when
Annual ordering cost = Annual holding cost
D
Q
Co = Ch
Q
2
Thus
Solving for Q
Q 2Ch = 2DCo
2DCo
Q =
Ch
2
2DCo
EOQ = Q =
Ch
*
2DCo
Q=
Ch
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Finding the EOQ (2 of 2)
• Equation summary
Annual ordering cost =
D
Co
Q
Annual holding cost =
Q
Ch
2
2DCo
EOQ = Q =
Ch
*
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Sumco Pump Company (1 of 5)
• Sells pump housings to other companies
• Reduce inventory costs by finding optimal order quantity
Annual demand = 1,000 units
Ordering cost = $10 per order
Average carrying cost per unit per year = $0.50
2DCo
2(1,000)(10)
Q =
=
= 40,000 = 200 units
Ch
0.50
*
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Sumco Pump Company (2 of 5)
• Total cost
TC =
=
D
Q
Co + Ch
Q
2
1,000
200
(10) +
(0.5)
200
2
= $50 + $50 = $100
Number of orders per year = (D ÷ Q ) = 5
Average inventory (Q 2) = 100
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Sumco Pump Company (3 of 5)
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Sumco Pump Company (4 of 5)
Program 6.1A Input Data and Excel QM Formulas for the
Sumco Pump Company Example
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Sumco Pump Company (5 of 5)
Program 6.1B Excel QM Solution for the Sumco Pump
Company Example
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Purchase Cost of Inventory Items (1 of 2)
• Total inventory cost can be written to include the cost of
purchased items
– Annual purchase cost is constant at D × C no matter
the order policy, where
C is the purchase cost per unit
D is the annual demand in units
• The average dollar level of inventory
(CQ )
Average dollar level =
2
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Purchase Cost of Inventory Items (2 of 2)
• Carrying cost often expressed as an annual percentage of
the unit cost or price of the inventory
• New variable
I = (Annual inventory holding charge as a percentage of unit price or cost)
Cost of storing inventory for one year = Ch = IC
Thus
2DCo
Q =
IC
*
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Sensitivity Analysis with the EOQ
Model (1 of 2)
• The EOQ model assumes all values are know and fixed
over time
• Values are estimated or may change
• Sensitivity analysis determines the effects of these
changes
• Because the EOQ is a square root, changes in the inputs
result in relatively minor changes in the order quantity
2DCo
EOQ = Q =
Ch
*
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Sensitivity Analysis with the EOQ
Model (2 of 2)
• Sumco Pump example
2(1,000)(10)
EOQ =
= 200 units
0.50
• Increase Co to $40
EOQ =
2(1,000)(40)
= 400 units
0.50
• In general, the EOQ changes by the square root of the
change to any of the inputs
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Reorder Point: Determining When To
Order
• Next decision is when to order
• Time between placing an order and its receipt is called the
lead time (L) or delivery time
• On hand and on order inventory must be available to meet
demand during lead, the inventory position
• Generally expressed as a reorder point (ROP)
ROP = (Demand per day)× (Lead time for a new order in days)
= d ×L
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Reorder Point Graphs
Figure 6.4 Reorder Point Graphs
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Procomp’s Computer Chips (1 of 2)
• Annual demand = 8,000
• Daily demand = 40 units
• Delivery in three working days
ROP = d L = 40 units per day 3 days
= 120 units
• An order for the EOQ (400) is placed when the inventory
reaches 120 units
• The order arrives 3 days later just as the inventory is
depleted
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Procomp’s Computer Chips (2 of 2)
• Annual demand = 8,000
• Daily demand = 40 units
• Delivery in three now 12 working days
ROP = d L = 40 units per day 12 days
= 480 units
Inventory Inventory Inventory
=
+
position
on
hand
on
order
480 = 80 + 400
• New order placed when inventory = 80 and one order is in
transit
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EOQ Without Instantaneous Receipt
• When inventory accumulates over time, the
instantaneous receipt assumption does not apply
– Daily demand rate must be taken into account
– Production run model
Figure 6.5 Inventory Control and the Production Process
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Annual Carrying Cost for Production
Run Model (1 of 3)
• Setup cost replaces ordering cost
– Model variables
Q = number of pieces per order, or production run
Cs = setup cost
Ch = holding or carrying cost per unit per year
p = daily production rate
d = daily demand rate
t = length of production run in days
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Annual Carrying Cost for Production
Run Model (2 of 3)
• Maximum inventory level
(Total produced during the production run)
− (Total used during production run)
= (Daily production rate)(Number of days production)
− (Daily demand)(Number of days production)
= ( pt ) − (dt )
Q
Since Total produced = Q = pt and t =
p
Maximum inventory level = pt – dt = p
Q
Q
d
– d = Q 1–
p
p
p
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Annual Carrying Cost for Production
Run Model (3 of 3)
• Average inventory is one-half the maximum
Q
d
Average inventory = 1–
2
p
and
Q
d
Annual holding cost = 1– Ch
2
p
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Annual Setup Cost for Production
Run Model
• Setup cost replaces ordering cost
D
Annual setup cost = Cs
Q
becomes
D
Annual ordering cost = Co
Q
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Determining the Optimal Production
Quantity
• Set setup costs equal to holding costs and solve for the
optimal order quantity
Annual holding cost = Annual setup cost
Q
d
D
1– Ch = Cs
2
p
Q
Solving for Q, we get
2DCs
Q =
d
Ch 1–
p
*
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Production Run Model
• Equation summary
Annual holding cost =
Q
d
1
–
Ch
2
p
D
Annual setup cost = Cs
Q
2DCs
Optimal production quantity Q =
d
Ch 1 –
p
*
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Brown Manufacturing (1 of 4)
• Produces commercial refrigeration units in batches
Annual demand = D = 10,000 units
Setup cost = Cs = $100
Carrying cost = Ch = $0.50 per unit per year
Daily production rate = p = 80 units daily
Daily demand rate = d = 60 units daily
1. How many units should Brown produce in each batch?
2. How long should the production part of the cycle last?
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Brown Manufacturing (2 of 4)
1.
2DCs
Q =
d
Ch 1 –
p
*
Q* =
2 10,000 100
60
0.5 1 –
80
2.
Production cycle =
=
Q
p
4,000
= 50 days
80
2,000,000
=
= 16,000,000
0.5 1
4
( )
= 4,000 units
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Brown Manufacturing (3 of 4)
Program 6.2A Excel QM Formulas and Input Data for the
Brown Manufacturing Problem
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Brown Manufacturing (4 of 4)
Program 6.2B Excel QM Solutions for the Brown
Manufacturing Problem
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Quantity Discount Models (1 of 5)
• Quantity discounts are commonly available
• Basic EOQ model is adjusted by adding in the purchase
or materials cost
Total cost = Material cost + Ordering cost + Holding cost
Total cost = DC +
D
Q
Co + Ch
Q
2
where
D = annual demand in units
Co = ordering cost of each order
C = cost per unit
Ch = holding or carrying cost per unit per year
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Quantity Discount Models (2 of 5)
Holding cost per unit is based on cost, so
Ch = IC
where
I = holding cost as a percentage of the unit cost (C)
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Quantity Discount Models (3 of 5)
• Discount schedule and EOQs might not align
• Buying at the lowest unit cost may not result in lowest total
cost
Table 6.3 Quantity Discount Schedule
DISCOUNT
NUMBER
DISCOUNT
QUANTITY
DISCOUNT (%)
DISCOUNT
COST ($)
1
0 to 999
0
5.00
2
1,000 to 1,999
4
4.80
3
2,000 and over
5
4.75
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Quantity Discount Models (4 of 5)
Figure 6.6 Total Cost (TC) Curve for the Quantity Discount
Model
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Quantity Discount Models (5 of 5)
• Steps in the process
1. For each discount price (C), compute EOQ =
2DCo
IC
2. If EOQ < Minimum for discount, adjust the quantity to
Q = Minimum for discount
3. For each EOQ or adjusted Q, compute
Total cost = DC +
D
Q
Co + Ch
Q
2
4. Choose the lowest-cost quantity
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Brass Department Store (1 of 5)
• Toy race cars
• Quantity discounts available
Step 1 – Compute EOQs for each discount
EOQ1 =
(2)(5,000)(49)
= 700 cars per order
(0.2)(5.00)
EOQ2 =
(2)(5,000)(49)
= 714 cars per order
(0.2)(4.80)
EOQ3 =
(2)(5,000)(49)
= 718 cars per order
(0.2)(4.75)
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Brass Department Store (2 of 5)
Step 2 – Adjust quantities below the allowable discount
range
– The EOQ for discount 1 is allowable
– The EOQs for discounts 2 and 3 are outside the
allowable range, adjust to the possible quantity
closest to the EOQ
Q1 = 700
Q2 = 1,000
Q3 = 2,000
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Brass Department Store (3 of 5)
Step 3 – Compute total cost for each quantity
Table 6.4 Total Cost Computations for Brass Department
Store
ANNUAL
MATERIAL
COST
ANNUAL
ORDERING
COST
ANNUAL
CARRYING
COST
Left parenthesis, dollar, right parenthesis, equals, left parenthesis, D divided by Q, right parenthesis, C sub o.
Left parenthesis, dollar, right parenthesis, equals, left parenthesis, Q divided by 2, right parenthesis, C sub h.
DISCOUNT
NUMBER
UNIT
PRICE
(C)
ORDER
QUANTITY
(Q)
1
$5.00
700
25,000
350.00
350.00
25,700.
00
2
4.80
1,000
24,000
245.00
480.00
24,725.
00
3
4.75
2,000
23,750
122.50
950.00
24,822.
50
Left parenthesis, dollar, right parenthesis, equals D C.
($) = DC
($) = (D ÷ Q )Co ($) = (Q ÷ 2)Ch
TOTAL
($)
Step 4 – Choose the alternative with the lowest total cost
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Brass Department Store (4 of 5)
Program 6.3A Excel QM Formulas and Input Data for the
Brass Department Store Quantity Discount Problem
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Brass Department Store (5 of 5)
Program 6.3B Excel QM Solutions for the Brass Department
Store Quantity Discount Problem
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Use of Safety Stock (1 of 4)
• If demand or the lead time are uncertain, the exact ROP
will not be known with certainty
• Safety stock can help prevent stockouts
• Can be implemented by adjusting the ROP
ROP = (Average demand during lead time) + Safety stock
ROP = (Average demand during lead time) + SS
where
SS = safety stock
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Use of Safety Stock (2 of 4)
Figure 6.7 Use of Safety Stock
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Use of Safety Stock (3 of 4)
• Objective is to choose a safety stock amount the
minimizes total holding and stockout costs
• If variation in demand and holding and stockout costs are
known, payoff/cost tables could be used to determine
safety stock
• More general approach is to choose a desired service
level based on satisfying customer demand
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Use of Safety Stock (4 of 4)
• Set safety stock to achieve a desired service level
Service level = 1− Probability of a stockout
or
Probability of a stockout = 1− Service level
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Safety Stock with the Normal
Distribution
ROP = (Average demand during lead time) + ZsdLT
where
Z = number of standard deviations for a given service level
sdLT = standard deviation of demand during the lead time
Thus
Safety stock = ZsdLT
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Hinsdale Company (1 of 8)
Item A3378 has normally distributed demand during lead
time
Mean = 350 units, standard deviation = 10
• Stockouts should occur only 5% of the time
μ = Mean demand = 350
σ dLT = Standard deviation = 10
X = Mean demand + Safety stock
SS = Safety stock = X − μ = Zσ
X −μ
Z=
σ
Figure 6.8 Safety Stock and the Normal Distribution
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Hinsdale Company (2 of 8)
• From Appendix A we find Z = 1.65 =
X − μ SS
=
σ
σ
ROP = (Average demand during lead time)+ ZsdLT
= 350 +1.65(10)
= 350 +16.5 = 366.5 units (or about 367 units)
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Calculating Lead Time Demand and
Standard Deviation (1 of 4)
• Three situations to consider
– Demand is variable but lead time is constant
– Demand is constant but lead time is variable
– Both demand and lead time are variable
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Calculating Lead Time Demand and
Standard Deviation (2 of 4)
1. Demand is variable but lead time is constant
(
ROP = dL + Z σ d L
)
where
d = average daily demand
σ d = standard deviation of daily demand
L = lead time in days
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Calculating Lead Time Demand and
Standard Deviation (3 of 4)
2. Demand is constant but lead time is variable
ROP = dL + Z ( dσ L )
where
L = average lead time
σ L = standard deviation of lead time
d = daily demand
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Calculating Lead Time Demand and
Standard Deviation (4 of 4)
3. Both demand and lead time are variable
ROP = dL + Z Lσ d2 + d 2σ L2
– The most general case
– Can be simplified to the earlier equations
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Hinsdale Company (3 of 8)
• Determine safety stock for three other items
• For SKU F5402, d = 15, σ d = 3, L = 4
• Desired service level = 97%
– For a 97% service level, Z = 1.88
(
ROP = dL + Z σ d L
)
= 15(4) + 1.88(3 4 ) = 15(4) + 1.88(6)
= 60 + 11.28 = 71.28
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Hinsdale Company (4 of 8)
• For SKU B7319, d = 25, L = 6, σ L = 3
• Desired service level = 98%
– For a 98% service level, Z = 2.05
ROP = dL + Z ( dσ L )
= 25(6) + 2.05(25)(3) = 150 + 2.05(75)
= 150 + 153.75 = 303.75
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Hinsdale Company (5 of 8)
• For SKU F9004, d = 20, σ d = 4, L = 5, σ L = 2
• Desired service level = 94%
– For a 94% service level, Z = 1.55
ROP = dL + Z Lσ d2 + d 2σ L2
= (20)(5) + 1.55 5(4)2 + (20)2 (2)2
= 100 + 1.55 1680
= 100 + 1.55(40.99) = 100 + 63.53 = 163.53
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Service Levels, Safety Stock, and
Holding Costs
• As service levels increase
– Safety stock increases at an increasing rate
• As safety stock increases
– Annual holding costs increase
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Hinsdale Company (6 of 8)
Table 6.5 Safety Stock for SKU A3378 at Different Service
Levels
SERVICE LEVEL
(%)
Z VALUE FROM NORMAL CURVE
TABLE
SAFETY STOCK
(UNITS)
90
1.28
12.8
91
1.34
13.4
92
1.41
14.1
93
1.48
14.8
94
1.55
15.5
95
1.65
16.5
96
1.75
17.5
97
1.88
18.8
98
2.05
20.5
99
2.33
23.3
99.99
3.72
37.2
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Calculating Annual Holding Cost with
Safety Stock
• Under standard assumptions of EOQ
– Average inventory = Q ÷ 2
– Annual holding cost = (Q ÷ 2)Ch
• With safety stock
Holding cost of
Total annual
Holding cost of
regular
+
holding
cost
safety
stock
inventory
Q
THC = Ch + (SS)Ch
2
Where
THC = total annual holding cost
Q = order quantity
Ch = holding cost per unit per year
SS = safety stock
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Hinsdale Company (7 of 8)
Figure 6.9 Service Level Versus Annual Carrying Costs
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Hinsdale Company (8 of 8)
Figure 6.9 Service Level Versus Annual Carrying Costs
This graph was developed for a specific case, but the
general shape of the curve is the same for all cases
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Using Excel QM for Safety Stock
Problems (1 of 2)
Program 6.4A Excel QM Formulas and Input Data for the
Hinsdale Safety Stock Problem
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Using Excel QM for Safety Stock
Problems (2 of 2)
Program 6.4B Excel QM Solutions for the Hinsdale Safety
Stock Problem
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Single-Period Inventory Models
• Some products have no future value beyond the current
period
– News vendor problems or single-period inventory
models
• Marginal analysis uses marginal profit (MP) and
marginal loss (ML)
– With a manageable number of states of nature and
alternatives, use discrete distributions
– When there are a large number of alternatives or
states of nature, use normal distribution
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Marginal Analysis with Discrete
Distributions (1 of 2)
• Stock an additional unit only if the expected marginal profit
for that unit exceeds the expected marginal loss
P = probability that demand will be greater than or
equal to a given supply (or the probability of selling
at least one additional unit)
1 − P = probability that demand will be less than supply
(or the probability that one additional unit will not
sell)
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Marginal Analysis with Discrete
Distributions (2 of 2)
• The expected marginal profit = P(MP)
• The expected marginal loss = (1 − P )(ML )
• The optimal decision rule
Stock the additional unit if P (MP) (1 − P ) ML
• With some basic manipulation
P (MP) ML − P (ML)
P (MP) + P (ML ) ML
P (MP + ML ) ML
or
ML
P
ML + MP
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Steps of Marginal Analysis with
Discrete Distributions
1. Determine the value of
ML
for the problem
ML + MP
2. Construct a probability table and add a cumulative
probability column
3. Keep ordering inventory as long as the probability (P) of
selling at least one
ML
additional unit is greater than
ML + MP
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Café du Donut (1 of 5)
• Café buys donuts each day for $4 per carton of 2 dozen
donuts
• Cartons not sold are thrown away at the end of the day
• If a carton is sold, the total revenue is $6
• The marginal profit per carton is
MP = Marginal profit = $6 − $4 = $2
• Marginal loss ML = $4 since doughnuts cannot be returned
or salvaged
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Café du Donut (2 of 5)
Table 6.6 Café du Donut’s Probability Distribution
DAILY SALES
(CARTONS OF DOUGHNUTS)
4
PROBABILITY (P) THAT DEMAND
WILL BE AT THIS LEVEL
0.05
5
0.15
6
0.15
7
0.20
8
0.25
9
0.10
10
0.10
Blank
Total 1.00
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Café du Donut (3 of 5)
Step 1. Determine the value of
ML
for the decision rule
ML + MP
ML
$4
4
P
=
=
= 0.67
ML + MP
$4 + $2
6
P 0.67
Step 2.Add a new column to the table to reflect the
probability that doughnut sales will be at each level or
greater
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Café du Donut (4 of 5)
Table 6.7 Marginal Analysis for Café du Donut
DAILY SALES
(CARTONS OF
DOUGHNUTS)
PROBABILITY (P) THAT
DEMAND WILL BE AT
THIS LEVEL
PROBABILITY (P) THAT
DEMAND WILL BE AT THIS
LEVEL OR GREATER
4
0.05
1.00 0.66
5
0.15
0.95 0.66
6
0.15
0.80 0.66
7
0.20
0.65
8
0.25
0.45
9
0.10
0.20
10
0.10
0.10
Blank
1.00 greater than or equal to 0.66.
0.95 greater than or equal to 0.66.
0.80 greater than or equal to 0.66.
Blank
Total 1.00
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Café du Donut (5 of 5)
Step 3. Keep ordering additional cartons as long as the
probability of selling at least one additional carton is
greater than P, the indifference probability
P at 6 cartons = 0.80 0.67
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Marginal Analysis with the Normal
Distribution
• Find four values
1. The average or mean sales for the product, μ
2. The standard deviation of sales, σ
3. The marginal profit for the product, MP
4. The marginal loss for the product, ML
X * = optimal stocking level
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Steps of Marginal Analysis with the
Normal Distributions
ML
1. Determine the value of
for the problem
ML + MP
2. Locate P on the normal distribution (Appendix A) and find
the associated Z-value
X* – μ
3. Find X * using the relationship Z =
σ
to solve for the resulting stocking policy
X * = μ + Zσ
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Newspaper Example (1 of 7)
• Chicago Tribune m = 60 papers a day, s = 10
• Marginal loss ML = 20 cents
• Marginal profit MP = 30 cents
Step 1. Stock the Tribune as long as the probability of
selling the last unit is at least ML ÷ (ML + MP)
ML
20 cents
20
=
=
= 0.40
ML + MP
20 cents + 30 cents
50
Let P = 0.40
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Newspaper Example (2 of 7)
Step 2. Using the normal distribution in Figure 6.10 find the
appropriate Z value
Z = 0.25 standard deviations from the mean
Figure 6.10 Joe’s Stocking Decision for the Chicago Tribune
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Newspaper Example (3 of 7)
• Chicago Tribune m = 60 papers a day, s = 10
Step 3.
X * – 60
0.25 =
10
or
X * = 60 + 0.25(10) = 62.5, or 62 newspapers
Joe should order 62 newspapers since the probability of
selling 63 newspapers is slightly less than 0.40
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Newspaper Example (4 of 7)
Figure 6.10 Joe’s Stocking Decision for the Chicago Tribune
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Newspaper Example (5 of 7)
• The procedure is the same when P > 0.50
• Chicago Sun-Times
• ML = 40 cents, MP = 10 cents, m = 100, s = 10
Step 1.
ML
40 cents
40
=
=
= 0.80
ML + MP
40 cents + 10 cents
50
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Newspaper Example (6 of 7)
Step 2. Using the normal distribution in Figure 6.11 find the
appropriate Z value for 0.80 and multiply by −1
Z = −0.84 standard deviations from the mean
Figure 6.11 Joe’s Stocking Decision for the Chicago SunTimes
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Newspaper Example (7 of 7)
• Chicago Sun-Times m = 100 papers a day, s = 10
Step 3.
X * – 100
−0.84 =
10
or
X * = 100 − 0.84(10) = 91.6, or 91 newspapers
Joe should order 91 copies of the Chicago Sun-Times every
day
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ABC Analysis (1 of 3)
• The purpose is to divide the inventory into three groups
based on the overall inventory value of the items
• Group A items account for the major portion of inventory
costs
– Typically 70% of the dollar value but only 10% of the
quantity of items
– Forecasting and inventory management must be done
carefully
– Mistakes can be expensive
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ABC Analysis (2 of 3)
• Group B items are more moderately priced
– May represent 20% of the cost and 20% of the quantity
– Moderate levels of control
• Group C items are very low cost but high volume
– It is not cost effective to spend a lot of time managing
these items
– Simple control policies and loose control
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ABC Analysis (3 of 3)
Table 6.8 Summary of ABC Analysis
INVENTORY
GROUP
DOLLAR
USAGE (%)
INVENTORY
ITEMS (%)
ARE QUANTITATIVE CONTROL
TECHNIQUES USED?
A
70
10
Yes
B
20
20
In some cases
C
10
70
No
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Dependent Demand: The Case for
Material Requirements Planning (1 of 2)
• Inventory models discussed so far have assumed item demand
was independent
• In many situations items demand is dependent on demand for
one or more other items
• In these situations material requirements planning (MRP)
can be employed effectively
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Dependent Demand: The Case for
Material Requirements Planning (2 of 2)
• Benefits of MRP
1. Increased customer service and satisfaction
2. Reduced inventory costs
3. Better inventory planning and scheduling
4. Higher total sales
5. Faster response to market changes and shifts
6. Reduced inventory levels without reduced customer
service
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Material Structure Tree (1 of 4)
• The first step is to develop a bill of materials (BOM)
• BOM identifies components, descriptions, and the number
required for production of one unit of the final product
• Material structure tree developed from the BOM
– Demand for product A is 50 units
– Each A requires 2 units of B and 3 units of C
– Each B requires 2 units of D and 3 units of E
– Each C requires 1 unit of E and 2 units of F
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Material Structure Tree (2 of 4)
Figure 6.12 Material Structure Tree for Item A
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Material Structure Tree (3 of 4)
• The demand for B, C, D, E, and F is completely
dependent on the demand for A
• The material structure tree has three levels
• Items above a level are called parents
• Items below any level are called components
• The number in parenthesis beside each item shows how
many are required for each unit of the parent
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Material Structure Tree (4 of 4)
• We can use the material structure tree and the demand
for Item A to compute demands for the other items
Part B: 2 × number of A’s = 2 × 50 = 100
Part C: 3 × number of A’s = 3 × 50 = 150
Part D: 2 × number of B’s = 2 × 100 = 200
Part E: 3 × number of B’s + 1× number of C’s
= 3 ×100 +1×150 = 450
Part F: 2 × number of C’s = 2×150 = 300
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Gross and Net Material Requirements
Plan (1 of 4)
• A gross material requirements plan is constructed after
the materials structure tree is complete
– Time schedule
– Shows when an item must be ordered
– Shows when there is no inventory on hand
– Shows when the production of an item must be started
in order to satisfy the demand for the finished product
at a particular date
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Gross and Net Material Requirements
Plan (2 of 4)
• Lead times are required for each item
Item A – 1 week
Item D – 1 week
Item B – 2 weeks
Item E – 2 weeks
Item C – 1 week
Item F – 3 weeks
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Gross Material Requirements Plan
Figure 6.13 Gross Material Requirements Plan for 50 Units
of A
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Gross and Net Material Requirements
Plan (3 of 4)
• A net material requirements plan can be constructed from
the gross materials requirements plan and on-hand
inventory information
Table 6.9 On-Hand Inventory
ITEM
ON-HAND INVENTORY
A
10
B
15
C
20
D
10
E
10
F
5
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Gross and Net Material Requirements
Plan (4 of 4)
• Using this data we can construct a plan that includes:
– Gross requirements
– On-hand inventory
– Net requirements
– Planned-order receipts
– Planned-order releases
• The net requirements plan is constructed like the gross
requirements plan
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Net Material Requirements Plan (1 of 7)
Figure 6.14(a) Net Material Requirements Plan for 50 Units of A
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Net Material Requirements Plan (2 of 7)
Figure 6.14(b) Net Material Requirements Plan for 50 Units of A
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Net Material Requirements Plan (3 of 7)
Figure 6.14(c) Net Material Requirements Plan for 50 Units of A
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Two or More End Products (1 of 2)
• Most manufacturing companies have more than one end
item
• A second product AA has
this material structure tree
• If we require 10 units of AA,
the gross requirements for
parts D and F are
Part D: 3 × number of AA’s = 3 × 10 = 30
Part F:
2 × number of A A’s = 2 × 10 = 20
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Two or More End Products (2 of 2)
• The lead time for AA is one week
• The gross requirement for AA is 10 units in week 6 and
there are no units on hand
• This new product can be added to the MRP process
– The addition of AA will only change the MRP
schedules for the parts contained in AA
• MRP can also schedule spare parts and components
– These have to be included as gross requirements
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Net Material Requirements Plan (4 of 7)
Figure 6.15a Net Material Requirements Plan, Including AA
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Net Material Requirements Plan (5 of 7)
Figure 6.15b Net Material Requirements Plan, Including AA
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Net Material Requirements Plan (6 of 7)
Figure 6.15c Net Material Requirements Plan, Including AA
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Net Material Requirements Plan (7 of 7)
Figure 6.15d Net Material Requirements Plan, Including AA
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Just-in-Time Inventory Control (1 of 2)
• Organizations have tried to have less in-process inventory
on hand to achieve greater efficiency in the production
process
• This is known as JIT inventory
– The inventory arrives just in time to be used during the
manufacturing process
• One technique of implementing JIT is a manual procedure
called kanban
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Just-in-Time Inventory Control (2 of 2)
• Kanban in Japanese means “card”
• Dual-card kanban system
– Conveyance, or C-kanban
– Production, or P-kanban
• Simple systems but require considerable discipline
• Little inventory to cover variability
• Schedule must be followed exactly
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The Kanban System
Figure 6.16 The Kanban System
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4 Steps of Kanban (1 of 2)
1. Full containers along with their
C-kanban card are taken from
the storage area to a user area,
typically on a manufacturing line
(arrow 1).
2. During the manufacturing process, parts in the container
are used up by the user. When the container is empty,
the empty container along with the same C-kanban card
is taken back to the storage area (arrow 2). Here the user
picks up a new full container, detaches the P-kanban
card from it, attaches his or her C-kanban card to it, and
returns with it to the user area.
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4 Steps of Kanban (2 of 2)
3. The detached P-kanban card
is then attached to an empty
container in the storage area
and then—and only then—is
the empty container taken back to the upstream producer
area (arrow 3).
4. This empty container is then refilled with parts and taken
with its P-kanban card back to the storage area (arrow 4).
This kanban process continuously cycles throughout the
day. Kanban is sometimes known as a “pull” production
system.
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Some Kanban Rules
• Minimum of two containers are required
• No containers are filled without the appropriate P-kanban
• Each container must hold exactly the specified number of
parts or items
• Only those parts that are needed are produced
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Enterprise Resource Planning (1 of 3)
•
MRP has evolved to include
– Labor hours
– Material cost
– Other resources related to production
– Refered to as MRP II and resource replaces the word
requirements
• Sophisticated software was developed, systems became
known as enterprise resource planning (ERP) systems
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Enterprise Resource Planning (2 of 3)
• Objective of ERP System is to reduce costs by integrating
all of the operations of a firm
– From supplier of materials needed through the
organization to invoicing the customer
– Data entered only once
– Central database quickly and easily accessed by
anyone in the organization
• Benefits include
– Reduced transaction costs
– Increased speed and accuracy of information
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Enterprise Resource Planning (3 of 3)
• Drawbacks to ERP
– Software is expensive to buy and costly to customize
– The implementation of an ERP system may require a
company to change its normal operations
– Employees are often resistant to change
– Training employees on the use of the new software
can be expensive
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Copyright
This work is protected by United States copyright laws
and is provided solely for the use of instructors in
teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including
on the World Wide Web) will destroy the integrity of the
work and is not permitted. The work and materials from
it should never be made available to students except by
instructors using the accompanying text in their
classes. All recipients of this work are expected to abide
by these restrictions and to honor the intended
pedagogical purposes and the needs of other
instructors who rely on these materials.
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Quantitative Analysis for Management
Thirteenth Edition
Chapter 7
Linear Programming Models:
Graphical and Computer
Methods
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Learning Objectives
After completing this chapter, students will be able to:
7.1 Identify the basic assumptions and properties of linear
programming (LP).
7.2 Formulate a linear programming problem algebraically.
7.3 Graphically solve any LP problem that has only two variables
by both the corner point and isoprofit line methods.
7.4 Use Excel spreadsheets to solve L P problems.
7.5 Understand the difference between minimization and
maximization objective functions.
7.6 Understand special issues in LP such as infeasibility,
unboundedness, redundancy, and alternative optimal
solutions.
7.7 Understand the role of sensitivity analysis.
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Chapter Outline
7.1 Requirements of a Linear Programming Problem
7.2 Formulating LP Problems
7.3 Graphical Solution to an LP Problem
7.4 Solving Flair Furniture’s LP Problem using Q M for
Windows, Excel 2016, and Excel QM
7.5 Solving Minimization Problems
7.6 Four Special Cases in LP
7.7 Sensitivity Analysis
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Introduction
• Many management decisions involve making the most
effective use of limited resources
• Linear programming (LP)
– Widely used mathematical modeling technique
– Planning and decision making relative to resource
allocation
• Broader field of mathematical programming
– Here programming refers to modeling and solving a
problem mathematically
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Requirements of a Linear Programming
Problem
• Four properties in common
– Seek to maximize or minimize some quantity (the
objective function)
– Restrictions or constraints are present
– Alternative courses of action are available
– Linear equations or inequalities
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LP Properties and Assumptions
Table 7.1 LP Properties and Assumptions
PROPERTIES OF LINEAR PROGRAMS
1. One objective function
2. One or more constraints
3. Alternative courses of action
4. Objective function and constraints are linear – proportionality
and divisibility
5. Certainty
6. Divisibility
7. Nonnegative variables
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Formulating LP Problems (1 of 2)
• Developing a mathematical model to represent the
managerial problem
• Steps in formulating a LP problem
1. Completely understand the managerial problem being
faced
2. Identify the objective and the constraints
3. Define the decision variables
4. Use the decision variables to write mathematical
expressions for the objective function and the
constraints
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Formulating LP Problems (2 of 2)
• Common LP application – product mix problem
• Two or more products are produced using limited
resources
• Maximize profit based on the profit contribution per unit of
each product
• Determine how many units of each product to produce
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Flair Furniture Company (1 of 6)
• Flair Furniture produces inexpensive tables and chairs
• Processes are similar, both require carpentry work and
painting and varnishing
– Each table takes 4 hours of carpentry and 2 hours of
painting and varnishing
– Each chair requires 3 of carpentry and 1 hour of
painting and varnishing
– There are 240 hours of carpentry time available and
100 hours of painting and varnishing
– Each table yields a profit of $70 and each chair a profit
of $50
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Flair Furniture Company (2 of 6)
• The company wants to determine the best combination of
tables and chairs to produce to reach the maximum profit
Table 7.2 Flair Furniture Company Data
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Flair Furniture Company (3 of 6)
• The objective is
Maximize profit
• The constraints are
1. The hours of carpentry time used cannot exceed 240
hours per week
2. The hours of painting and varnishing time used
cannot exceed 100 hours per week
• The decision variables are
T = number of tables to be produced per week
C = number of chairs to be produced per week
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Flair Furniture Company (4 of 6)
• Create objective function in terms of T and C
Maximize profit = $70T + $50C
• Develop mathematical relationships for the two constraints
– For carpentry, total time used is
(4 hours per table)(Number of tables produced) +
(3 hours per chair)(Number of chairs produced)
– First constraint is
Carpentry time used Carpentry time available
4T + 3C 240 ( hours of carpentry time )
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Flair Furniture Company (5 of 6)
• Similarly
Painting and varnishing time used
Painting and varnishing time available
2 T + 1C 100 (hours of painting and varnishing time)
This means that each table produced requires two
hours of painting and varnishing time
– Both of these constraints restrict production capacity
and affect total profit
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Flair Furniture Company (6 of 6)
• The values for T and C must be nonnegative
T 0 (number of tables produced is greater than or equal
to 0)
C 0 (number of chairs produced is greater than or equal
to 0)
The complete problem stated mathematically
Maximize profit = $70T + $50C
subject to
4T + 3C 240 (carpentry constraint)
2T + 1C 100
T, C 0
(painting and varnishing constraint)
(nonnegativity constraint)
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Graphical Solution to an LP Problem
• Easiest way to solve a small LP problems is graphically
• Only works when there are just two decision variables
– Not possible to plot a solution for more than two
variables
• Provides valuable insight into how other approaches work
• Nonnegativity constraints mean that we are always
working in the first (or northeast) quadrant of a graph
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Graphical Representation of
Constraints (1 of 11)
Figure 7.1 Quadrant Containing All Positive Values
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Graphical Representation of
Constraints (2 of 11)
• The first step is to identify a set or region of feasible
solutions
• Plot each constraint equation on a graph
• Graph the equality portion of the constraint equations
4T + 3C = 240
• Solve for the axis intercepts and draw the line
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Graphical Representation of
Constraints (3 of 11)
• When Flair produces no tables, the carpentry constraint
is:
4 ( 0 ) + 3C = 240
3C = 240
C = 80
• Similarly for no chairs:
4T + 3 ( 0 ) = 240
4T = 240
T = 60
– This line is shown on the following graph
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Graphical Representation of
Constraints (4 of 11)
Figure 7.2 Graph of Carpentry Constraint Equation
4T + 3C = 240
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Graphical Representation of
Constraints (5 of 11)
Figure 7.3 Region that
Satisfies the Carpentry
Constraint
• Any point on or below the
constraint plot will not
violate the restriction
• Any point above the plot
will violate the restriction
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Graphical Representation of
Constraints (6 of 11)
• The point (30, 20) lies below the line and satisfies the
constraint
4 ( 30 ) + 3 ( 20 ) = 180
• The point (70, 40) lies above the line and does not satisfy
the constraint
4 ( 70 ) + 3 ( 40 ) = 400
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Graphical Representation of
Constraints (7 of 11)
Figure 7.4 Region that Satisfies the Painting and Varnishing
Constraint
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Graphical Representation of
Constraints (8 of 11)
• To produce tables and chairs, both departments must be
used
• Find a solution that satisfies both constraints
simultaneously
• A new graph shows both constraint plots
• The feasible region is where all constraints are satisfied
– Any point inside this region is a feasible solution
– Any point outside the region is an infeasible solution
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Graphical Representation of
Constraints (9 of 11)
Figure 7.5 Feasible Solution Region for the Flair Furniture
Company Problem
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Graphical Representation of
Constraints (10 of 11)
• For the point (30, 20)
Constraint
Carpentry
constraint
Production time
4T + 3C 240 hours available
Feasibility
A checkmark.
( 4 )( 30 ) + ( 3 )( 20 ) = 180 hours used
4 T plus 3 C less than or equal to 240 hours available.
Left parenthesis 4 right parenthesis times left parenthesis 30 right parenthesis plus left parenthesis 3 right parenthesis times left parenthesis 20 right parenthesis equals 180 hours used.
A checkmark.
Painting
constraint
2T + 1 C 100 hours available
( 2 )( 30 ) + (1)( 20 ) = 80 hours used
2 T plus 1 C less than or equal to 100 hours available.
Left parenthesis 2 right parenthesis times left parenthesis 30 right parenthesis plus left parenthesis 1 right parenthesis times left parenthesis 20 right parenthesis equals 80 hours used.
• For the point (70, 40)
Constraint
Carpentry
constraint
Production time
4T + 3C 240 hours available
Feasibility
A checkmark.
( 4 )( 70 ) + ( 3 )( 40 ) = 400 hours used
4 T plus 3 C less than or equal to 240 hours available.
Left parenthesis 4 right parenthesis times left parenthesis 70 right parenthesis plus left parenthesis 3 right parenthesis times left parenthesis 40 right parenthesis equals 400 hours used.
A checkmark.
Painting
constraint
2T + 1 C 100 hours available
( 2 )( 70 ) + (1)( 40 ) = 180 hours used
2 T plus 1 C less than or equal to 100 hours available.
left parenthesis 2 right parenthesis times left parenthesis 70 right parenthesis plus left parenthesis 1 right parenthesis times left parenthesis 40 right parenthesis equals 180 hours used.
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Graphical Representation of
Constraints (11 of 11)
• For the point (50, 5)
Constraint
Production time
4T + 3C 240 hours available
4 T plus 3 C less than or equal to 240 hours available.
Feasibility
A checkmark.
4 times 50 plus 3 times 5 equals 215 hours used.
Carpentry
constraint
( 4 )( 50 ) + ( 3 )( 5 ) = 215 hours used
2T + 1 C 100 hours available
2 T plus 1 C less than or equal to 100 hours available.
A cross mark.
2 times 50 plus 1 time 5 equals 105 hours used.
Painting
constraint
( 2 )( 50 ) + (1)( 5 ) = 105 hours used
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Isoprofit Line Solution Method (1 of 7)
• Find the optimal solution from the many possible solutions
• Speediest method is to use the isoprofit line
• Starting with a small possible profit value, graph the
objective function
• Move the objective function line in the direction of
increasing profit while maintaining the slope
• The last point it touches in the feasible region is the
optimal solution
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Isoprofit Line Solution Method (2 of 7)
• Choose a profit of $2,100
• The objective function is
$2,100 = 70T + 50C
• Solving for the axis intercepts, draw the graph
• Obviously not the best possible solution
• Further graphs can be created using larger profits
– The further we move from the origin, the larger the
profit
( $4,100 ) will be generated when the
isoprofit line passes through the point ( 30, 40 )
• The highest profit
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Isoprofit Line Solution Method (3 of 7)
Figure 7.6 Profit line of $2,100 Plotted for the Flair Furniture Company
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Isoprofit Line Solution Method (4 of 7)
Figure 7.7 Four Isoprofit Lines Plotted for the Flair Furniture
Company
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Isoprofit Line Solution Method (5 of 7)
Figure 7.7 Four Isoprofit Lines Plotted for the Flair Furniture
Company
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Isoprofit Line Solution Method (6 of 7)
Figure 7.7 Four Isoprofit Lines Plotted for the Flair Furniture
Company
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Isoprofit Line Solution Method (7 of 7)
Figure 7.8 Optimal Solution to the Flair Furniture Problem
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Corner Point Solution Method (1 of 4)
• The corner point method for solving LP problems
• Look at the profit at every corner point of the feasible
region
• Mathematical theory is that an optimal solution must lie at
one of the corner points or extreme points
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Corner Point Solution Method (2 of 4)
Figure 7.9 Four Corner Points of the Feasible Region
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Corner Point Solution Method (3 of 4)
• Solve for the intersection of the two constraint lines
• Using the elimination method to solve simultaneous
equations method, select a variable to be eliminated
• Eliminate T by multiplying the second equation by −2
and add it to the first equation
−2 ( 2T + 1C = 100 ) = − 4T − 2C = − 200
4T + 3C =
240
−4T − 2C = − 200
C=
( carpentry )
(painting)
40
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Corner Point Solution Method (4 of 4)
• Substitute C = 40 into either equation to solve for T
4T + 3 ( 40 ) = 240
4T + 120 = 240
4T = 120
T = 30
Thus the corner point is ( 30, 40 )
Table 7.3 Feasible Corner Points and Profits for Flair Furniture
Number of Tables (T)
Number of Chairs (C)
Profit = $70T + $50C
Profit equals 70 dollars T plus 50 dollars C.
0
0
$0
50
0
0
80
30
40
$3,500
$4,000
$4,100
0 dollars.
3,500 dollars.
4,000 dollars.
4,100 dollars.
Highest profit – Optimal Solution
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Slack and Surplus (1 of 3)
• Slack is the amount of a resource that is not used
– For a less-than-or-equal constraint
Slack = ( Amount of resource available )
− ( Amount of resource used )
– Flair decides to produce 20 tables and 25 chairs
4 ( 20 ) + 3 ( 25 ) = 155
(carpentry time used)
240 =
(carpentry time available)
240 − 155 = 85
(Slack time in carpentry)
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Slack and Surplus (2 of 3)
At the optimal solution, slack is 0 as all 240 hours are used
• Slack is the amount of a resource that is not used
– For a less-than-or-equal constraint
Slack = ( Amount of resource available )
− ( Amount of resource used )
– Flair decides to produce 20 tables and 25 chairs
4 ( 20 ) + 3 ( 25 ) = 155
(carpentry time used)
240 =
(carpentry time available)
240 − 155 = 85
(Slack time in carpentry)
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Slack and Surplus (3 of 3)
• Surplus is used with a greater-than-or-equal-to constraint
to indicate the amount by which the right-hand side of the
constraint is exceeded
Surplus = ( Actual amount ) − (Minimum amount )
– New constraint
T + C 42
– If T = 20 and C = 25, then
20 + 25 = 45
Surplus = 45 − 42 = 3
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Summaries of Graphical Solution
Methods
Table 7.4 Summaries of Graphical Solution Methods
Isoprofit Method
1. Graph all constraints and find the feasible region.
2. Select a specific profit (or cost) line and graph it to find the slope.
3. Move the objective function line in the direction of increasing profit (or decreasing cost) while
maintaining the slope. The last point it touches in the feasible region is the optimal solution.
4. Find the values of the decision variables at this last point and compute the profit (or cost).
Corner Point Method
1. Graph all constraints and find the feasible region.
2. Find the corner points of the feasible reason.
3. Compute the profit (or cost) at each of the feasible corner points.
4. Select the corner point with the best value of the objective function found in Step 3. This is the
optimal solution.
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Solving Flair Furniture’s LP Problem
• Most organizations have access to software to solve big
LP problems
• There are differences between software
implementations, the approach is basically the same
• With experience with computerized LP algorithms, it is
easy to adjust to minor changes
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Using QM for Windows (1 of 4)
• Select the Linear Programming module
• Specify the number of constraints (non-negativity is
assumed)
• Specify the number of decision variables
• Specify whether the objective is to be maximized or
minimized
• For Flair Furniture there are two constraints, two decision
variables, and the objective is to maximize profit
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Using QM for Windows (2 of 4)
Program 7.1A QM for Windows Linear Programming Input
Screen
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Using QM for Windows (3 of 4)
Program 7.1B QM for Windows Data Input for Flair
Furniture Problem
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Using QM for Windows (4 of 4)
Program 7.1C QM for Windows Output and Graph for Flair
Furniture Problem
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Using Excel’s Solver
• The Solver tool in Excel can be used to find solutions to
– LP problems
– Integer programming problems
– Noninteger programming problems
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Using Solver (1 of 10)
• Recall the model for Flair Furniture is
Maximize profit = $70T + $50C
Subject to
4T + 3C 240
2T + 1C 100
• To use Solver, it is necessary to enter data and formulas
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Using Solver (2 of 10)
1. Enter problem data
– Variable names, coefficients for the objective function
and constraints, RHS values for each constraint
2. Designate specific cells for the values of the decision
variables
3. Write a formula to calculate the value of the objective
function
4. Write a formula to compute the left-hand sides of each of
the constraints
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Using Solver (3 of 10)
Program 7.2A Excel Data Input for Flair Furniture Example
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Using Solver (4 of 10)
Program 7.2B Formulas for Flair Furniture Example
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Using Solver (5 of 10)
Program 7.2C Excel Spreadsheet for Flair Furniture
Example
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Using Solver (6 of 10)
Program 7.2D Starting Solver in Excel 2016
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Using Solver (7 of 10)
Program 7.2E Solver Parameters Dialog Box
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Using Solver (8 of 10)
Program 7.2F Solver Add Constraint Dialog Box
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Using Solver (9 of 10)
Program 7.2G Solver Results Dialog Box
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Using Solver (10 of 10)
Program 7.2H Flair Furniture Solution Found by Solver
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Using Excel QM (1 of 3)
Program 7.3A Using Excel QM in Excel 2016 for the Flair
Furniture Example
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Using Excel QM (2 of 3)
Program 7.3B Excel QM Data Input for the Flair Furniture
Example
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Using Excel QM (3 of 3)
Program 7.3C Excel QM Output for the Flair Furniture
Example
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Solving Minimization Problems
• Many LP problems involve minimizing an objective such
as cost
• Minimization problems can be solved graphically
– Set up the feasible solution region
– Use either the corner point method or an isocost line
approach
– Find the values of the decision variables (e.g., X1
and X 2 ) that yield the minimum cost
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Holiday Meal Turkey Ranch (1 of 10)
• The Holiday Meal Turkey Ranch is considering buying two
different brands of turkey feed and blending them to
provide a good, low-cost diet for its turkeys
Table 7.5 Holiday Meal Turkey Ranch data
Blank
COMPOSITION
OF EACH POUND
OF FEED (OZ.)
Blank
Blank
INGREDIENT
BRAND 1 FEED
BRAND 2 FEED
MINIMUM MONTHLY
REQUIREMENT PER
TURKEY (OZ.)
A
5
10
90
B
4
3
48
C
0.5
0
1.5
Blank
Cost per pound
2 cents
3 cents
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Holiday Meal Turkey Ranch (2 of 10)
• Let
X1 = number of pounds of brand 1 feed purchased
X 2 = number of pounds of brand 2 feed purchased
Minimize cost ( in cents ) = 2 X1 + 3 X 2
subject to:
5 X1 + 10 X 2 90 ounces
(ingredient A constraint)
4 X1 + 3 X 2 48 ounces
(ingredient B constraint)
0.5 X1 1.5 ounces
(ingredient C constraint)
X1 0
(nonnegativity constraint)
X2 0
(nonnegativity constraint)
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Holiday Meal Turkey Ranch (3 of 10)
Figure 7.10 Feasible Region for the Holiday Meal Turkey
Ranch Problem
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Holiday Meal Turkey Ranch (4 of 10)
• Solve for the values of the three corner points
– Point a is the intersection of ingredient constraints C
and B
4 X1 + 3 X 2 = 48
X1 = 3
– Substituting 3 in the first equation, we find X 2 = 12
– Solving for point b we find X1 = 8.4 and X 2 = 4.8
– Solving for point c we find X1 = 18 and X 2 = 0
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Holiday Meal Turkey Ranch (5 of 10)
• Substituting these values back into the objective function
we find
Cost = 2 X1 + 3 X 2
Cost at point a = 2 ( 3 ) + 3 (12 ) = 42
Cost at point b = 2 ( 8.4 ) + 3 ( 4.8 ) = 31.2
Cost at point c = 2 (18 ) + 3 ( 0 ) = 36
– The lowest cost solution is to purchase 8.4 pounds of
brand 1 feed and 4.8 pounds of brand 2 feed for a
total cost of 31.2 cents per turkey
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Holiday Meal Turkey Ranch (6 of 10)
• Solving using an isocost line
• Move the isocost line toward the lower left
• The last point touched in the feasible region will be the
optimal solution
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Holiday Meal Turkey Ranch (7 of 10)
Figure 7.11 Graphical Solution to the Holiday Meal Turkey
Ranch Problem Using the Isocost Line
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Holiday Meal Turkey Ranch (8 of 10)
Program 7.4 Solution to the Holiday Meal Turkey Ranch
Problem in QM for Windows
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Holiday Meal Turkey Ranch (9 of 10)
Program 7.5A Excel 2016 Solution for Holiday Meal Turkey
Ranch Problem
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Holiday Meal Turkey Ranch (10 of 10)
Program 7.5B Excel 2016 Formulas for Holiday Meal
Turkey Ranch Problem
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Four Special Cases in LP
• Four special cases and difficulties arise at times when
using the graphical approach
1. No feasible solution
2. Unboundedness
3. Redundancy
4. Alternate Optimal Solutions
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No Feasible Solution (1 of 2)
• No solution to the problem that satisfies all the constraint
equations
• No feasible solution region exists
• A common occurrence in the real world
• Generally one or more constraints are relaxed until a
solution is found
• Consider the following three constraints
X1 + 2 X 2 6
2X 1 + X 2 8
X1 7
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No Feasible Solution (2 of 2)
X1 + 2 X 2 6
2X 1 + X 2 8
X1 7
Figure 7.12 A problem
with no feasible solution
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Unboundedness (1 of 2)
•
Sometimes a linear program will not have a finite solution
•
In a maximization problem
– One or more solution variables, and the profit, can be
made infinitely large without violating any constraints
•
In a graphical solution, the feasible region will be open
ended
•
Usually means the problem has been formulated
improperly
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Unboundedness (2 of 2)
Maximize profit =
Blank
+ $5X 2
$3X 1
3 dollars X sub 1.
Plus 5 dollars X sub 2.
blank
subject to
X1
X sub 1.
blank
blank
X2
X sub 2.
blank
X1
X sub 1.
blank
5
Greater than or equal to 5.
+ 2X 2
Plus 2 X sub 2.
blank
X1 , X 2
X sub 1, X sub 2.
10
Less than or equal to 10.
10
Greater than or equal to 10.
0
Greater than or equal to 0.
Figure 7.13 A Feasible
Region That Is
Unbounded to the Right
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Redundancy (1 of 2)
• A redundant constraint is one that does not affect the
feasible solution region
• One or more constraints may be binding
• This is a very common occurrence in the real world
• Causes no particular problems, but eliminating redundant
constraints simplifies the model
Maximize profit =
subject to
blank
$1X 1
+ $2X 2
X1
+ X2
2X1
+ X2
1 Dollar X Sub 1.
X sub 1.
2 X sub 1.
blank
X1
X sub 1.
Plus 2 dollars X sub 2.
Plus X sub 2.
20
Less than or equal to 20.
30
Plus X sub 2.
Less than or equal to 30.
blank
Less than or equal to 25.
blank
blank
Blank
X1 , X 2
X sub 1, X sub 2.
25
0
Greater than or equal to 0.
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Redundancy (2 of 2)
Maximize profit =
$1X 1
+ $2X 2
X1
+ X2
Less than or equal to 20.
2X1
+ X2
Plus X sub 2.
Less than or equal to 30.
blank
Less than or equal to 25.
1 dollar X sub 1.
subject to
X sub 1.
Plus 2 dollars X sub 2.
Plus X sub 2.
blank
2 X sub 1.
blank
X1
X sub 1.
blank
blank
X1 , X 2
X sub 1, X sub 2.
Blank
20
30
25
0
Greater than or equal to 0.
Figure 7.14 Problem
with a Redundant
Constraint
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Alternate Optimal Solutions (1 of 2)
• Occasionally two or more optimal solutions may exist
• Graphically this occurs when the objective function’s
isoprofit or isocost line runs perfectly parallel to one of the
constraints
• Allows management great flexibility in deciding which
combination to select as the profit is the same at each
alternate solution
Maximize profit =
subject to
$3X 1
+ $2X 2
6X1
+ 4 X2
3 dollars X sub 1.
6 X sub 1.
Plus 2 dollars X sub 2.
Plus 4 X sub 2.
blank
X1
X sub 1.
blank
blank
blank
X1 , X 2
X sub 1, X sub 2.
Blank
24
Less than or equal to 24.
3
Less than or equal to 3.
0
Greater than or equal to 0.
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Alternate Optimal Solutions (2 of 2)
Maximize profit =
$3X 1
+ $2X 2
subject to
6X1
+ 4 X2
Plus 4 X sub 2.
Less than or equal to 24.
blank
Less than or equal to 3.
3 dollars X sub 1.
6 X sub 1.
Plus 2 dollars X sub 2.
Blank
24
blank
X1
X sub 1.
3
blank
blank
X1 , X 2
X sub 1, X sub 2.
0
Greater than or equal to 0.
Figure 7.15 Example of
Alternate Optimal
Solutions
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Sensitivity Analysis (1 of 2)
• Optimal solutions to LP problems thus far have been
found under deterministic assumptions
– We assume complete certainty in the data and
relationships of a problem
• Real world conditions are dynamic
• Analyze how sensitive a deterministic solution is to
changes in the assumptions of the model
• This is called sensitivity analysis, postoptimality
analysis, parametric programming, or optimality analysis
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Sensitivity Analysis (2 of 2)
• Involves a series of what-if? questions concerning
constraints, variable coefficients, and the objective
function
• Trial-and-error method
– Values are changed and the entire model is resolved
• Preferred way is to use an analytic postoptimality analysis
– After a problem has been solved, we determine a
range of changes in problem parameters that will not
affect the optimal solution or change the variables in
the solution
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High Note Sound Company (1 of 3)
• The company manufactures quality speakers and stereo
receivers
• Products require a certain amount of skilled artisanship
which is in limited supply
• Product mix LP model
Maximize profit =
$50X 1
50 dollars X sub 1.
2 X sub 1.
subject to
blank
2X1
3 X sub 1.
3X1
blank
blank
+ $120X 2
Plus 120 dollars X sub 2.
Plus 4 X sub 2.
+ 4 X2
Plus 1 X sub 2.
+ 1X 2
X sub 1, X sub 2.
X1 , X 2
blank
blank
80
(hours of electricians’
time available)
60
(hours of audio
technicians’ time
available)
Less than or equal to 80.
Less than or equal to 60.
0
Greater than or equal to 0.
blank
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High Note Sound Company (2 of 3)
Figure 7.16 High Note Sound Company Graphical Solution
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High Note Sound Company (3 of 3)
• Electrician hours used are
2X1 + 4X2 = 2 ( 0 ) + 4 ( 20 ) = 80
– All hours are utilized so slack = 0
– Additional units of a binding constraint will generally
increase profits
• Technician hours used are
3X1 + 1X2 = 3 ( 0 ) + 1( 20 ) = 20
– Available hours = 60 so slack = 60 − 20 = 40
– Additional units of a nonbinding constraint will only
increase slack
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Changes in the Objective Function
Coefficient (1 of 2)
• Contribution rates in the objective functions fluctuate
– The feasible solution region remains exactly the same
– The slope of the isoprofit or isocost line changes
• Modest increases or decreases in objective function
coefficients may not change the current optimal corner
point
• Know how much an objective function coefficient can
change before the optimal solution would be at a different
corner point
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Changes in the Objective Function
Coefficient (2 of 2)
Figure 7.17 Changes in the Receiver Contribution
Coefficients
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QM for Windows (1 of 2)
Program 7.6A Input to QM for Windows High Note Sound
Company Data
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QM for Windows (2 of 2)
Program 7.6B High Note Sound Company Sensitivity
Analysis Output
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Excel Solver (1 of 3)
Program 7.7A Excel 2016 Spreadsheet for High Note
Sound Company
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Excel Solver (2 of 3)
Program 7.7B Excel 2016 Solution and Solver Results
Window for High Note Sound Company
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Excel Solver (3 of 3)
Program 7.7C Excel 2016 Sensitivity Report for High Note
Sound Company
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Changes in the Technological
Coefficients (1 of 2)
• Changes in the technological coefficients often reflect
changes in the state of technology
• If the amount of resources needed to produce a product
changes, coefficients in the constraint equations will
change
• Objective function does not change
• May produce significant change in the shape of the
feasible region
• May cause a change in the optimal solution
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Changes in the Technological
Coefficients (2 of 2)
Figure 7.18 Change in the Technological Coefficients for the
High Note Sound Company
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Changes in Resources or RightHand-Side Values (1 of 3)
•
Right-hand-side values of the constraints often represent
resources available to the firm
• Additional resources may lead to higher total profit
• Sensitivity analysis about resources helps answer
questions about
– How much should be paid for additional resources
– How much more of a resource would be useful
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Changes in Resources or RightHand-Side Values (2 of 3)
• Changing the RHS will change the feasible region, unless
the constraint is redundant
• Often changes the optimal solution
• The dual price or dual value
– The amount of change in the objective function value
that results from a unit change in one of the resources
– The dual price for a constraint is the improvement in
the objective function value that results from a one-unit
increase in the right-hand side of the constraint
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Changes in Resources or RightHand-Side Values (3 of 3)
• The amount of possible increase in the RHS is limited
• If the RHS is increased beyond the upper bound, then the
objective function would no longer increase by the dual
price
• There would be excess (slack) resources or the objective
function may change by an amount different from the dual
price
• The dual price is relevant only within limits
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Changes in the Electricians’ Time
Resource (1 of 3)
Figure 7.19 Changes in the Electricians’ Time Resource for
the High Note Sound Company
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Changes in the Electricians’ Time
Resource (2 of 3)
Figure 7.19 Changes in the Electricians’ Time Resource for
the High Note Sound Company
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Changes in the Electricians’ Time
Resource (3 of 3)
Figure 7.19 Changes in the Electricians’ Time Resource for
the High Note Sound Company
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QM for Windows
Program 7.6B High Note Sound Company Sensitivity
Analysis Output
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Excel Solver
Program 7.7C Excel 2016 Sensitivity Report for High Note
Sound Company
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Copyright
This work is protected by United States copyright laws
and is provided solely for the use of instructors in
teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including
on the World Wide Web) will destroy the integrity of the
work and is not permitted. The work and materials from
it should never be made available to students except by
instructors using the accompanying text in their
classes. All recipients of this work are expected to abide
by these restrictions and to honor the intended
pedagogical purposes and the needs of other
instructors who rely on these materials.
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Quantitative Analysis for Management
Thirteenth Edition
Chapter 8
Linear Programming Applications
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Learning Objectives
After completing this chapter, students will be able to:
8.1 Formulate and solve LP problems with Excel Solver in
marketing.
8.2 Formulate and solve LP problems with Excel Solver in
production.
8.3 Formulate and solve LP problems with Excel Solver in
the scheduling of employees.
8.4 Formulate and solve LP problems with Excel Solver in
finance.
8.5 Formulate and solve LP problems with Excel Solver in
the blending of ingredients.
8.6 Formulate and solve LP problems with Excel Solver in
revenue management.
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Chapter Outline
8.1 Marketing Applications
8.2 Manufacturing Applications
8.3 Employee Scheduling Applications
8.4 Financial Applications
8.5 Ingredient Blending Applications
8.6 Other Linear Programming Applications
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Introduction
• The graphical method of LP is useful for understanding
how to formulate and solve small LP problems
• Many types of problems can be solved using LP
• Principles developed here are applicable to larger
problems
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Marketing Applications
• Linear programming models have been used in the
advertising field as a decision aid in selecting an effective
media mix
• Media selection LP problems can be approached from two
perspectives
– Maximize audience exposure
– Minimize advertising costs
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Win Big Gambling Club (1 of 4)
• Club promotes gambling junkets to the Bahamas
– $8,000 per week to spend on advertising
– Goal is to reach the largest possible high-potential
audience
– Media types and audience figures shown below
– Place at least five radio spots per week
– No more than $1,800 can be spent on radio advertising
each week
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Win Big Gambling Club (2 of 4)
• Advertising options
AUDIENCE
REACHED PER
AD
COST PER
AD ($)
MAXIMUM ADS
PER WEEK
T V spot (1 minute)
5,000
800
12
Daily newspaper (fullpage ad)
8,500
925
5
Radio spot (30
seconds, prime time)
2,400
290
25
Radio spot (1 minute,
afternoon)
2,800
380
20
MEDIUM
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Win Big Gambling Club (3 of 4)
•
Problem formulation
X1 = number of 1-minute T V spots taken each week
X 2 = number of daily newspaper ads taken each week
X 3 = number of 30-second prime-time radio spots taken each week
X 4 = number of 1-minute afternoon radio spots taken each week
Objective:
Maximize audience coverage = 5,000 X1 + 8,500 X 2 + 2,400 X 3 + 2,800 X 4
Subject to
X1 12
(max T V spots/wk)
X2 5
(max newspaper ads/wk)
(max 30-sec radio spots/wk)
X 3 25
(max 1-min radio spots/wk)
X 4 20
800 X1 + 925 X 2 + 290 X 3 + 380 X 4 $8,000
X3 + X 4 5
(weekly advertising budget)
(min radio spots contracted)
290 X 3 + 380 X 4 $1,800 (max dollars spent on radio)
X1 , X 2 , X 3 , X 4 0
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Win Big Gambling Club (4 of 4)
Solution
X1 =1.97 T V spots
X2 = 5
newspaper ads
X 3 = 6.2 30-second radio spots
X4 = 0
1-minute radio spots
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Solution in Excel 2016 (1 of 19)
Program 8.1 Win Big Solution in Excel 2016
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Solution in Excel 2016 (2 of 19)
Program 8.1 Win Big Solution in Excel 2016
blank
Solver Parameter Inputs and Selections
Key Formulas
blank
A screenshot shows the entry Total audience in cell F 5 and a key formula entered in cell F 6: equals SUM PRODUCT, left parenthesis, dollar sign B dollar sign 5 colon dollar sign E dollar sign 5 comma B 6 colon E 6, right parenthesis.
Set Objective: F6
By Changing cells: B5:E5
To: Max
Subject to the Constraints:
Copy F6 to F9:F15
F9 :F14 < =H9 :H14
F15 > =H15
F 9 colon F 14 less than or equal to H 9 colon H 14.
F 15 greater than or equal to H 15.
Solving Method: Simplex LP
A checked box.
Make Variables Non-Negative
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Management Sciences Association (1 of 5)
• MSA is a marketing research firm
• Several requirements for a statistical validity
1. Survey at least 2,300 U.S. households
2. Survey at least 1,000 households whose heads are
30 years old
3. Survey at least 600 households whose heads are
between 31 and 50
4. Ensure that at least 15% of those surveyed live in a
state that borders Mexico
5. Ensure that no more than 20% of those surveyed who
are 51 years of age or over live in a state that borders
Mexico
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Management Sciences Association (2 of 5)
• MSA decides to conduct all surveys in person
• Estimates of the costs of reaching people in each age and
region category
• Goal is to meet the sampling requirements at the least
possible cost
COST PER PERSON SURVEYED ($)
AGE 30
Age less than or equal to 30
AGE 31- 50
AGE 51
State bordering Mexico
$7.50
$6.80
$5.50
State not bordering Mexico
$6.90
$7.25
$6.10
REGION
Age 31 to 50
Age greater than or equal to 51
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Management Sciences Association (3 of 5)
• Decision variables
X1 = number of 30 or younger and in a border state
X 2 = number of 31-50 and in a border state
X 3 = number 51 or older and in a border state
X 4 = number 30 or younger and not in a border state
X 5 = number of 31-50 and not in a border state
X 6 = number 51 or older and not in a border state
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Management Sciences Association (4 of 5)
Objective function
Minimize total interview costs = $7.50 X1 + $6.80 X 2 + $5.50 X 3
+ $6.90 X 4 + $7.25 X 5 +$6.10 X 6
subject to
X1 + X 2 + X 3 + X 4 + X 5 + X 6 2,300
X4
1,000
(households 30 or younger)
X2 + X5
600
(households 31-50)
X2 + X3
0.15( X1 + X 2 + X 3 + X 4 + X 5 + X 6 ) (border states)
X3
0.20( X 3 + X 6 )
X1 +
X1 +
(total households)
(limit on age group 51+ who
can live in border state)
X1 , X 2 , X 3 , X 4 , X 5 , X 6 0
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Management Sciences Association (5 of 5)
• Optimal solution will cost $15,166
REGION
State bordering Mexico
State not bordering Mexico
AGE 30
AGE 31- 50
AGE 51
0
600
140
1,000
0
560
Age less than or equal to 30.
Age 31 to 50.
Age greater than or equal to 51.
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Solution in Excel 2016 (3 of 19)
Program 8.2 MSA Solution in Excel 2016
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Solution in Excel 2016 (4 of 19)
Program 8.2 MSA Solution in Excel 2016
blank
Solver Parameter Inputs and
Selections
Key Formulas
blank
Set Objective: H5
By Changing cells: B4:G4
To: Min
Subject to the Constraints:
A screenshot shows the entry Total cost in cell H 4 and a key formula entered in cell H 5: equals SUM PRODUCT, left parenthesis, dollar sign B dollar sign 4 colon dollar sign G dollar sign 4 comma B 5 colon G 5, right parenthesis.
Copy H5 to H8:H12
H8 : H11< = J8 : J11
H 8 colon H 11 less than or equal to J 8 colon J 11.
H12 > = J12
H 12 greater than or equal to J 12.
Solving Method: Simplex LP
Make Variables Non-Negative
A checked box.
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Manufacturing Applications (1 of 2)
• Production Mix
– LP can be used to plan the optimal mix of products to
manufacture
– Company must meet a myriad of constraints
▪ Financial concerns
▪ Sales demand
▪ Material contracts
▪ Union labor demands
– Primary goal is to generate the largest profit possible
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Fifth Avenue Industries (1 of 6)
• Produces four varieties of ties
– Expensive all-silk
– All-polyester
– Two are polyester-cotton or silk-cotton blends
• Cost and availability of the three materials used in the
production process
MATERIAL
Silk
Polyester
Cotton
COST PER YARD
($)
24
6
9
MATERIAL AVAILABLE
PER MONTH (YARDS)
1,200
3,000
1,600
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Fifth Avenue Industries (2 of 6)
• The firm has contracts with several major department store
chains
– Contracts require a minimum number of ties
– May be increased if demand increases
• Goal is to maximize monthly profit
• Decision variables
X1 = number of all-silk ties produced per month
X 2 = number all-polyester ties
X 3 = number of blend 1 polyester-cotton ties
X 4 = number of blend 2 silk-cotton ties
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Fifth Avenue Industries (3 of 6)
Table 8.1 Data for Fifth Avenue Industries
VARIETY
OF TIE
MONTHL
SELLIN
Y
MONTH
G PRICE CONTRA
LY
PER TIE
CT
DEMAN
($)
MINIMUM
D
MATERIAL
REQUIRED
PER TIE
(YARDS)
MATERIAL
REQUIREMENTS
All silk
19.24
5,000
7,000
0.125
100% silk
All
polyester
8.70
10,000
14,000
0.08
100% polyester
Poly-cotton
blend 1
9.52
13,000
16,000
0.10
50% polyester – 50%
cotton
Silk-cotton
blend 2
10.64
5,000
8,500
0.11
60% silk – 40% cotton
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Fifth Avenue Industries (4 of 6)
• Establish profit per tie
PROFIT
POLYESTER
COTTON
MATERIAL SELLING
COST
REQ’D
COST REQ’D COST
COST
PRICE
SILK REQ’D
All-silk X1
blank
blank
blank
blank
blank
blank
blank
blank
blank
blank
blank
blank
$3.00
$19.24
$16.24
blank
blank
blank
$0.48
$8.70
$8.22
blank
blank
blank
$0.75
$9.52
$8.77
blank
blank
blank
$1.98
$10.64
$8.66
X sub 1.
$24.00
0.125
All-polyester X2
blank
blank
blank
0.08
$6
blank
blank
blank
blank
Blank
blank
X sub 2.
blank
blank
Poly-cotton blend X3
blank
blank
blank
0.05
$9
X sub 3.
blank
blank
0.05
$6
Silk-cotton blend
blank
blank
blank
blank
blank
blank
blank
0.044
$9
X4
X sub 4.
0.066
$24.00
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Fifth Avenue Industries (5 of 6)
Objective function
Maximize profit = $16.24 X1 + $8.22 X 2 + $8.77 X 3 + $8.66 X 4
Subject to 0.125 X1 + 0.066 X 4 1200
(yds of silk)
0.08 X 2 + 0.05 X 3 3,000
(yds of polyester)
0.05 X 3 + 0.44 X 4 1,600
(yds of cotton)
X1 5,000
(contract min for silk)
X1 7,000
(contract min)
X 2 10,000 (contract min for all polyester)
X 2 14,000 (contract max)
X 3 13,000 (contract min for blend 1)
X 3 16,000 (contract max)
X 4 5,000
(contract min for blend 2)
X 4 8,500
(contract max)
X1, X 2 , X 3 , X 4 0
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Fifth Avenue Industries (6 of 6)
• Optimal solution will result in a profit of $412,028 per
month
TIE
QUANTITY PER MONTH
All-Silk
5,112
All-Polyester
14,000
Poly-Silk
16,000
Silk-Cotton
8,500
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Solution in Excel 2016 (5 of 19)
Program 8.3 Fifth Avenue Solution in Excel 2016
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Solution in Excel 2016 (6 of 19)
Program 8.3 Fifth Avenue Solution in Excel 2016
blank
Solver Parameter Inputs and Selections
Key Formulas
blank
A screenshot shows the entry Total profit in cell F 5 and a key formula entered in cell F 6: equals SUM PRODUCT, left parenthesis, dollar sign B dollar sign 5 colon dollar sign E dollar sign 5 comma B 6 colon E 6, right parenthesis.
Set Objective: F6
By Changing cells: B5:E5
To: Max
Copy F6 to F9:F19
Subject to the Constraints:
F9 : F15 < = H9 :H15
F 9 colon F 15 less than or equal to H 9 colon H 15.
F16 : F19 > = H16 :H19
Solving Method: Simplex L P
F 16 colon F 19 greater than or equal to H 16 colon H 19.
A checked box.
Make Variables Non-Negative
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Manufacturing Applications (2 of 2)
• Production Scheduling
– Low-cost production schedule
▪ Period of weeks or months
– Important factors include
▪ Labor capacity
▪ Inventory and storage costs
▪ Space limitations
▪ Product demand
▪ Labor relations
– With more than one product, the scheduling process
can be quite complex
– The problem resembles the product mix model for each
time period in the future
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (1 of 12)
• Manufactures two different electric motors for sale under
contract to Drexel Corp
– Orders placed three times a year for four months at a
time
– Demand varies month to month
– Develop a production plan for the next four months
Table 8.2 Four-Month Order Schedule for Electrical Motors
MODEL
JANUARY
FEBRUARY
MARCH
APRIL
GM3A
800
700
1,000
1,100
GM3B
1,000
1,200
1,400
1,400
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (2 of 12)
• Production planning must consider four factors
1. Produce the required number of motors each month
and ensure the desired ending inventory
2. Desire to keep inventory carrying costs down
3. No-lay-off policy, minimize fluctuations in production
levels
4. Warehouse limitations
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (3 of 12)
• Basic data
MOTOR
ENDING INV
CARRYING
COST
LABOR HRS
REQ’D
PRODUCTION COST
PER UNIT
GM3A
450
$0.36
1.3
$20
GM3B
300
$0.26
0.9
$15
2,240 Desired labor hrs per month 2,560
Maximum total inventory space available = 3,300 units
Labor cost increases 10% March 1
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (4 of 12)
Model formulation
Objective
Minimize total cost (production plus inventory carrying cost)
Constraints
4 demand constraints (1 constraint for each of 4 months) for GM3A
4 demand constraints (1 constraint for each of 4 months) for GM3B
2 constraints (1 for GM3A and 1 for GM3B) for the inventory at the
end of April
4 constraints for minimum labor hours (1 constraint for each month)
4 constraints for maximum labor hours (1 constraint for each
month)
4 constraints for inventory storage capacity each month
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (5 of 12)
• Objective function – production costs
Number of model GM3A motors produced in month i
Ai = (i = 1, 2, 3, 4 for January – April)
Bi = Number of model GM3B motors produced in month I
A sub i equals.
B sub i equals.
Cost of production =$20 A1 + $20 A2 +$22 A3 + $22 A4
+ $15B1 + $15B2 + $16.50B3 + $16.50B4
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (6 of 12)
• Objective function – inventory carrying costs
Iai =
Units of GM3A left in inventory at the end of month i (i =
1, 2, 3, 4 for January – April)
Ibi =
Units of GM3B left in inventory at the end of month i (i =
1, 2, 3, 4 for January – April)
I a sub i equals.
I b sub i equals.
Cost of carrying inventory = $0.36IA1 + $0.36IA2 + $0.36IA3
+0.36IA4 + $0.26IB1 + $0.26IB2 + $0.26IB3 + $0.26IB4
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (7 of 12)
• Complete objective function
Minimize costs = $20 A1 + $20 A2 + $22 A3 + $22 A4
+ $15B1 + $15B2 + $16.50B3 +
$16.50B4 + $0.36IA1 + $0.36IA2 +
$0.36IA3 + 0.36IA4 + $0.26IB1 +
$0.26IB2 + $0.26IB3 + $0.26IB4
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (8 of 12)
• End of month inventory is calculated using
Inventory at Inventory at Current Sales to
the
end
of
=
the
end
of
+
month’s
−
Drexel
this
this month last month production month
• Rearranged to create a standard format for a constraint
equation
Inventory at Current Inventory at Sales to
the
end
of
+
month’s
−
the
end
of
=
Drexel
this
last month production this month month
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved
Greenberg Motors (9 of 12)
• The demand constraints
A1 − IA1 = 800
(demand for GM3A in Jan)
IA1 + A2 − IA2 = 700
(demand for GM3A in Feb)
IA2 + A3 − IA3 = 1,000 (demand for GM3A in Mar)
IA3 + A4 − IA4 = 1,000 (demand for GM3A in Apr)
B1 − IB1 = 1,000 (demand for GM3B in Jan)
IB1 + B2 − IB2 = 1,200 (demand for GM3B in Feb)
IB2 + B3 − IB3 = 1,400 (demand for GM3B in…
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