BUSINESS ANALYTICS
LP Formulation - Solutions
Learning Objectives:
 Develop an understanding of how LP problems are formulated.
 Ability to recognize situations in which LP might be used effectively.
From a managerial perspective, LP formulation is the key to understanding LP applications. This
module will help you develop (or improve) your understanding of linear programming. We will
cover a variety of problems for all areas of business with varying levels of difficulty. It is
important that you thoroughly understand the problems.
In formulating a LP, answer the following three questions:
1. What does the model seek to determine? These variables can be controlled and are called
decision variables.
2. What is the goal that needs to be achieved? This is called the objective function. If larger
values of a measure of performance are more desirable (such as profit) then we want to
maximize the measure. On the other hand, if lower values of a measure of performance
are more desirable (such as cost) then we want to minimize the measure.
3. What limitations must be imposed on the variables to satisfy the problem? These are
called constraints.
Recommendation:
 Review the problems done in class.
 Read the problems in this module carefully before attempting to formulate.
An important class of problems faced by many companies is called the product-mix problems.
Broadly speaking, the problem is to determine the optimal mix of products to make, given the
availability and usage of resources. Solutions to these problems can provide useful insights to the
decision maker. Several problems in this module deal with this type of problems.
Problem 1
Par, Inc., a small manufacturing of golf-equipment and supplies whose management has decided to move
into the market for medium and high priced golf bags. Par’s distributor is enthusiastic about the new
product line and has agreed to buy all the golf bags Par produces over the next three months.
Production requirements (hrs)
Cutting
Product
Dyeing
7/10
Std Bag
1
Deluxe Bag
630
Total time
available
Sewing
Finishing
1/2
5/6
600
1
2/3
708
Inspection
Packaging
1/10
1/4
135
Profit
10
9
Problem 1 - Solution
Decision variables:
X1: number of standard bags to make
X2: number of deluxe bags to make
Formulation:
Max
10 X1 + 9 X2
st
7/10 X1 + X2  630
1/2 X1 + 5/6X2  600
X1 + 2/3X2  708
1/10 X1 + 1/4X2  135
X1, X2  0
(Total Profit Contribution)
(Capacity C&D)
(Capacity sewing)
(Capacity finishing)
(Capacity insp. & pack)
(Nonnegativity)
Problem 2
M & D Chemicals produces two products that are sold as raw materials to companies manufacturing both
soaps, laundry detergents, and other soap products. Management has specified that the total production for
products 1 and 2, combined must be as least 350 gallons. A major customer’s order for 125 gallons of
product 1 must also be satisfied. Product 1 requires 2 hrs of processing time per gallon and product 2
requires 1 hr of processing time per gallon. A total of 600 hrs of processing time are available. M & D’s
objective is to satisfy the above requirements at a minimum total production cost. Production costs are $2
per gallon for product 1, and $3 per gallon for product 2.
Problem 2 - Solution
Decision variables:
X1: number of gallons of product 1 to be produced
X2: number of gallons of product 2 to be produced
Formulation:
Min
2 X1 + 3 X2
st
X1  125
X1 + X2  350
2X1 + X2  600
X1, X2  0
(Total production cost)
(Demand for major customer)
(Minimum production of two products)
(Production capacity)
(Nonnegativity)
Problem 3
Relax-and-Enjoy have provided BP & J with an advertising budget of $30,000 for the first month’s
campaign. In addition, Relax-and-Enjoy have imposed the following restrictions on how BP & J may
allocate these funds: at least 10 television commercials must be used, at least 50,000 potential purchasers
must be reached during the month, and no more than $18,000 may be spent on television advertisements.
What advertising media selection plan should the advertising firm recommend?
Advertising
Media
1. Daytime TV
(1 min) section
WKLA
2. Evening TV
(30 sec) section
WKLA
3. Daily
newspaper
The Morning
Journal
4. Sunday
newspaper
The Sunday
Press
5. Radio 8am or
5pm station
KNOP
Number of
Potential
Customers
Reached
1,000
Cost per
Advertisement
Maximum Times
Available
Per Month
Exposure
Units
$1,500
15
65
2,000
$3,000
10
90
1,500
$400
25
40
2,500
$1,000
4
60
300
$100
30
20
Problem 3 – Solution
Decision variables:
X1: number of time daytime TV is used
X2: number of time evening TV is used
X3: number of time daily newspaper is used
X4: number of time Sunday newspaper is used
X5: number of time radio is used
Formulation:
Max
65 X1 + 90 X2 + 40 X3 + 60 X4 + 20 X5
(Total effectiveness of ads)
st
1500 X1 + 3000 X2 + 400 X3 + 1000 X4 + 100 X5  30,000 (Budget)
X1 + X2  10
(Number of TV commercials)
1500 X1 + 3000 X2  18,000
(TV budget)
1000 X1 + 2000 X2 + 1500 X3 + 2500 X4 + 300 X5 > 50,000 (Potential purchasers)
X1  15
(Availability of daytime TV)
X2  10
(Availability of evening TV)
X3  25
(Availability of daily paper)
X4  4
(Availability of Sunday paper)
X5  30
(Availability of radio)
X1, X2, X3, X4, X5  0
(Nonnegativity)
Problem 4
Consider the case of Welte Mutual Funds located in New York City. Welte has just obtained $10 million
by converting industrial bonds to cash. They are now looking for investment opportunities for these funds.
Considering Welte’s current investments, the firm’s top financial analyst recommends that all new
investments should be made in the oil or steel industry, or in government bonds. Specifically, the analyst
has identified five investment opportunities and projected their annual rates of return. The investments and
rates of return are shown below.
Investment
Rate of Return (%)
Atlantic Oil
Pacific Oil
Midwest Steel
Huber Steel
Government Bonds
7.3
10.3
6.4
7.5
4.5
Management of Welte has imposed the following investment guidelines:
 Neither industry (oil or steel) should receive more than 50% of the total new investment
 Government bonds should be at least 25% of the steel industry investment
 The investment in Pacific Oil, the high return but high risk investment cannot be more than
60% of the total oil industry investment.
What portfolio recommendations should be made assuming they have $10 million available for new
investment?
Problem 4 - Solution
Decision variables:
X1: $ invested in Atlantic Oil
X2: $ invested in Pacific Oil
X3: $ invested in Midwest steel
X4: $ invested in Huber steel
X5: $ invested in Govt. Bond
Formulation:
Max
0.073 X1 + 0.103 X2 + 0.064 X3 + 0.075 X4 + 0.045 X5
st
X1 + X2 + X3 + X4 + X5  10,000,000
X1 + X2  0.5 (X1 + X2 + X3 + X4 + X5)
X3 + X4  0.5 (X1 + X2 + X3 + X4 + X5)
X5  0.25 (X3 + X4)
X2  0.60 (X1 + X2)
X1, X2, X3, X4, X5  0
(Total return)
(Max available for investment)
(Oil industry less than 50%)
(Steel industry less than 50%)
(Govt. bonds at least 50%)
(Pacific Oil no more than 60%)
(Nonnegativity)
Problem 5
A hospital wants to maximize its margin while optimally using its resources. To meet this objective,
management needs to know the exact number of cases it should service from each diagnosis related group
(DRG) classification. For the purpose of simplicity, only four DRGs are considered. The table below gives
contribution margin, resource requirement, minimum weekly demand for each DRG, and the availability of
the resources.
Hours of diagnostic services
Bed days
Hours of nursing care
Dollar amount of pharmacy ($)
Service demand (number of
patients)
Contribution margin ($)
DRG1
DRG2
DRG3
DRG4
7
5
30
800
10
4
2
10
500
15
2
1
5
150
40
1
1
50
160
2,000
1,500
500
300
Maximu
m
available
480
170
1,000
50,000
For example, DRG1 requires seven hours of diagnostic services, five bed days, 30 hours of nursing care,
$800 of pharmaceuticals.
Problem 5 – Solution
Decision variables:
X1: number of patients to treat in DRG 1
X2: number of patients to treat in DRG 2
X3: number of patients to treat in DRG 3
X4: number of patients to treat in DRG 4
Formulation:
Max
2000 X1 + 1500 X2 + 500 X3 + 300 X4
st
7 X1 + 4 X2 + 2 X3 + X4 < 480
5X1 + 2 X2 + X3 < 170
30 X1 + 10 X2 + 5 X3 + X4 < 1000
800 X1 + 500 X2 + 150 X3 + 50 X4 < 50000
X1 > 10
X2 > 15
X3 > 40
X4 > 160
X1, X2, X3, X4  0
(Total contribution margin)
(Diagnostic hours)
(Bed days)
(Nursing hours)
(Pharmacy dollars)
(Min demand DRG1)
(Min demand DRG2)
(Min demand DRG3)
(Min demand DRG4)
(Nonnegativity)
Linear programming is extensively used in staffing and workforce scheduling. Next two problems
deal with this class of problems.
Problem 6
In a typical hospital, often the nursing labor costs are the largest expense component of the hospital budget.
Consider a hospital where the nursing staff is made up of registered nurses (RNs), licensed practical nurses
(LPNs), and non-licensed personnel such as nursing assistants (NAs). In the nursing hierarchy, the RNs are
the most capable and the NAs represent the least. The LPNs are somewhere in the middle. Generally, the
RNs devote much of their time to assessment, care planning, coordination of care and patient education.
Quantitative assessments, mobility, hygiene, and clerical tasks are delegated to NAs. LPNs administer oral
medications and perform complex treatments such as dressing changes, in addition to the tasks carried out
by NAs. Team nursing is used to deliver patient care, in which RN functions as the leader – he or she
assesses each patient, and organizes and implements the plan of care for each patient in the group. It
typically involves delegating tasks to team members consisting of NAs and LPNs. Problem is to find the
optimal mix of nurses in a hospital with three shifts. Hourly salary RN ($50), LPN ($25), and NA ($15).
Total patient care hours required in a 24-hour period is estimated to be 150 hours apportioned as follows:
57, 51, and 42 hours for the morning, afternoon and night shifts respectively. Hourly capability coefficient
of RNs is 1.0, those of LPNs and NAs are estimated to be 0.90 and 0.75 respectively. It is also estimated
that LPNs can do 85% of all tasks and the NAs can do 50% of all tasks. It is also required that at least 50%
of the total staff must be RNs to maintain a required level of quality.
Problem 6 - Solution
Decision variables:
Xni: # of RNs working in shift i (i= 1, 2, 3)
Xli: # of LPNs working in shift i (i= 1, 2, 3)
Xai: # of NAs working in shift i (i= 1, 2, 3)
Formulation:
Min
8(50 Xn1+ 25 Xl1 + 15 Xa1 + 50 Xn2+ 25 Xl2
+ 15 Xa2 + 50 Xn3+ 25 Xl3 + 15 Xa3)
st
Xni > 0.5 (Xni + Xli + Xai)
( for i=1,2,3)
8 (Xn1 + 0.90 Xl1 + 0.75 Xa1) > 57
8 (Xn2 + 0.90 Xl2 + 0.75 Xa2) > 51
8 (Xn3 + 0.90 Xl3 + 0.75 Xa3) > 42
0.90Xli < 0.85 (Xni + 0.90 Xli + 0.75 Xai) (for i =1,2,3)
0.75Xai < 0.50 (Xni + 0.90 Xli + 0.75 Xai) (for i =1,2,3)
All variables  0
(Total salary)
(50% must be RNs)
(Shift 1 hours)
(Shift 2 hours)
(Shift 3 hours)
(LPN can do 85% tasks)
(NAs can do 50% tasks)
(Nonnegativity)
Problem 7
A post office requires different numbers of full-time employees on different days of the week. The number
of full-time employees required each day is given below.
Weekday
Minimum Employees Required
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
17
13
15
19
14
16
11
Union rules state that each full-time employee work five consecutive days and then receive two days off.
For example, an employee who works Monday to Friday must be off on Saturday and Sunday. The post
office wants to meet its daily requirement using only full-time employees. Its objective is to minimize the
number of full-time employees that must be hired.
Problem 7 - Solution
Decision variables:
Xi: number of employees starting work each day
1 = MON, 2 = TUE, 3 = WED, 4 = TH, 5 = FRI, 6 = SAT, 7 = SUN
Formulation:
Min
X1 + X2 + X3 + X4 + X5 + X6 + X7
st
X1 + X4 + X5 + X6 + X7  17
X1 + X2 + X5 + X6 + X7  13
X1 + X2 + X3 + X6 + X7  15
X1 + X2 + X3 + X4 + X7  19
X1 + X2 + X3 + X4 + X5  14
X2 + X3 + X4 + X5 + X6  16
X3 + X4 + X5 + X6 + X7  11
X1, X2, X3, X4, X5, X6, X7  0
(Total number of employees)
(Employees required on Monday)
(Employees required on Tuesday)
(Employees required on Wednesday)
(Employees required on Thursday)
(Employees required on Friday)
(Employees required on Saturday)
(Employees required on Sunday)
(Nonnegativity)
Next problem deals with the so called diet problem. It is one of the earliest LP problems studied
by researchers. In simple terms, the problem involves finding the minimum cost diet that meets all
nutritional requirements. George Stigler (who won the Nobel Prize in 1982) framed a diet
problem involving 77 different foods in 1939 and was able to obtain an approximate solution
manually yielding an annual cost of $39.93 in 1939 prices. With the advent of computers in 1947,
this problem was again one of the first LP problems to be solved on a computer. Optimal solution
turned out to be only slightly better - annual cost of $39.67 in 1939 dollars. George Dantzig also
worked on his diet problem.
Problem 8
The Petgrow Pet Food Company is developing a new dog food, to be produced by combining meat by
products, fish meal, and cereal. To maximize profits, Petgrow wishes to develop the least-cost salable
product. Government standards require that dog food contain at least 70 percent meat or fish by weight.
The nutritional standards printed on the label require that there be at least 1/2 pound of protein in each 1
pound can. Meat by products is 60 percent protein by weight, fish meal is 70 percent, and cereal is 30
percent. To make the product more desirable, (to the owner, not the dog), the advertising campaign will be
based on the “meaty” appearance of the dog food. This requires a red “meaty” color. A pound of meat by
products averages 80 units of redness, a pound of fish meal averages 30, and a pound of cereal averages 40.
The advertising people believe that there should be a minimum of 50 units of redness per pound in the
product. Meat by product cost $0.50 per pound, fish meal costs $0.30 per pound, and cereal costs $0.20 per
pound. Develop an LP model to determine how to make the new dog food.
Problem 8 - Solution
Decision variables:
M: amount of meat products in 1 pound of dog food
F: amount of fish meal in 1 pound of dog food
C: amount of cereal in 1 pound of dog food
Formulation:
Min
0.50 M + 0.30 F + 0.20 C
st
M + F > 0.70
0.60 M + 0.70 F + 0.30 C > 0.50
80 M + 30 F + 40C > 50
M+F+C = 1.0
M, F, C > 0
(Total cost per pound of dog food)
(Minimum meat and fish content)
(Protein requirement)
(Minimum redness)
(One lb of dog food)
(Nonnegativity)
Problem 9
Agriculture is another area where linear programming has been used very successfully. LP was first
applied to agriculture problems in 1950s. .
Farmer Brown has a 600-arce farm. He is trying to determine how to allocate the land among three
possible uses. He can plant hay or corn, or he can use the land as pasture for raising beef cattle. An acre
planted in hay will produce 5 tons of hay and will require 0.25 ton of fertilizer. An acre planted in corn
will yield 100 bushels of corn and require 0.50 ton of fertilizer. Each cow requires 4 acres of pasture, each
acre of which requires 0.25 ton of fertilizer. In addition, each cow will consume 200 bushels of corn and
10 tons of hay. Farmer Brown will not buy hay or corn, but he can sell any that he does not feed to a cow
for a net gain of $3 a ton for the hay and $4 per bushel for the corn. A cow will sell for a net gain of $700.
There are 200 tons of fertilizers available. Develop an LP model to determine how Farmer Brown should
allocate his land.
Problem 9 - Solution
Decision variables:
X1: acres for hay to be sold
X2: acres for corn to be sold
X3: number of cattle
X4: acres for hay to be fed to cows
X5: acres for corn to be fed to cows
Formulation:
Max
15 X1 + 400 X2 + 700 X3
st
X1 + X2 + 4X3 + X4 + X5  600
0.25 X1 + 0.5 X2 + X3 + 0.25 X4 + 0.5 X5  200
X3  1/2 X4
X3  1/2 X5
X1, X2, X3, X4, X5  0
(Total contribution in $)
(Land)
(Fertilizers)
(Hay for cows)
(Corn for cows)
(Nonnegativity)
One area where management science techniques have lately found lot of applications and
acceptance by the decision maker is marketing.
Problem 10
Chandler Enterprises produces two competing products, A and B. The company wants to sell these
products to two groups of customers, group 1 and group 2. The values each customer places on a unit of A
and B products are shown below.
Value
A
B
Group 1 Customer
$10
$8
Group 2 Customer
$12
$15
Each customer will buy either product A or product B, but not both. A customer is willing to buy product
A if he/she believes that
Premium of product A >= Premium of product B
and
Premium of product A >= 0
Here the premium of a product is its value minus its price. Similarly, a customer is willing to buy product
B if he/she believes that
Premium of product B >= Premium of product A
and
Premium of product B >= 0
Group 1 has 1,000 members, and group 2 has 1,500 members. Chandler wants to set prices for each
product to ensure that group 1 members purchase product A and group 2 members purchase product B.
Determine how Chandler can maximize its revenue.
Problem 10 – Solution
Decision variables:
PA: price charged for product A
PB: price charged for product B
Formulation:
Max
1000 PA + 1500 PB
st
10 - PA  8 – PB
10 - PA  0
15 - PB  12 - PA
15 - PB  0
PA, PB  0
(Total revenue)
(Group 1: premium of
(Group 1: premium of
(Group 2: premium of
(Group 2: premium of
(Nonnegativity)
A > premium of B)
A > 0)
B > premium of A)
B > 0)
Problem 11
The manager of a department store in Seattle is attempting to decide on the types and amounts of
advertising the store should use. He has invited representatives from the local radio station, television
station, and newspaper to make presentations in which they describe their audiences. The television station
representative indicates that a TV commercial, which costs $15,000 would reach 25,000 potential
customers. The breakdown of the audience is as follows.
Old
Young
Male
5,000
5,000
Female
5,000
10,000
The newspaper representative claims to be able to provide an audience of 10,000 potential customers at a
cost of $4,000 per ad. The breakdown of the audience is as follows.
Old
Young
Male
4,000
2,000
Female
3,000
1,000
The radio station representative says that the audience for one of the station's commercials, which costs
$6,000 is 15,000 customers. The breakdown of the audiences is as follows.
Old
Young
Male
1,500
4,500
Female
1,500
7,500
The store has the following advertising policy:
(a) Use at least twice as many radio commercials as newspaper ads.
(b) Reach at least 100,000 customers.
(c) Reach at least twice as many young people as old people.
(d) Make sure that at least 30% of the audience is women.
(e) Available space limits the number of newspaper ads to 7.
The store wants to know the optimal number of each type of advertising to purchase to minimize total cost.
Problem 11 - Solution
Decision variables:
X1: # of TV commercial
X2: # of Newspaper Ads
X3: # of radio commercial
Formulation:
Min
15,000 X1 + 4,000 X2 + 6,000 X3
st
X3  2 X2
25,000 X1 + 10,000 X2 + 15,000 X3  100,000
15,000 X1 + 3,000 X2 + 12,000 X3  2(10,000 X1 + 7,000 X2 + 3,000 X3)
15,000 X1 + 4,000 X2 + 9,000 X3  0.3(25,000 X1 +10,000 X2 +15,000X3)
X2  7
X1, X2, X3  0
(Total cost)
(Policy a)
(Policy b)
(Policy c)
(Policy d)
(Policy e)
(Nonnegativity)
Operations and supply chain management has always been a rich area for LP application. Many
ERP and SCM software have several built-in LP applications.
Problem 12
The Ferguson Paper Company produces rolls of paper for use in adding machines, desk calculators, and
cash registers. The rolls, which are 200 feet long, are produced in widths of 1.5, 2.5, and 3.5 inches. The
production process provides 200 feet rolls in 10-inch width only. The firm must therefore cut the rolls to
the desired final product sizes. The minimum requirements for the three products are:
Roll width (inches)
Units
1.5
1,000
2.5
2,000
3.5
4,000
(a) If the company wants to minimize the number of 10-inch rolls that must be manufactured. How many
rolls are required, and what is the total waste?
(b) If the company wants to minimize the waste generated, how many 10-inch units will be processed?
What is the total waste?
Problem 12 - Solution
In this problem, identifying the decision variables requires a little bit of pre-processing. We first need to
find out the number of ways in which a 10” roll can be cut (using 1.5”, 2.5” and 3.5” rolls). The following
table lists all the possible ways of cutting a 10” roll.
Number of Rolls
Cutting Alternative
1
2
3
4
5
6
7
8
9
10
1.5"
6
0
2
0
1
1
4
5
3
2
2.5"
0
4
0
1
3
2
0
1
2
1
3.5"
0
0
2
2
0
1
1
0
0
1
Waste(" )
1
0
0
0.5
1
0
0.5
0
0.5
1
Our decision variables are the number of 10” rolls to be cut according to each cutting alternative.
Decision variables:
Xi: number of 10" rolls processed using cutting alternative i, (i = 1,2,….,10)
Formulation (a):
Our first formulation minimizes the number of 10” rolls to be cut to meet the demand.
Min
st
X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10
(Total number of rolls)
6 X1 + 2 X3 + X5 + X6 + 4 X7 + 5 X8 + 3 X9 + 2 X10  1000
4 X2 + X4 + 3 X5 + 2 X6 + X8 + 2 X9 + X10  2000
2 X3 + 2 X4 + X6 + X7 + X10  4000
X1, X2, X3, X4, X5, X6, X7, X8, X9, X10  0
(Demand for 1.5” rolls)
(Demand for 2.5” rolls)
(Demand for 3.5” rolls)
(Non negativity)
Formulation (b):
Second formulation minimizes the total waste generated in the cutting process. Waste is anything less than
1.5”.
Min
st
X1 +0.5X4 + X5 + 0.5X7 + 0.5X9 + X10
(Total waste)
6 X1 + 2 X3 + X5 + X6 + 4 X7 + 5 X8 + 3 X9 + 2 X10  1000
4 X2 + X4 + 3 X5 + 2 X6 + X8 + 2 X9 + X10  2000
2 X3 + 2 X4 + X6 + X7 + X10  4000
X1, X2, X3, X4, X5, X6, X7, X8, X9, X10  0
(Demand for 1.5” rolls)
(Demand for 2.5” rolls)
(Demand for 3.5” rolls)
(Non negativity)
Problem 13
Multi-period production planning deals with finding the best production and marketing plan over a
planning horizon to maximize the overall profit or to minimize the production cost.
A production manager is attempting to determine a production schedule for the next five months for a
product. From past production records the manager knows that 2,000 units can be produced per month.
Also, ad additional 600 units can be produced monthly on an overtime basis. The unit cost of items
produced is $10 on a regular time basis and $15 on an overtime basis. Contracted sales per month are as
follows:
Month
1
2
3
4
5
Contracted
sales
1,200
2,100
2,400
3,000
4,000
Inventory carrying costs are $2 per unit per month. The manager does not want any inventory carried over
past the fifth month. The manager wants to know the monthly production that will minimize total
production and inventory costs.
Problem 13 - Solution
Decision variables:
Xi: regular time production in month i, (i = 1, 2, 3, 4, 5)
Yi: overtime production in month i, (i = 1, 2, 3, 4, 5)
Wi: inventory at end of month i, (i = 1, 2, 3, 4)
Formulation:
Min
10(X1 + X2 + X3 + X4 + X5) + 15(Y1 + Y2 + Y3 + Y4 + Y5)
+ 2(W1 + W2 + W3 + W4)
(Total production and holding cost)
st
Xi  2000
(i = 1, 2, 3, 4, 5)
(Regular time capacity)
Yi  600 for
(i = 1, 2, 3, 4, 5
(Overtime capacity)
X1 + Y1 - W1 = 1200
(Demand month 1)
X2 + Y2 + W1 - W2 = 2100
(Demand month 2)
X3 + Y3 + W2 -W3 = 2400
(Demand month 3)
X4 + Y4 + W3 -W4 = 3000
(Demand month 4)
X5 + Y5 + W4 = 4000
(Demand month 5)
All variables > 0
(Nonnegativity)
Problem 14
A supply chain network usually consists of several stages, including suppliers, plants, warehouses and
markets. There may also be other intermediate facilities such as consolidation centers or transit points.
The following figure shows a typical supply network:
Suppliers
Plants
Warehouses
Markets
Suppose that a manufacturer of telecommunication equipment has 4 major markets, 2 production facilities
and 3 warehouse locations. The company has used forecasting methods for estimating demands at each
market and has capacity information for both the warehouses and production facilities. The problem is to
determine how to ship product from the plants to the warehouses and from the warehouses to the markets in
order to minimize total transportation and production costs.
Production and transportation cost per thousand units and capacities (thousand units) for plants and
warehouses
Plant 1
Plant 2
Capacity
Warehouse 1
1,675
1,460
12
Warehouse 2
1,630
1,355
23
Warehouse 3
1,160
1,200
15
Capacity
23
27
Transportation cost per thousand units and demand (thousand units)
Warehouse 1
Warehouse 2
Warehouse 3
Demand
Market 1
400
380
622
10
Market 2
570
343
700
12
Write down the linear programming formulation of this problem
Market 3
300
665
311
18
Market 4
495
508
650
10
Problem 14 - Solution
Decision variables:
Xij = number of (thousand) units shipped from plant i to warehouse j
Yjk = number of (thousand) units shipped from warehouse j to market k
Formulation:
Minimize
1675X11 + 1630X12 + 1160X13 + 1460X21 + 1355X22 + 1200X23 +
400Y11 + 570Y12 + 300Y13 + 495Y14 + 380Y21 + 343Y22 + 665Y23 +
508Y24 + 622Y31 + 700Y32 + 311Y33 + 650Y34
st
X11 + X12 + X13  23
X21 + X22 + X23  27
X11 + X21  12
X12 + X22  23
X13 + X23  15
Y11 + Y21 + Y31  10
Y12 + Y22 + Y32  12
Y13 + Y23 + Y33  18
Y14 + Y24 + Y34  10
X11 + X21 = Y11 + Y12 + Y13 + Y14
X12 + X22 = Y21 + Y22 + Y23 + Y24
X13 + X23 = Y31 + Y32 + Y33 + Y34
All variables > 0
(Plant capacity 1)
(Plant capacity 2)
(Warehouse capacity 1)
(Warehouse capacity 2)
(Warehouse capacity 3)
(Market demand 1)
(Market demand 2)
(Market demand 3)
(Market demand 4)
(Conservation of flow at warehouses 1)
(Conservation of flow at warehouses 2)
(Conservation of flow at warehouses 3)
Next two problems (15 and 16) are somewhat more difficult for us to formulate mathematically at
this stage. However, these two problems capture many of the complexities of real-world
problems. Just read them to get an idea about complexity of real problems.
Problem 15
At the beginning of month 1, JE Capital has 50 million accounts. Of those, 40 million are paid up (0-due),
4 million are one month overdue (1-due), 4 million are two months overdue (2-due), and 2 million are three
months overdue (3-due). Once an account is more than three months overdue, it is written off as a bad
debt. For each overdue account JE Capital can phone the cardholder, send a letter, or do nothing. A letter
requires an average of 0.05 hour of labor, while a phone call requires an average of 0.10 hour of labor.
Each month 500,000 hours of labor are available. We assume that an average amount of monthly payment
is $30. Thus, if a 2-due account remains 2-due, it means that one month's payment ($30) has been received,
and if a 2-due account becomes 0-due, it means that three months payments ($90) have been received. By
examining thousands of accounts, DMMs (Delinquency Movement Matrices) have been estimated and
shown below.
DMM if Account Receives a Letter
0-due
1-due
1-due
0.60
0.10
2-due
0.15
0.15
3-due
0.00
0.00
Bad Debt
0.00
0.00
2-due
0.30
0.40
0.20
0.00
3-due
0.00
0.30
0.60
0.00
Bad Debt
0.00
0.00
0.20
1.00
DMM if Account Receives a Phone Call
0-due
1-due
1-due
0.70
0.20
2-due
0.30
0.25
3-due
0.00
0.00
Bad Debt
0.00
0.00
2-due
0.10
0.30
0.50
0.00
3-due
0.00
0.15
0.40
0.00
Bad Debt
0.00
0.00
0.10
1.00
DMM for Do Nothing Action
0-due
1-due
1-due
0.90
0.10
2-due
0.20
0.25
3-due
0.00
0.00
Bad Debt
0.00
0.00
2-due
0.00
0.55
0.10
0.00
3-due
0.00
0.00
0.40
0.00
Bad Debt
0.00
0.00
0.50
1.00
Determine how to allocate your workforce over the next four months in order to maximize the expected
collection of revenue received during that time.
Problem 16
Blending is concerned with mixing different ingredients so that the resulting product meets all the
specifications. Oil and chemical companies deal a lot with blending related issues.
Ole Oil produces three products: heating oil, gasoline, and jet fuel. The average octane levels must be at
least 4.5 for heating oil, 8.5 for gas, and 7.0 for jet fuel. To produce these products, Ole purchases two
types of oil: crude 1 (at $12 per barrel) and crude 2 (at $10 per barrel). Each day at most 10,000 barrels of
each type of oil can be purchased. Before crude can be used to produce products for sale, it must be
distilled. It costs 10 cents to distill a barrel of oil. The result of distillation is shown below.
Crude 1
Crude 2
Naphtha
0.6
0.4
Distilled 1
0.3
0.2
Distilled 2
0.1
0.4
Distilled naphtha can be used only to produce gasoline or jet fuel. Distilled oil can be used produce heating
oil or it can be sent through the catalytic cracker (at a cost of 15 cents per barrel). Each day at most 5,000
barrels of distilled oil can be sent through the cracker, with the results shown below.
Distilled 1
Distilled 2
Cracked 1
0.8
0.7
Cracked 2
0.2
0.3
Cracked oil can be used to produce gasoline and jet fuel but not heating oil. The octane level of each type
of oil is as follows: naphtha, 8; distilled 1, 4; distilled 2, 5; cracked 1, 9; cracked 2, 6. All heating oil
produced can be sold at $14 per barrel; all gasoline produced, $18 per barrel; all jet fuel produced, $16 per
barrel. Marketing considerations dictate that at least 3,000 barrels of each product must be produced daily.
Determine how Ole can maximize its profits.