ProductMix Problem 1

1
Product-Mix Problem 1

• Two products are manufactured by passing
  sequentially through three machines. Time per
  machine allocated to the two products is limited
  to 10 hours per day. The production time and
  profit per unit of each product are given below.
  Find the optimal mix of the two products.
• Minutes
  per Unit
• 
  
• Product Machine1 Machine 2
  Machine 3 Profit
• I 10 6
  8 2
• II 5
  20 15
  3

2
Product-Mix Problem 1
Verbal Summary of the Model A company seeks
the determination of the daily number of units to
be produced ( variables) of each product in
order to maximize the total profit (objective)
provided that time per machine allocated to the
two products is limited to 10 hours per day
(constraints
In defining the variables, we need to ask
ourselves what it is that we wish the model to
determine. In this case, we need to know how much
of each product to produce. We will use the units
that are specified namely, minutes. These
observations lead us to believe the following
might be appropriate variable definitions
X
X
Y
3
Product-Mix Problem 1
We are interested in maximizing our profits, so
our objective function should express our total
profits in terms of our variables. We can write
our profits as  We can simplify this expression
to Profit 2 X 3Y Therefore our objective
function is
Z
2
3
4
Product-Mix Problem 1
Our only three constraints in this problem arise
from the availability limitations for three
machines. Time per machine allocated to the two
products is limited to 10 hours (600 minutes) per
day. We can express these restrictions with the
following inequalities
Machine 1
Machine 2
Machine 3
5
Product-Mix Problem 1
And finally, we need to determine if it is
necessary to impose sign restrictions on our
variables. In this example, both our variables
must be positive. Therefore we impose the
following non-negativity conditions
6
Product-Mix Problem 1

• Putting all the pieces together, we have the
  following LP formulation

Z ( objective function )
2
3
Subject to
( Constraints )
( sign restrictions )
7
Product-Mix Problem 2

• The XYZ company produces three different items.
  The production process
• utilizes three operations. Below figure shows
  sequence the producing
• items 1,2, and 3. Item 2 does not pass through
  operation 2 and item 3
• goes through operation 1 and 2 only. The
  processing times per unit of each
• item are shown in respective boxes. Since the
  same operations are used in
• the companys other production activities, the
  daily usage of operations
• 1,2, and 3 by all three items are limited to 430,
  460, and 420 minutes,
• respectively.A market study shows that the
  expected per unit profits for
• items 1, 2 and 3 are 3, 2, 5 . What is the
  best daily production level for
• each item ?

8
Product-Mix Problem 2
Verbal Summary of the Model A company seeks
the determination of the daily number of units to
be produced ( variables) of each item in order to
maximize the total profit (objective) provided
that the daily usage of each operation does not
exceed its maximum daily capacity (constraints)
In defining the variables, we need to ask
ourselves what it is that we wish the model to
determine. In this case, we need to know how much
of each item to produce. These observations lead
us to believe the following might be appropriate
variable definitions
daily number of units to be produced item 1
daily number of units to be produced item 2
daily number of units to be produced item 3
9
Product-Mix Problem 2
We are interested in maximizing our profits, so
our objective function should express our total
profits in terms of our variables. We can write
our profits as  We can simplify this expression
to Profit 3X 2Y 5Z Therefore our
objective function is
R
10
Product-Mix Problem 2
Our only three constraints in this problem arise
from the the daily usage of operations 1,2, and 3
by all three items are limited to 430, 460, and
420 minutes, respectively We can express these
restrictions with the following inequalities
Operation 1
Operation 2
Operation 3
11
Product-Mix Problem 2
And finally, we need to determine if it is
necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions
12
Product-Mix Problem 2

• Putting all the pieces together, we have the
  following LP formulation

( objective function )
R
Subject to
( Constraints )
( sign restrictions )
13
Blending or Mixing Problem

• We illustrate this class with the problem of
  determining the optimum
• amounts of three ingredients to include in an
  animal feed mix. The final
• product must satisfy several nutrient
  restrictions. The possible ingredients,
• their nutritive contents (in kilograms of
  nutrient per kilograms of ingredient)
• and the unit cost are shown in the following
  table. The mixture must meet the
• following restrictions
• Calcium at least 0.8 but not more
  than 1.2.
• Protein at least 22.
• Fiber at most 5.
• The problem is to find the composition of the
  feed mix that satisfies these
• constraints while minimizing cost.
•  

14
Blending or Mixing Problem
Verbal Summary of the Model A farm needs to
determine the number of pounds of the three
ingredients ( variables ) at minumum cost
(objective) provided that the nutritional needs
and quantity of feed mix are realized. (
constraints )
In defining the variables, we need to ask
ourselves what it is that we wish the model to
determine. In this case, we need to know what
percentage of a ton of the final blend should
come from each ingredient. Therefore, we will use
the following variable definitions
fraction of a kg to be blended from limestone
fraction of a kg to be blended from corn
fraction of a kg to be blended soybean meal
15
Blending or Mixing Problem
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
minimizing our costs. For each kg we blend in a
particular ingredient, we incur the blending
cost. For each fraction of a kg we blend in that
ingredients, we incur that fraction of the cost.
Therefore, the following objective function
captures our total per kg cost
16
Blending or Mixing Problem
We will now concentrate on our constraints.
First, we have our constraints for the maximum
requirements for each of the three elements.
These constraints can be expressed with the
following expressions
Minumum Calcium
Maximum Calcium
Minumum Protein
Maximum Fiber
In addition to these, we have an implied
constraint on our variables that we must include
in the formulation. We have defined our variables
as percentages and we need them to sum to one.
This constraint is easy to miss since it isn't
part of our problem definition, but rather a
by-product of our variable definition. As a
constraint, this is written
Blend
17
Blending or Mixing Problem
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions. Note that these
conditions, together with the blend constraint,
assure that our variables will be between 0 and 1
which is what we want.
18
Blending or Mixing Problem

• Putting all the pieces together, we have the
  following LP formulation

( objective function )
Subject to
( Constraints )
( sign restrictions )
19
Personnel Scheduling Problem

• A PO requires different numbers of employees on
  different days of the
• week. Union rules state each employee must work 5
  consecutive days and
• then receive two days off. Find the minimum
  number of employees
• needed.
• Mon Tue
  Wed Thur Fri Sat Sun
• Staff Needed 17 13 15 19 14 16 11

20
Personnel Scheduling Problem
Verbal Summary of the Model It is required to
determine different number of employees during
different days ( variables ) that will meet the
minumum number of employees during different days
( constraints ) while minimizing the total number
of employees.
X1 of employees starting on monday
X2 of employees starting on tuesday
X3 of employees starting on wednesday
X4 of employees starting on thursday
X5 of employees starting on friday
X6 of employees starting on saturday
X7 of employees starting on sunday
21
Personnel Scheduling Problem
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
minimizing the total number of employees
Minimize
Z X1 X2 X3 X4 X5 X6 X7
22
Personnel Scheduling Problem
We know that employee will work 5 consecutive
days and then receive two days off but we dont
know which day start to work.
X1 X4 X5 X6 X7 17
(started thu,fri,sat,sun,mon)
X1 X2 X5 X6 X7 13
(started fri,sat,sun,mon,tue)
X1 X2 X3 X6 X7 15
(started sat,sun,mon,tue,wed)
X1 X2 X3 X4 X7 19
(started sun,mon,tue,wed)
X1 X2 X3 X4 X5 14
(started mon,tue,wed,thu,fri)
X2 X3 X4 X5 X6 16
(started tue,wed,thu,fri,sat)
X3 X4 X5 X6 X7 11
(started wed,thu,fri,sat,sun)
23
Personnel Scheduling Problem
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions.
X1 , X2 , X3 , X4 , X5 , X6 , X7 0
24
Personnel Scheduling Problem

• Putting all the pieces together, we have the
  following LP formulation

Minimize
Z X1 X2 X3 X4 X5 X6 X7 (Objective
funtion)
Subject to
X1 X4 X5 X6 X7 17

X1 X2 X5 X6 X7 13
X1 X2 X3 X6 X7 15
( Constraints )
X1 X2 X3 X4 X7 19
X1 X2 X3 X4 X5 14
X2 X3 X4 X5 X6 16
X3 X4 X5 X6 X7 11
( sign restrictions )
X1 , X2 , X3 , X4 , X5 , X6 , X7 0
25
A Diet Problem

• Suppose the only foods available in your local
  store are
• potatoes and steak. The decision about how much
  of each food
• to buy is to made entirely on dietary and
  economic
• considerations. We have the nutritional and cost
  information in
• the following table
•  

26
A Diet Problem
Verbal Summary of the Model The problem is to
find a diet (a choice of the numbers of units of
the two foods) that meets all minimum nutritional
requirements at minimal cost.
X of units potatoes to be selected
Y of units steak to be selected
27
A Diet Problem
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
minimizing the total cost of foods.(potatoes,
steak)
Minimize
Z 25 X 50Y
28
A Diet Problem
The first constraint represents the minimum
requirement for carbohydrates, which is 8 units
per some unknown amount of time. 3 units can be
consumed per unit of potatoes and 1 unit can be
consumed per unit of steak. The second
constraint represents the minimum requirement for
vitamins, which is 19 units. 4 units can be
consumed per unit of potatoes and 3 units can be
consumed per unit of steak. The third
constraint represents the minimum requirement for
proteins, which is 7 units. 1 unit can be
consumed per unit of potatoes and 3 units can be
consumed per unit of steak
3X Y 8
4X 3Y 19
X 3Y 7
29
A Diet Problem
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. because we can't buy negative
quantities Therefore we impose the following
non-negativity conditions.
X 0
Y 0
30
A Diet Problem

• Putting all the pieces together, we have the
  following LP formulation

Minimize
Z 25 X 50Y (Objective funtion)
Subject to
3X Y 8
( Constraints )
4X 3Y 19
X 3Y 7
( sign restrictions )
X, Y 0
31
Customer Service Level Problem (1)

• CSL services computers. Its demand (hours) for
  the time of skilled
• technicians in the next 5 months is
• t Jan Feb Mar Apr May
• dt 6000 7000 8000 9500 11000
• It starts with 50 skilled technicians at the
  beginning of January. Each
• technician can work 160 hrs/month. To train a new
  technician they must
• be supervised for 50 hrs by an experienced
  technician. Each
• experienced technician is paid 2K/mth and a
  trainee is paid 1K/mth.
• Each month 5 of the skilled technicians leave.
  CSL needs to meet
• demand and minimize costs.

32
Customer Service Level Problem (1)
Verbal Summary of the Model It is required to
determine experienced tech. to be trained in
month t ( variables ) that will meet the demand
in month t ( constraints ) while minimizing the
total cost which is paid for technician and
trainee.
Xt experienced tech. at start of t th month
t 1,2,3,4,5
Yt to be trained in month t
33
Customer Service Level Problem (1)
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
minimizing the total cost that is paid for
technician and trainee
Minimize
Z 2000(X1X2X3X4X5) 1000 (Y1Y2Y3Y4Y5)
34
Customer Service Level Problem (1)
Each technician can work 160 hrs/month. To train
a new technician they must be supervised for 50
hrs by an experienced technician. Then first
constraint will be
dt demand during month t
160 Xt - 50 Yt dt where t
1,2,3,4,5
It starts with 50 skilled technicians at the
beginning of January. Then second constraint
will be
X1 50

Each month 5 of the skilled technicians leave.
And then third constraint will be
Xt 0.95Xt-1 Yt-1 where t 2,3,4,5
35
Customer Service Level Problem (1)
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions.
Xt , Yt 0 where t 1,2,3,4,5
36
Customer Service Level Problem (1)

• Putting all the pieces together, we have the
  following LP formulation

Minimize
Z 2000(X1X2X3X4X5) 1000 (Y1Y2Y3Y4Y5)
Subject to
160 Xt - 50 Yt dt t
1,2,3,4,5
( Constraints )
X1 50
Xt 0.95Xt-1 Yt-1 t 2,3,4,5
( sign restrictions )
Xt , Yt 0 t 1,2,3,4,5
37
Aggregate Planning Problem

• A company wants a high level, aggregate
  production plan for the next 6 months.
• Projected orders for the company's product are
  listed in the table. Over the 6-
• month period, units may be produced in one month
  and stored in inventory to
• meet some later month's demand. Because of
  seasonal factors, the cost of
• production is not constant, as shown in the
  table.

The cost of holding an item in inventory for 1
month is 4/unit-mo. Items produced and sold in
the same month are not put in inventory. The
maximum number of units that can be held in
inventory is 250. The inventory level at the
beginning of the planning horizon is 200 units
the inventory level at the end of the planning
horizon is to be 100. The problem is to determine
the optimal amount to produce in each month so
that demand is met while minimizing the total
cost of production and inventory. Shortages are
not permitted
38
Aggregate Planning Problem
Verbal Summary of the Model The problem is to
determine the optimal amount to produce in each
month so that demand is met while minimizing the
total cost of production and inventory.
Pt production level in month t
t 1,2,3,4,5,6
It inventory level at the end of month t
Parameters
dt demand in month t t
1,2,3,4,5,6
39
Aggregate Planning Problem
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
minimizing the total cost of production and
inventory
Minimize
Z 100P1105P2110P3115P4110P5110P6
4(I1I2I3I4I5I5)
40
Aggregate Planning Problem
Conservation of flow A basic requirement in
production planning problems is that product or
material must be conserved. In our case, this
leads to the following production constraints.
It-1 Pt
It dt for t 1,2,3,4,5,6 The demand in
month t must be met by the production in month t
plus the net reduction in inventory.
Maximum inventory This is simply an upper bound
constraint on the inventory levels.
It 250 for t
1,2,3,4,5,6
Initial and final conditions I0 200 I6
100 Although I(0) and I(6) have constant values
because of these constraints, we leave them as
variables in the model. Aggregate planning
models, as well as many others, are meant to be
solved over and over again as time advances and
as parameters change. It is easier to treat the
initial and final values as constraints rather
than replace the two variables by their
equivalent values.
41
Aggregate Planning Problem
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions.
Pt , It 0 where t 1,2,3,4,5,6
42
Aggregate Planning Problem

• Putting all the pieces together, we have the
  following LP formulation

Minimize
Z 100P1105P2110P3115P4110P5110P6
4(I1I2I3I4I5I5)
(Objective Function)
Subject to
It-1 Pt It Dt for t
1,2,3,4,5,6
( Constraints )
It 250 for t 1,2,3,4,5,6
I0 200 I6 100
( sign restrictions )
Pt , It 0 for t 1,2,3,4,5
43
Customer Service Level Problem (2)

• An electronics company has a contract to
  deliver 20,000 radios within the next four weeks.
  The client is willing to pay 20 for each radio
  delivered by the end of the first week, 18 for
  those delivered by the end of the second week,
  16 by the end of the third week, and 14 by the
  end of the fourth week. Since each worker can
  assemble only 50 radios per week, the company
  cannot meet the order with its present labor
  force of 40 hence it must hire and train
  temporary help. Any of the experienced workers
  can be taken off the assembly line to instruct a
  class of 3 trainees after one week of
  instruction, each of the trainees can either
  proceed to the assembly line or instruct
  additional new classes. At present, the company
  has no other contracts hence some workers may
  become idle once the delivery is completed. All
  of them, whether permanent or temporary, must be
  kept on the payroll 'til the end of the fourth
  week. The weekly wages of a worker, whether
  assembling, instructing, or being idle, are 200
  the weekly wages of a trainee are 100. The
  production costs, excluding the worker's wages,
  are 5 per radio

44
Customer Service Level Problem (2)
Verbal Summary of the Model It is required to
determine total of experienced , of
experinced workers which is assembling, of
trainees to be trained in week i ( variables )
that will meet the total demand (20000 radio )for
four weeks ( constraints ) while maximizing the
total profit. .
Ei of experinced workers in week i Ai
of experinced workers assembling in week i Ti
of trainees to be trained in week i Ii
of experinced workers is idle in week i
i 1,2,3,4
45
Customer Service Level Problem (2)
We need to define our objective function in terms
of the variables that we defined for this
problem. In this problem, we are interested in
maximizing the total profit. Total Profit
Total Selling Price Total Production cost -
Total Labor Total Selling Price (2050) A1
(1850) A2 (1650) A3 (1450) A4 Total Prod.
cost ( 550 ) A1 ( 5-50 ) A2 ( 550
) A3 ( 550 ) A4 Total Labor 200
( E1 E2 E3 E4 ) 100 ( T1 T2 T3
T4 )
Profit The total profit over the four week
period is
Z 750 A1 650 A2 550 A3 450 A4
- 200 ( E1 E2 E3 E4 ) - 100 ( T1
T2 T3 T4 )
Maximize
46
Customer Service Level Problem (2)
Number of Experienced Employees The relationship
between the number of experienced employees
between weeks is E1 40 ( 40
experinced workers available in week 1 )
E2 E1 T1 E3 E2 T2 E4 E3
T3 Number of Idle Employees The number of
idle employees during each week is I1
E1 A1 T1/3 I2 E2 A2 T2 /3
I3 E3 A3 T3 /3 I4 E4 A4 T4
/3 Total Number of Radios The total number of
radios produced is      20,000  50  (
A1  A2  A3  A4 )

47
Customer Service Level Problem (2)
And as a final step, we need to determine if it
is necessary to impose sign restrictions on our
variables. In this example, our variables must be
positive. Therefore we impose the following
non-negativity conditions.
Ei 0 , Ai 0 , Ti 0 , Ii 0
where i 1,2,3,4
48
Customer Service Level Problem (2)

• Putting all the pieces together, we have the
  following LP formulation

Maximize
Z 750 A1 650 A2 550 A3 450 A4
(Objective Function) - 200 ( E1 E2
E3 E4 ) - 100 ( T1 T2 T3 T4 )
Subject to
Number of Experienced Employees E1 40
E2 E1 T1 E3 E2 T2 E4
E3 T3 Number of Idle Employees I1
E1 A1 T1 I2 E2 A2 T2 I3
E3 A3 T3 I4 E4 A4 T4 The total
number of radios produced is       20,000 
50  ( A1  A2  A3  A4 )
( Constraints )
Ei 0 , Ai 0 , Ti 0 , Ii 0
where i 1,2,3,4 ( sign restrictions )