Production and Operations Management: Manufacturing and Services

1
Resource Allocation
Class 7 3/9/11
2
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

3
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)

4
Network Scheduling Example

• Single product
• Two 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.

5
Unit Distribution Costs
Facility warehouse p1 p2 c1 c2 c3
w1 0 4 3 4 5
w2 5 2 2 1 2
6
Heuristic 1Choose the Cheapest Warehouse to
Source Demand
D 50,000
2 x 50,000
D 100,000
5 x 140,000
1 x 100,000
2 x 60,000
Cap 60,000
D 50,000
2 x 50,000
Total Costs 1,120,000
7
Heuristic 2Choose the warehouse where the
total delivery costs to and from the warehouse
are the lowestConsider inbound and outbound
distribution costs
0
D 50,000
3
P1 to WH1 3 P1 to WH2 7 P2 to WH1 7 P2 to WH
2 4
4
2
5
D 100,000
5
P1 to WH1 4 P1 to WH2 6 P2 to WH1 8 P2 to WH
2 3
4
1
2
Cap 60,000
D 50,000
2
P1 to WH1 5 P1 to WH2 7 P2 to WH1 9 P2 to WH
2 4
Market 1 is served by WH1, Markets 2 and 3 are
served by WH2
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Heuristic 2Choose the warehouse where the
total delivery costs to and from the warehouse
are the lowestConsider inbound and outbound
distribution costs
0 x 50,000
D 50,000
3 x 50,000
Cap 200,000
P1 to WH1 3 P1 to WH2 7 P2 to WH1 7 P2 to WH
2 4
D 100,000
5 x 90,000
P1 to WH1 4 P1 to WH2 6 P2 to WH1 8 P2 to WH
2 3
1 x 100,000
2 x 60,000
Cap 60,000
D 50,000
2 x 50,000
P1 to WH1 5 P1 to WH2 7 P2 to WH1 9 P2 to WH
2 4
Total Cost 920,000
9
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?

10
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).

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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.

12
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.

13
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

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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.

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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.

16
ASSUMPTIONS

• Non-negativity All decision variables must take
  on positive or zero values.

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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.

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LEGO EXAMPLE Introduction
Pictures of a table and a chair are shown below.
Table
Chair
19
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.

20
Using Solver

• Solver in Excel can be used to obtain the
  solution and will now be demonstrated
• The problem formulation is
• MAX 20X116X2
• Subject to
• 2X11X2lt6
• 2X12X2lt8
• How to enter this formulation into Solver?

21
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.

22
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.

23
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
  (X1gt1). The original optimal solution is still
  the best. Since X12, we produce one surplus
  table.

24
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 X12, X22, profit 72
• New solution X13, X21, profit 76.
• You would be willing to spend up to 4 (76-72)
  for one additional large Lego.

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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 X12, X22, profit 72
• New solution X14, X20, profit 80.
• You would be willing to spend up to 8 (80-72)
  for two additional large Legos.

26
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 X11, X23, profit68,
  so we lose 4.

27
RIGHT HAND SIDE CHANGES

• What happens if your supplier can only provide
  four large legos each day?
• The optimal solution is X10, X24, profit64,
  so we lose another 4.
• What happens if your supplier can only provide
  three large legos each day?
• The optimal solution is X10, X23, profit48,
  so we lose an additional 16, and not 4.

28
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.

29
SHADOW PRICES

• What is the shadow price of the small legos?
• With two additional small legos the new solution
• X11, X24, 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).

30
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.

31
REDUCED COSTS

• What happens if the profit of tables increases to
  35?
• The optimal solution is X13, X20, profit
  105. Note that no chairs are being produced.

32
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.

33
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

34
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

35
SENSITIVITY ANALYSIS PROBLEM

• AVAILABLE TIME
• MACHINE TYPE (machine hours / week)
• Milling machine 500
• Lathe 350
• Grinder 150

36
SENSITIVITY ANALYSIS PROBLEM

• The number of machine hours required for each
  unit of the respective products is
• MACHINE TYPE P1 P2 P3
• Milling machine 9 3 5
• Lathe 5 4 0
• Grinder 3 0 2

37
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.

38
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

39
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)

40
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.

41
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.

42
SENSITIVITY ANALYSIS PROBLEM

• 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 (10285.714) which is greater
  than the cost increase of 1600 (10160).
  Therefore the milling capacity should be
  increased b y 10 hours.

43
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.

44
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 Demand
• Warehouse 1 200 Store 1 100
• Warehouse 2 150 Store 2 200
• Warehouse 3 300 Store 3 125
• 650 Store 4 225 650

45
TRANSPORTATION PROBLEM

• The shipping cost per hundred TV sets for each
  route is given below
• To
• From Store 1 Store 2 Store 3 Store 4
• warehouse 1 10 5 12 3
• warehouse 2 4 9 15 6
• warehouse 3 15 8 6 11

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(No Transcript)
47
TRANSPORTATION PROBLEM

• What are the decision variables?
• XIJnumber 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)

48
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 10X115X1212X13 3X14
• 4 X219X2215X23 6X24
• 15X318X32 6X3311X34

49
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
• X11X12X13X14200 SUPPLY CONSTRAINT FOR
  WAREHOUSE 1
• X21X22X23X24150 SUPPLY CONSTRAINT FOR
  WAREHOUSE 2
• X31X32X33X34300 SUPPLY CONSTRAINT FOR
  WAREHOUSE 3

50
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
• X11X21X31100 DEMAND CONSTRAINT FOR STORE 1
• X12X22X32200 DEMAND CONSTRAINT FOR STORE 2
• X13X23X33125 DEMAND CONSTRAINT FOR STORE 3
• X14X24X34225 DEMAND CONSTRAINT FOR STORE 4

51
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

52
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

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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 lt
• Referring to the original problem, suppose store
  3 needs 150 TV sets. How should the original
  problem be modified?
• The demand constraints are now lt

54
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
• X320

55
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 FIXED COST CAPACITY
• Warehouse 1 125,000 300
• Warehouse 2 185,000 525
• Warehouse 3 100,000 325

56
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

57
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
• 125000Y1185000Y2100000Y3
• Why cant 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 1s capacity by Y1.

58
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

59
Finally, back to the motivating problem. . .
60
Network Scheduling Example

• Single product
• Two 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.

61
Unit Distribution Costs
Facility warehouse p1 p2 c1 c2 c3
w1 0 4 3 4 5
w2 5 2 2 1 2
62
The Network
D 50,000
D 100,000
Cap 60,000
D 50,000
63
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

64
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