Transcript Production and Operations Management: Manufacturing and

Resource Allocation
Class 7: 3/9/11
3.1 WHY NETWORK PLANNING?
Find the right balance between inventory,
transportation and manufacturing costs,
 Match supply and demand under uncertainty by
positioning and managing inventory effectively,
 Utilize resources effectively by sourcing products
from the most appropriate manufacturing facility

RESOURCE ALLOCATION

In operations and supply chain management
there are several problems related to resource
allocation:
Production mix: how many units of each product
or service should be produced given their profitability
and constraints on available resources and their
usage for each product
 Network location and sourcing:

where to locate facilities, including manufacturing plants,
distribution centers, and warehouses
 Given a network of facilities, how to best service my
customer mix considering transportation and distribution
costs (sourcing decision)

NETWORK SCHEDULING EXAMPLE
 Single
 Two

product
plants p1 and p2
Plant p2 has an annual capacity of 60,000
units.
 The
two plants have the same production
costs.
 There
are two warehouses w1 and w2
with identical warehouse handling costs.
 There
are three markets areas c1,c2 and
c3 with demands of 50,000, 100,000 and
50,000, respectively.
UNIT DISTRIBUTION COSTS
Facility
warehouse
p1
p2
c1
c2
c3
w1
0
4
3
4
5
w2
5
2
2
1
2
HEURISTIC #1:
CHOOSE THE CHEAPEST WAREHOUSE TO SOURCE
DEMAND
D = 50,000
$2 x 50,000
$5 x 140,000
Cap = 60,000
$2 x 60,000
D = 100,000
$1 x 100,000
$2 x 50,000
Total Costs = $1,120,000
D = 50,000
HEURISTIC #2:
CHOOSE THE WAREHOUSE WHERE THE TOTAL DELIVERY
COSTS TO AND FROM THE WAREHOUSE ARE THE LOWEST
[CONSIDER INBOUND AND OUTBOUND DISTRIBUTION
COSTS]
$0
D = 50,000
$3
$5
$4
$2
$5
$3
$7
$7
$4
D = 100,000
$4
Cap = 60,000
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
$1
$2
$2
$4
$6
$8
$3
D = 50,000
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
Market #1 is served by WH1, Markets 2 and 3
are served by WH2
$5
$7
$9
$4
HEURISTIC #2:
CHOOSE THE WAREHOUSE WHERE THE TOTAL
DELIVERY COSTS TO AND FROM THE WAREHOUSE ARE
THE LOWEST
[CONSIDER INBOUND AND OUTBOUND DISTRIBUTION
COSTS]
$0 x 50,000
D = 50,000
$3 x 50,000
Cap = 200,000
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
$5 x 90,000
D = 100,000
$1 x 100,000
Cap = 60,000
$3
$7
$7
$4
$2 x 60,000
$2 x 50,000
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
D = 50,000
P1 to WH1
P1 to WH2
P2 to WH1
P2 to WH 2
Total Cost = $920,000
$4
$6
$8
$3
$5
$7
$9
$4
OPTIMIZATION APPROACH
•The problem described earlier can be
framed as a linear programming problem.
•A much better solution is found – total cost
= $740,000!
•How does optimization work?
LINEAR PROGRAMMING
LP deals with the problem of allocating limited
resources among competing activities
For example, consider a company that makes
tables and chairs (competing activities) using
a limited amount of large and small Legos
(limited resources).
LINEAR PROGRAMMING
The objective of LP is to select the best or
optimal solution from the set of feasible
solutions (those that satisfy all of the
restrictions on the resources).
Suppose profit for each table is $20 while
profit for each chair is $16.
We may choose to identify the number of
tables and chairs to produce to maximize
profit while not using more Legos than are
available.
COMPONENTS OF AN LP
Decision Variables: factors which are
controlled by the decision maker.
x1 = the number of tables produced per day
x2 = the number of chairs produced per day
Objective function: profit, cost, time, or
service must be optimized.
The objective may be to optimize profit.
COMPONENTS OF AN LP
Constraints: restrictions which limit the
availability and manner with which
resources can be used to achieve the
objective
It takes 2 large and 2 small Legos to
produce a table and 1 large and 2 smalls to
produce a chair
We may only have 6 large and 8 small Legos
available each day
ASSUMPTIONS
Linearity: Linear objective function and
linear constraints.
This implies proportionality and additivity.
For example, it takes 2 large Legos to
produce 1 table and 4 to produce 2 tables.
It takes 3 large Legos to produce 1 table and
1 chair.
ASSUMPTIONS
Divisibility: The decision variables can take
on fractional values.
The optimal solution may tell us to produce
2.5 tables each day.
Certainty: The parameters of the model are
known or can be accurately estimated.
For example, we assume that the
profitability information is accurate.
ASSUMPTIONS
Non-negativity: All decision variables must
take on positive or zero values.
LEGO PRODUCTS, INC.
Lego Products, Inc. manufactures tables and
chairs.
Profit for each table is $20 while each chair
generates $16 profit.
Each table is made by assembling two large and
two small legos. Each chair requires one large
and two small legos.
Currently, Lego Products has six large and
eight small legos available each day.
LEGO EXAMPLE: INTRODUCTION
Pictures of a table and a chair are shown below.
Table
Chair
LEGO EXAMPLE: OPTIMAL SOLUTION
How many tables and chairs should we
produce to maximize daily profit?
producing 3 tables generates a daily profit of
$60,
producing 4 chairs generates a daily profit of
$64, however,
producing 2 tables and 2 chairs generates the
optimal daily profit of $72.
USING SOLVER
Solver in Excel can be used to obtain the solution
and will now be demonstrated
The problem formulation is:
MAX 20*X1+16*X2
Subject to:
2*X1+1*X2<=6
2*X1+2*X2<=8
How to enter this formulation into Solver?
LEGO EXAMPLE: UNUSED LEGOS
Do we have any unused large or small legos
for all of the solutions that you just found?
There are no unused large or small legos for
the optimal solution.
There are 2 unused small legos if 3 tables
are made.
There are 2 unused large legos if 4 chairs
are made.
LEGO EXAMPLE: SLACK AND SURPLUS
The difference between the available
resources and resources used is either slack
or surplus.
Slack is associated with each less than or
equal to constraint, and represents the
amount of unused resource.
Surplus is associated with each greater than
or equal to constraint, and represents the
amount of excess resource above the stated
level.
LEGO EXAMPLE: SLACK AND SURPLUS
We have two slack values - one for large legos
and one for small legos - and no surplus
values.
The slack or surplus section shows that both
constraints have zero slack.
Suppose we must produce at least one table
(X1>=1). The original optimal solution is
still the best. Since X1=2, we produce one
surplus table.
RIGHT HAND SIDE CHANGES
Now we are ready to illustrate the key
concepts of sensitivity analysis.
How much would you be willing to spend
for one additional large Lego?
One additional large Lego is worth $4.
Original solution: X1=2, X2=2, profit =
$72;
New solution: X1=3, X2=1, profit = $76.
You would be willing to spend up to $4
(76-72) for one additional large Lego.
RIGHT HAND SIDE CHANGES
How much would you be willing to spend for
two additional large Legos?
Two additional large Legos are worth $8.
Original solution: X1=2, X2=2, profit = $72;
New solution: X1=4, X2=0, profit = $80.
You would be willing to spend up to $8 (8072) for two additional large Legos.
RIGHT HAND SIDE CHANGES
How much would you be willing to spend for
three more large legos?
The third large lego is not worth anything
since the optimal solution remains
unchanged.
What happens if your supplier can only
provide five large legos each day?
The optimal solution is: X1=1, X2=3,
profit=68, so we lose $4.
RIGHT HAND SIDE CHANGES
What happens if your supplier can only
provide four large legos each day?
The optimal solution is: X1=0, X2=4,
profit=64, so we lose another $4.
What happens if your supplier can only
provide three large legos each day?
The optimal solution is: X1=0, X2=3,
profit=48, so we lose an additional $16, and
not $4.
SHADOW PRICES
This last set of exercises enables us to
determine the shadow price for a resource
constraint (large legos).
The shadow price, for a particular constraint, is
the amount the objective function value will
increase (decrease) if the right hand side
value of that constraint is increased
(decreased) by one unit.
We found that the shadow price for large legos
is $4.
SHADOW PRICES
What is the shadow price of the small legos?
With two additional small legos the new solution:
X1=1, X2=4, profit = $84.
You would be willing to spend up to $12 (84-72) for
two additional small legos, so the shadow price is $6
(12/2).
SHADOW PRICES
In general, the shadow prices are meaningful
if one right hand side (RHS) value of a
constraint is changed,
and all other parameters of the model remain
unchanged.
REDUCED COSTS
What happens if the profit of tables increases
to $35?
The optimal solution is X1=3, X2=0, profit =
$105. Note that no chairs are being
produced.
REDUCED COSTS
When a decision variable has an optimal value
of zero, the allowable increase for the
objective function coefficient is also called
the reduced cost.
The reduced cost of a decision variable is the
amount the corresponding objective function
coefficient would have to change before the
optimal value would change from zero to
some positive value.
REDUCED COSTS
The reduced cost for tables is zero in the
original formulation. Why is this the case?
We are already producing tables.
What is the reduced cost for X2 and what
does it mean?
The reduce cost is -1.5 meaning that if I
force production of 1 chair profit will drop
by 1.5
SENSITIVITY ANALYSIS PROBLEM
A manufacturing firm has discontinued
production of a certain unprofitable product
line thus creating considerable excess
production capacity.
Management is considering devoting this
excess capacity to one or more of three
products; call them products 1, 2, and 3.
The available capacity on the machines that
might limit output is summarized below:
SENSITIVITY ANALYSIS PROBLEM
MACHINE TYPE
week)
Milling machine
Lathe
Grinder
AVAILABLE TIME
(machine hours /
500
350
150
SENSITIVITY ANALYSIS PROBLEM
The number of machine hours required for
each unit of the respective products is:
MACHINE TYPE
Milling machine
Lathe
Grinder
P1
9
5
3
P2
3
4
0
P3
5
0
2
SENSITIVITY ANALYSIS PROBLEM
The sales department indicates that the sales
potential for products 1 and 2 exceeds the
maximum production rate and that the sales
potential for product 3 is 20 units per week.
The unit profit would be $3000, $1200, and
$900, respectively, for products 1, 2, and 3.
SENSITIVITY ANALYSIS PROBLEM
Solver is used to determine the optimal
solution
a.
What are the optimal weekly production
levels for each of the three products?
Product 1 = 45.23
Product 2 = 30.95
Product 3 = 0
SENSITIVITY ANALYSIS PROBLEM
b.
What profit will be obtained if the
optimal solution is implemented?
$172,857.10 per week
c.How much unused capacity exists on the
milling machine, the lathe, and the grinder?
Milling = 0;
Lathe = 0;
Grinder = 14.28
(See SLACK entries)
SENSITIVITY ANALYSIS PROBLEM
d.
How much would the objective function
change if the amount of available time on
the grinder increased from 150 hours per
week to 250 hours?
Will the objective function increase or
decrease?
Currently the grinder has 14.28 hours of
slack time so its shadow price is 0.
Increasing the available hours from 150 to
250 will not change the total profit.
SENSITIVITY ANALYSIS PROBLEM
e.
The profit for product 3 is $900 per unit
and the current production level is zero.
How much would the profit per unit have to
change before it would be profitable to
produce product 3's?
The profit per unit would have to increase
by its reduced cost of $528.57.
SENSITIVITY ANALYSIS PROBLEM
f.
The milling machine capacity can be
increased at a cost of $160 per hour. Is it
economic to increase capacity by 10 hours?
The shadow price per hour (285.7143) is greater than
the cost (160), so it is worth increasing milling
capacity on the margin. However, since the shadow
price might change with increasing capacity we need
to rerun to see the full effect of increasing capacity by
10 hours. Profit does increase by 2857 (10*285.714)
which is greater than the cost increase of 1600
(10*160). Therefore the milling capacity should be
increased b y 10 hours.
TRANSPORTATION PROBLEM
 Mathematical
programming has been
successfully applied to important supply
chain problems.
 These
problems address the movement of
products across links of the supply chain
(supplier, manufacturers, and customers).
 We
now focus on supply chain applications in
transportation and distribution planning.
TRANSPORTATION PROBLEM
A manufacturer ships TV sets from three warehouses to four
retail stores each week. Warehouse capacities (in hundreds)
and demand (in hundreds) at the retail stores are as follows:
Capacity
Warehouse 1
Warehouse 2
Warehouse 3
200
150
300
650
Demand
Store 1 100
Store 2 200
Store 3 125
Store 4 225
650
TRANSPORTATION PROBLEM
The shipping cost per hundred TV sets for each route
is given below:
From
warehouse 1
warehouse 2
warehouse 3
To
Store 1 Store 2
$10
5
4
9
15
8
Store 3
12
15
6
Store 4
3
6
11
TRANSPORTATION PROBLEM
What are the decision variables?
XIJ=number of TV sets (in cases) shipped
from warehouse I to store J
I is the index for warehouses (1,2,3)
J is the index for stores (1,2,3,4)
TRANSPORTATION PROBLEM
What is the objective?
Minimize the total cost of transportation which
is obtained by multiplying the shipping cost by
the amount of TV sets shipped over a given
route and then summing over all routes
OBJECTIVE FUNCTION ;
MIN = 10*X11+5*X12+12*X13+ 3*X14+
4* X21+9*X22+15*X23+ 6*X24+
15*X31+8*X32+ 6*X33+11*X34
TRANSPORTATION PROBLEM
How are the supply constraints expressed?
For each warehouse the amount of TV sets
shipped to all stores must equal the
capacity at the warehouse
X11+X12+X13+X14=200; SUPPLY
CONSTRAINT FOR WAREHOUSE 1
X21+X22+X23+X24=150; SUPPLY
CONSTRAINT FOR WAREHOUSE 2
X31+X32+X33+X34=300; SUPPLY
CONSTRAINT FOR WAREHOUSE 3
TRANSPORTATION PROBLEM
How are the demand constraints expressed?
For each store the amount of TV sets shipped
from all warehouses must equal the demand
of the store
X11+X21+X31=100; DEMAND
CONSTRAINT FOR STORE 1
X12+X22+X32=200; DEMAND
CONSTRAINT FOR STORE 2
X13+X23+X33=125; DEMAND
CONSTRAINT FOR STORE 3
X14+X24+X34=225; DEMAND
CONSTRAINT FOR STORE 4
TRANSPORTATION PROBLEM
Since partial shipment cannot be made,
the decision variables must be integer –
valued
However, if all supplies and demands are
integer-valued, the values of our
decision variables will be integer
valued
TRANSPORTATION PROBLEM
After solution in Solver:
The total shipment cost is $3500, and the
optimal shipments are: warehouse 1 ships
25 cases to store 2 and 175 to store 4;
warehouse 2 ships 100 to store 1 and 50
to store 4, and warehouse 3 ships 175 to
store 2 and 125 to store 3.
The reduced cost of X11 is 9, so the cost of
shipping from warehouse 1 to store 1
would have to be reduced by $9 before
this route would be used
UNBALANCED PROBLEMS
Suppose warehouse 2 actually has 175 TV
sets. How should the original problem be
modified?
Since total supply across all warehouses is
now greater than total demand, all supply
constraints are now “<=“
Referring to the original problem, suppose
store 3 needs 150 TV sets. How should the
original problem be modified?
The demand constraints are now “<=“
RESTRICTED ROUTE
Referring to the original problem, suppose there
is a strike by the shipping company such that
the route from warehouse 3 to store 2 cannot
be used.
How can the original problem be modified to
account for this change?
Add the constraint:
X32=0
WAREHOUSE LOCATION
Suppose that the warehouses are currently not open,
but are potential locations.
The fixed cost to construct warehouses and their
capacity values are given as:
WAREHOUSES
Warehouse 1
Warehouse 2
Warehouse 3
FIXED COST CAPACITY
125,000
300
185,000
525
100,000
325
WAREHOUSE LOCATION
How do we model the fact that the
warehouses may or may not be open?
Define a set of binary decision variables
YI, I =1,2,3, where warehouse I is open if
YI = 1 and warehouse I is closed if YI = 0
WAREHOUSE LOCATION
How must the objective function change?
Additional terms are added to the objective
function which multiply the fixed costs of
operating the warehouse by YI and
summing over all warehouses I:
125000*Y1+185000*Y2+100000*Y3
Why can’t we use the current capacity
constraints?
Product cannot be shipped from a
warehouse if it is not open. Since the
capacity is available only if the warehouse
is open, we multiply warehouse 1’s capacity
by Y1.
WAREHOUSE LOCATION
Also, we must make the YI variables
binary integer
Total fixed and shipping costs are
$289,100; warehouses 2 and 3 are
open; warehouse 2 ships 100 to store
1, 225 to store 4; and warehouse 3
ships 200 to store 2 and 125 to store 4
FINALLY, BACK TO THE MOTIVATING
PROBLEM. . .
NETWORK SCHEDULING EXAMPLE
 Single
 Two

product
plants p1 and p2
Plant p2 has an annual capacity of 60,000
units.
 The
two plants have the same production
costs.
 There
are two warehouses w1 and w2
with identical warehouse handling costs.
 There
are three markets areas c1,c2 and
c3 with demands of 50,000, 100,000 and
50,000, respectively.
UNIT DISTRIBUTION COSTS
Facility
warehouse
p1
p2
c1
c2
c3
w1
0
4
3
4
5
w2
5
2
2
1
2
THE NETWORK
D = 50,000
D = 100,000
Cap = 60,000
D = 50,000
THE OPTIMIZATION MODEL
This problem can be framed as the following
linear programming problem:
Let
 x(p1,w1), x(p1,w2), x(p2,w1) and x(p2,w2) be the
flows from the plants to the warehouses.
 x(w1,c1), x(w1,c2), x(w1,c3) be the flows from the
warehouse w1 to customer zones c1, c2 and c3.
 x(w2,c1), x(w2,c2), x(w2,c3) be the flows from
warehouse w2 to customer zones c1, c2 and c3
OPTIMAL SOLUTION
Facility
warehouse
p1
p2
c1
c2
c3
w1
140,000
0
50,000
40,000
50,000
w2
0
60,000
0
60,000
0
Total cost for the optimal strategy is
$740,000