US Army Logistics Management College
Math
Programming
Intro to Optimization Modeling,
Linear Programming Models,
and Network Models
Math Programming - 1
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Introduction to LP Modeling
Producing Frames at the Monet Company
Math Programming - 2
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Background Information
Producing Frames at the Monet Company
• The Monet Company produces four type of picture
frames that differ with respect to size, shape,
materials used, labor required and selling price
• During the coming week Monet can purchase up to
4000 hours of skilled labor, 6000 ounces of metal,
and 10,000 ounces of glass
• Market constraints limit sales to no more than 1000
type 1 frames, 2000 type 2, frames, 500 type 3
frames, and 1000 type 4 frames
• The company wants to maximize its weekly profit
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Traditional Algebraic Method
• The resulting algebraic formulation is shown below:
Maximize
6x1 + 2x2 + 4x3 + 3x4 (profit objective)
Subject to
2x1 + x2 + 3x3 + 2x4  4000 (labor constraint)
4x1 + 2x2 + x3 + 2x4  6,000 (metal constraint)
6x1 + 2x2 + x3 + 2x4  10,000 (glass constraint)
x1  1000 (frame 1 sales constraints)
x2  2000 (frame 2 sales constraints)
x3  500 (frame 3 sales constraints)
x4  1000 (frame 4 sales constraints)
x1, x2, x3, x4  0 (non-negativity constraint)
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LP Spreadsheet Modeling
• There is no exact one way to develop an LP spreadsheet model.
• The common elements in all LP spreadsheet models are the
following:
Inputs: all numeric data needed to form the objective and the constraints.
Our convention is to enclose all inputs in a blue border with shading in the
upper left corner.
Changing cells: a set of designed cells that play the roles of the decision
variables. We will enclose them in a red border.
Target (objective) cell: a single cell that contains the value of the objective.
Our convention is to enclose the target cell in a double-line border.
Constraints: represented by formulas in cells and identified in a Solver
dialog box.
Non-negativity: specified by selecting an option in the Solver dialog box.
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LP Spreadsheet Modeling
In general, the complete solution of the problem
involves three stages:
1) Formulation: enter all the inputs, trial values for the
changing cells, and formulas relating these in a
spreadsheet.
2) Set Solver Settings: formally designate the objective cell,
the changing cells, the constraints, and selected options,
and we tell the Solver to find the optimal solution.
3) Sensitivity Analysis: see how the optimal solution changes
as we vary inputs. Also, a sensitivity analysis often gives us
important insights about how the model works.
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Model Formulation
Producing Frames at the Monet Company
• The first stage is to setup the spreadsheet
– Inputs: Enter numeric inputs in the shaded ranges.
– Production levels: Enter trial values in the range named
Produced
– Resources used: Enter formulas for labor, metal, & glass
=SUMPRODUCT (B9:E9,Produced) in cell B21 and copy
– Revenue: enter the formula =B12*B16 in cell B27 and copy
– Cost: enter the formula =$B4*B$16*B9 in cell B29 and copy
– Profit: enter the formula =B27-SUM(B29:B31) in cell B32 and
copy
– Calculate total revenues, costs, and profit in column F
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Model Formulation
Enter Trial Values
• Why should we input trial values in the changing cells?
– Verification: entering different sets of values in the changing
cells allow us to confirm that the formulas are working correctly
– Trial values help us to understand the model
– Good estimates of trial values can help find the solution in a
shorter number of iterations
• For example, it is tempting to guess that the frame
types with the highest profit margins should be
produced to the greatest extent possible
• Does this guarantee that this solution is the best
possible product mix? Unfortunately, it does not!
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Set Solver Settings
Producing Frames at the Monet Company
• The second stage is to estimate an initial solution and
specify the Solver settings (Solver is on Data Ribbon)
• To enter information, type cell references or point, click
and drag
– Objective: Select TotProfit cell as the target cell (set to
Maximize)
– Changing cells: Select the Produced range, (numbers of
frames produced)
– Constraints: Click on the Add button to add the following
constraints:
» Used<=Available (can’t use more than available resources)
» Produced<=MaxSales (produce no more than demand of
each product)
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Set Solver Settings
Producing Frames at the Monet Company
• Setting Solver Options
– Non-negativity: Make all changing cells non-negative
– Linear model: Uses a Simplex based algorithm to solve the model
» Changing cells cannot be multiplied by other changing cells
used as a Solver input
» Changing cells cannot be raised to a power or be a part of
other non-linear EXCEL functions
» The changing cell cannot be a part of the right hand side of a
constraint in the Solver constraint box
» If a model is not linear and assume linear model is checked, an
error message will be displayed
» If assume linear model is NOT checked, a non-linear solution
method will be used
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Set Solver Settings
Producing Frames at the Monet Company
• Optimize: Click on the Solve button in the Solver
dialog box
• Possible Solver Results
– Optimal solution found
– Infeasible solution (over-constrained or model error)
– Target Cell value does not converge (Unbounded problem –
check for model error)
– Assume Linear Model conditions not satisfied (problem has
not been set up properly to meet linear conditions)
– Other messages due to time/iteration/memory limits or
formula errors in the model
• Options for creating post-analysis reports
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Optimal Solution Results
Producing Frames at the Monet Company
• The optimal plan is to produce 1,000 type 1 frames,
800 type 2 frames, 400 type 3 frames, and no type 4
frames (earn $9200 profit)
• Use all of the available labor and metal (binding
constraints) but only 8,000 of the 10,000 available
ounces of glass (non-binding constraint)
• Think of binding constraints as “bottlenecks” (prevent
Monet from earning even higher profits)
• More profit could be made by producing more of
frames 2, 3, and 4 if more labor and metal were
available
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Sensitivity Analysis
Producing Frames at the Monet Company
• Sensitivity Analysis: changing one or more parameters
(numeric data) to see the impact on the solution
• To experiment with different inputs to this problem (unit
revenues or resource availabilities for example), change
the inputs and rerun Solver
• You do not have to change Solver settings as long as
you don’t change the model itself
• Example:
– Change the unit selling price for frame type 4 from $21.50 to
$26.50 (all other inputs remain the same)
– By making type 4 frames more profitable, they enter the optimal
solution mix
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Sensitivity Analysis
Producing Frames at the Monet Company
• Sensitivity Report: Provides insight into changes
made to objective coefficients and constraint right
hand sides
– Reduced Cost: the amount the objective coefficient would
need to change before that variable (changing cell) would be
in the optimal solution mix
– Shadow Price (Dual Price): the amount the objective
function (target cell) would change if we changed the right
hand side value of that constraint by one unit
– Allowable Increase/Decrease: the valid range of the
reduced cost/shadow price changes
• Can only change one parameter at a time!
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Sensitivity Analysis and the
Solver Table Add-In
• Solver Table Add-In allows analysis of one or two
parameter changes over a range specified by the
modeler
• Similar to data table, except will optimize each
cell entry in the table
• Provides for more flexibility and visual display of
sensitivity analysis than sensitivity analysis report
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Solver Table Add-In
Producing Frames at the Monet Company
• Setting up a Solver Table:
– Select Data/SolverTable menu item
– Select Oneway or Twoway table in the first dialog box
– Select input cell(s) (parameter you want to change)
– Select values for input cells (either min, max, increment or
list specific values)
– Select output cell(s) (typically target cell and others)
– Select location of table (specify upper left corner)
• Click OK and Solver solves a separate optimization
problem for each of the cells in the table and reports
the requested outputs
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Solver Table Add-In Example Producing
Frames at the Monet Company
•
Check the sensitivity of profit and the optimal product
mix to changes in the number of labor hours available
– Columns B through E show how the product mix changes as
more labor hours become available
– Extra labor hours increase total profit (col F)
– Profit increase as labor increases is in col G (not part of
Solver Table)
– Three different shadow prices over this range of labor hours
Shadow price:
$500/250 = $2
Shadow price:
$300/250 = $1.20
Shadow price:
$250/250 = $1
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Solver Table Add-In Example Producing
Frames at the Monet Company
• Notice the shadow price for labor is not constant.
• The line chart below illustrates how the shadow price
decreases as labor hours increase
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Solver Table Add-In Example
Producing Frames at the Monet Company
• Check the sensitivity of profit and the optimal product
mix to changes in the cost per ounce of metal
– The optimal product mix remains unchanged for a cost of
metal in the $.30 to $.70 range
– Within this range, the only thing that changes is the profit, and
it decreases only because metal gets more expensive
– Once the cost goes above $.70, the product mix changes (and
profit decreases)
– Intuitively, once metal becomes expensive enough, products
that use metal most heavily become less attractive
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Solver Table Add-In Example
Producing Frames at the Monet Company
• Check how
sensitive the
optimal profit
is to
simultaneous
changes in
the hourly
labor cost
and the total
labor hours
available
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Solution
Producing Frames at the Monet Company
• Total profit decreases in each row as the hourly labor
cost increases
• Total profit increases in each column as the available
labor increases
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Aggregate Planning Models
Worker and Production Planning at SureStep
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Background Information
Worker and Production Planning at SureStep
• During the next four months the SureStep Company
must meet (on time) the following demands for pairs
of shoes: 3,000 in month 1; 5,000 in month 2; 2,000
in month 3; and 1,000 in month 4
• At the beginning of month 1, 500 pairs of shoes are
on hand, and SureStep has 100 workers
• A worker is paid $1,500 per month and can work up
to 160 hours a month before he or she receives
overtime
• A worker may be required to work up to 20 hours of
overtime per month and is paid $13 per hour for
overtime labor
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Background Information
Worker and Production Planning at SureStep
• It takes 4 hours of labor and $15 of raw material to
produce a pair of shoes
• At the beginning of each month workers can be hired
or fired
• Each hired worker costs $1600, and each fired worker
cost $2000
• At the end of each month, a holding cost of $3 per pair
of shoes left in inventory is incurred
• SureStep wants to determine its optimal production
schedule and labor policy
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Model Formulation
Worker and Production Planning at SureStep
• To model SureStep’s problem with a spreadsheet,
we must keep track of the following:
– Number of workers hired, fired, and available during each
month.
– Number of pairs of shoes produced each month with regular
time and overtime labor
– Number of overtime hours used each month
– Beginning and ending inventory of shoes each month
– Monthly costs and the total costs
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Model Formulation
Worker and Production Planning at SureStep
• To develop this model, proceed as follows.
– Inputs: Enter the input data in the range B4:B14 and in the
Demand range.
– Production, hiring and firing plans: These values comprise the
changing cells in the problem
» Number of pairs of shoes produced each month
» Number of overtime hours used each month
» Number of workers hired and fired each month
– Workers available each month:
» Initial number of workers available (cell B17 =InitWorkers)
» Number of workers available in month 1 (cell B20 =B17+B18B19 and copy this formula to the range C20:E20)
» Number of workers at beginning = number of workers at end of
previous month (cell C17=B20 and copy to range D17:E17)
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Model Formulation
Worker and Production Planning at SureStep
– Overtime capacity: Each available worker can work up to 20
hours of overtime in a month (cell B25 =MaxOTHrs*B20 and
copy it to the range C25:E25)
– Production capacity:
» Each worker can work 160 regular-time hours per month
(cell B22 =StdRTHrs*B20 and copy it to the range
C22:E22)
» Sum the total monthly production hours available (cell B27
=SUM(B23:B24) and copy it to the range C27:E27)
» Calculate the production capacity using the labor hours for
shoes (cell B32 =B27/HrsPerPair and copy it to the range
C32:E32)
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Model Formulation
Worker and Production Planning at SureStep
– Monthly inventory:
» Inventory after production in month 1 (cell B34 =InitInv+B30)
» For other months, inventory after production is the previous
month’s ending inventory plus this month’s production (cell
C34 =B37+C30 and copy it to the range D34:E34)
» Ending inventory is inventory after production less demand
(cell B37 =B34-B36 and copy it to the range C37:E37)
– Monthly costs: Calculate the various costs shown in rows 40
through 45
(cell B40 =UnitHireCost*B18, cell B41 =UnitFireCost*B19,
cell B42 =RTWageRate*B20, cell B43 =OTWageRate*B23,
cell B44 =UnitMatCost*B30, cell B45 =UnitHoldCost*B37
and copy B40:B45 to the range C40:E45)
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Set Solver Settings
Worker and Production Planning at SureStep
• Invoking the Solver
– Objective: Target cell is minimize TotCost
– Changing cells: Hired, Fired, Production, and OTHrs
– Overtime constraints: OTHrs <= OTAvailable
– Production capacity constraint: Production<=ProdCap
– Demand constraint: OnHand>=Demand
– Integer constraints: Constrain the number hired and fired to
be integers
– Specify non-negativity, assume linear, and optimize
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Optimal Solution
Worker and Production Planning at SureStep
• The company produces slightly over 3700 pairs of
shoes during each of the first 2 months, 200 pairs in
month 3, and 1000 in month 4
• SureStep should never hire any workers
• Fire 6 workers in month 1, 1 worker in month 2, and
43 workers in month 3
• Eighty hours of overtime are used (only in month 2)
• Minimum total cost of $692,820
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Optimal Solution – Integer Considerations
Worker and Production Planning at SureStep
• It is important to ensure that the number of workers
hired and fired each month is an integer, given the small
number of workers involved
• The number of pairs of shoes produced each month
does not have to be solved as an integer (due to the
large quantity)
• To ensure that Solver finds the optimal solution in a
problem where some or all of the changing cells must
be integers, go into Options, then to Integer Options,
and set the tolerance to 0
• If the tolerance is set to a value other than 0, Solver
might stop when it finds a solution that is close to
optimal
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Model with Backlogging Allowed
Worker and Production Planning at SureStep
• In many situations backlogging is allowed, that is,
customer demand, can be met later than it occurs.
• We’ll modify this example to include the option of
backlogged demand.
• We assume that at the end of each month a cost of
$20 is incurred for each unit of demand that remains
unsatisfied at the end of the month.
• This is easily modeled by allowing a month’s ending
inventory to be negative. The last month, month 4,
should be nonnegative. This also ensures that all
demand will eventually be met by the end of the fourmonth horizon.
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Model with Backlogging Allowed
Worker and Production Planning at SureStep
• Model the previous problem to allow backlogging
(demand does not have to be met in the same month,
but can be met later at some additional cost)
• To allow backlogging, modify the monthly cost
computations to incorporate the costs due to
shortages
• Two possible modeling approaches:
– The first is the more “natural”, but it results in a nonlinear
model
– The second is more complicated, but allows for a linear
model
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Backlogging – Nonlinear Approach Worker
and Production Planning at SureStep
• Enter the per unit monthly shortage cost in the
UnitShortCost cell
• Remove <= signs for months 1-3
• Adjust holding and shortage cost formulas in rows 46
and 47:
– For holding costs:
cell B46 =IF(B38>0,UnitHoldCost*B38,0) and copy across
– For shortage costs:
cell B47 =IF(B38<0,-UnitShortCost*B38,0) and copy across
• Adjust Solver Settings so that inventory on hand
meets or exceeds demand for month 4 only
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Backlogging – Nonlinear Approach Worker and
Production Planning at SureStep
• Use of IF functions make the objective function
nonlinear (Do not check Assume Linear option)
• Problem with nonlinear approach: solution found not
guaranteed to be global optimal
• Tips on finding good nonlinear solutions:
– Start with initial trial values for your adjustable cells that
seem like a good solution
– Vary your initial trial values and see if the solution changes
– Read the help screen for the Solver Options dialog box to
understand settings associated with nonlinear problems
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Backlogging –Linear Approach
Worker and Production Planning at SureStep
• Enter per unit monthly shortage cost in the
UnitShortCost cell
• Create rows for excess, shortage, and net inventory
(B39:E41):
– The excess range in row 39 contains the amounts left in
inventory
– The shortage range in row 40 contains unmet demand
(backlog)
– The net Inventory is excess inventory – shortage inventory
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Backlogging –Linear Approach
Worker and Production Planning at SureStep
• Note that if we force net inventory to equal the ending
inventory (which equals inventory after production
less demand), then one of three conditions occur:
– Excess inventory > 0, shortage inventory = 0 (surplus)
– Excess inventory = 0, shortage inventory > 0 (shortage)
– Excess inventory = 0, shortage inventory = 0 (balanced)
• Adjust holding and shortage cost formulas in rows 51
and 52
–
–
–
–
For holding costs: cell B51 =UnitHoldCost*B39
For shortage costs: cell B52 =UnitShortCost*B40
Copy the range B51:B52 to the range C51:E52
Make sure the totals in row 53 and column F include the
shortage costs
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Backlogging –Linear Approach
Worker and Production Planning at SureStep
• The changes from the original Solver setup are as
follows:
– Extra changing cells: Add the Excess and Shortage ranges
as changing cells
– Constraint on last month’s inventory: Change the
constraints Onhand>=Demand to LastOnhand>=LastDemand
(this allows months 1 through 3 to have negative ending
inventory and ensures that all demand is met by the end of
month 4)
– Logical constraint on ending inventory: Add the constraints
Net=EndInv (ensures that net inventory is equal to the actual
ending inventory)
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Comparing Results
Backlog Models versus No Backlog
• The linear and nonlinear backlog solutions are the
same
• This solution is quite similar to the solution with no
backlogging allowed
– SureStep fires more workers in month 3 than before
– Purposely incurs shortages in months 2 and 3
• The company’s total cost cannot be any more than
when backlogging was not allowed (think of
backlogging as an addition option available)
• The cost savings with a backlog option are minor: from
$692,820 to $690,180.
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Sensitivity Analysis
Worker and Production Planning at SureStep
• There are many sensitivity analyses we could
perform on this final SureStep model
• Example: how will the total cost and the shortages
SureStep is willing to incur in months 1-3 vary with
the unit shortage cost
• To set up the SolverTable:
– Specify a one-way table
– Make UnitShortCost the input range (range between 0 and
35)
– Specify the TotCost cell and the range B40:D40 as the
output cells.
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Sensitivity Analysis
Worker and Production Planning at SureStep
• When the unit shortage cost is below $20,
SureStep is willing to incur large shortages – at a
significantly lower total cost
• Shortages become much less attractive when the
unit shortage cost increases
• No shortages are incurred at all when the unit
shortage cost is above $25
• When the shortage cost is above $25, we get the
same solution as when shortages are disallowed
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The Rolling Planning Horizon Approach
• In reality, an aggregate planning model is usually implemented via a
rolling planning horizon
• Let’s assume that SureStep works with a 4-month planning horizon
• To implement the SureStep model in the rolling planning horizon
context:
– View the “demands” as forecasts and solve a 4-month model with
these forecasts
– Implement only the month 1 production and work scheduling
recommendation
– Observe month 1’s actual demand and ending inventory and
workers available.
– Input inventory and worker results from month 1 in cells B4 & B5
– Replace demands in the Demands range with the updated
forecasts for the next 4 months and resolve
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The Rolling Planning Horizon Approach
Worker and Production Planning at SureStep
• Rolling Planning Horizon Example:
– Based on original demands/inputs, SureStep should hire no
workers, fire 6 workers, and produce 3760 pairs of shoes with
regular time labor in month 1
– Assume that month 1’s actual demand was 2950
– Begin month 2 with 1310 pairs of shoes (input in B4) and 94
workers (input in B5)
– Replace demands in the Demands range with the updated
forecasts for the next 4 months
– Rerun Solver and use the production levels and hiring and firing
recommendations in column B as the production level and
workforce policy for month 2
• Just like the caissons, the planning horizon goes rolling
along!
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Modeling Issues
Worker and Production Planning at SureStep
1. Hiring costs include training costs as well as the cost
of decreased productivity due to the fact that a new
worker must learn his or her job
2. Firing costs include severance costs and costs due
to loss of morale
3. Peterson and Silver recommend that when demand
is seasonal, the planning horizon should extend
beyond the next seasonal peak
4. Beyond a certain point, the cost of using extra hours
of overtime labor increases because workers
become less efficient
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Minimum Cost Network
Flow Models
Producing and Shipping Tomato
Products at RedBrand
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Background Information
Producing and Shipping Tomato Products at RedBrand
S:200
1
4
T:0
6 D:400
S:300 2
7 D:180
3
S:100
5 T:0
• The RedBrand Company produces tomato products at three plants
• These products can be shipped directly to their two customers or
they can first be shipped to the company’s two warehouses and
then to the customers
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Background Information
Producing and Shipping Tomato Products at RedBrand
• The cost of producing food at each plant is the same, so
RedBrand is concerned with minimizing the total
shipping cost incurred in meeting customer demands
• The network diagram shows the production capacity of
each plant (tons per year) and demand of each customer
• The cost of shipping a ton of food (In thousands of
dollars) between each pair of points is given in the
workbook
• At most 200 tons of food can be shipped between any
two nodes
• We must track amount shipped along each arc, inflows
and outflows into nodes, and total shipping cost
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Developing the Model
Producing and Shipping Tomato Products at RedBrand
• The steps are:
– Input data: Enter the unit shipping cost (in thousands of dollars),
the common arc capacity, the supply capacities, and the demands
in the appropriate cells
– Origin and destination indexes: Enter the indexes (1 to 7) for the
origins and destinations of the various arcs in the range A18:B43
– Shipping costs on arcs: cell C18 =INDEX(CostMatrix,A18,B18)
and copy (transfers data in the CostMatrix range to the UnitCosts
range)
– Flow on arcs: Enter initial values in D18:D43 (changing cells)
– Arc capacities: cell F18 =ArcCapacity and copy (common arc
capacity for all arcs)
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Developing the Model
Producing and Shipping Tomato Products at RedBrand
– Flow balance constraints: Nodes 1, 2, and 3 are supply nodes,
nodes 4 and 5 are transshipment points, and nodes 6 and 7 are
demand nodes
» Supply nodes: flow out cannot exceed supply
I19 =SUMIF(Origins,H19,Flows)-SUMIF(Dests,H19,Flows)
and copy down
» Transshipment nodes: flow in must equal flow out
Copy the supply node formula to cells I25 and I26
» Demand nodes: flow in must meet demand
I30 =SUMIF(Dests,H30,Flows)-SUMIF(Origins,H30,Flows)
and copy down (NOTE: >= since aggregate supply is greater
than aggregate demand)
– Total cost: TotCost cell =SUMPRODUCT(UnitCosts,Flows)
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Optimal Solution
Producing and Shipping Tomato Products at RedBrand
• Optimal solution total shipping cost = $3,260,000
• Plant 1 ships all of its production to plant 3
• Plant 2 ships some of its supply directly to customer 1
• All shipments from the warehouses go directly to customer 1 and
customer 1 ships 180 tons to customer 2
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Sensitivity Analysis
Producing and Shipping Tomato Products at RedBrand
• How much effect does the arc capacity have on optimal
solution?
– Currently, three arcs with positive flow are at the arc capacity of
200
– Use SolverTable to see how a change in arc capacity effects the
shipping pattern and total cost
– The single input cell is the ArcCapacity cell (vary from 150 to 300
in increments of 25)
– Track two outputs: total cost and number of arcs at capacity [arc
capacity: C47 =COUNTIF(Flows,ArcCapacity) ]
– As arc capacity decreases, more flows bump up against it, and
total cost increases
– Even at an arc capacity is 300, up to two flows are constrained
(multi-optimal solutions)
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Model Variation: Multi-product
Producing and Shipping Tomato Products at RedBrand
• Suppose RedBrand ships two products along the
given network
– Assume that the unit shipping costs are the same for
either product
– The arc capacity represents the maximum flow of both
products that can flow on any arc (the two products are
competing for arc capacity)
– Each plant has a separate production capacity for each
product and each customer has a separate demand for
each product
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Multi-product Model Formulation
Producing and Shipping Tomato Products at RedBrand
•
Very little needs to be changed from the original
model
•
If you rename cells appropriately, the Solver dialog
box does not have to be changed
•
Changes to the original model:
– have two columns of changing cells
– apply the previous logic to both products separately in
the flow balance constraints
– apply the arc capacities to the total flows in column F
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Model Variation: Flow loss Model
Producing and Shipping Tomato Products at RedBrand
•
A second variation of the model is appropriate for
perishable goods
•
Assume that there is a single product, but that some
percentage of the product that is shipped to
warehouses perishes and cannot be sent on to
customers
•
Having this flow loss means that the total inflow to a
warehouse is greater than the total outflow from the
warehouse
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Flow Loss Model Formulation
Producing and Shipping Tomato Products at RedBrand
• Changes to the original model:
– Add a new input cell that contains the “shrinkage factor”
(percentage that does not spoil in the warehouses)
– Incorporate shrinkage factor into the warehouse flow balance
constraints: I25 =SUMIF(Origins,H25,Flows) ShrinkFactor*SUMIF(Dests,H25,Flows) and copy down
– This formula says that what goes out is 90% of what goes in
• Shrinkage results in a larger total cost (about 50% larger)
– Some units are still sent to both warehouses, and the entire
capacity of all plants if now used
– A feasible solution exists even for a shrinkage factor of 0% - you
will send everything directly from plants to customers – at a steep
cost
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Math
Programming
Integer Programming and
Multi-objective Models
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Plant and Warehouse Location Models
Facility Location and Logistics
Planning at Huntco
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Background Information
Facility Location and Logistics Planning at Huntco
•
•
•
•
Huntco produces tomato sauce at five different plants.
Each plant has a finite supply capacity (in tons).
The tomato sauce is stored at one of three warehouses.
Huntco has four customers that receive shipments from the
warehouses based on their demand.
• Costs incurred include:
– production and shipping costs from plant to warehouse
– Shipping costs from warehouse to customer
– Plant and warehouse operating costs (annual fixed amount)
• Huntco’s goal is to minimize the annual cost of meeting
customer demands.
• The company wants to determine which plants and warehouses
to open, as well as the optimal shipping plan.
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Modeling Approach
Facility Location and Logistics Planning at Huntco
• To model Huntco’s problem, we need to track:
–
–
–
–
–
–
Shipments from plants to warehouses
Shipments from warehouses to customers
Fixed costs of operating plants and warehouses
Shipping and production costs from plants to warehouses
Shipping costs from warehouses to customers
Total amount shipped out of each plant
• We must also ensure that
– Huntco pays the fixed costs for all plants and warehouses that it
uses.
– The amount shipped into each warehouse equals the amount
received by each warehouse.
– Each customer receives the specified demand.
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Model Formulation
Facility Location and Logistics Planning at Huntco
To form the model, follow these steps:
• Inputs: Enter the given data in the shaded ranges.
• Shipments: Enter trial values for the shipments from
each plant to each warehouse in the Shipped1 range and
for the shipments from each warehouse to each customer
in the Shipped2 range.
• Binary fixed cost variables: Enter trial 0-1 values for the
plant fixed-cost variables in the UsePlants range and the
warehouse fixed-cost variables in the UseWhses range.
(1 = plant/warehouse is used, 0 =plant/warehouse not
used)
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Model Formulation
Facility Location and Logistics Planning at Huntco
• Amount shipped out of each plant: Enter the formula
=SUM(B30:D30) in cell E30 and copy
• Upper limit on amount shipped out of each plant: For
each plant we need a constraint of the form
Total shipped out of plant  Plant capacity * Fixed-cost
variable for plant
If Huntco uses a plant, then the company pays the plant’s
operating cost (the plant’s fixed-cost variable will equal 1)
If a plant is not used, the Solver is free to make this plant’s
fixed-cost variable 0, and no fixed cost for this plant will be
incurred.
Enter the formula =B21*H6 in cell G30 and copy
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Model Formulation
Facility Location and Logistics Planning at Huntco
• Amount shipped into and out of each warehouse: For each
warehouse, we need “flow balance”
Total shipments into warehouse = Total shipments out of
warehouse
Total shipments into warehouse: enter formula
=SUM(B30:B34) in cell B35 and copy
Total shipments out of warehouses: enter formula
=SUM(B42:E42) in cell F42 and copy
Put shipment out into a row: select the range B37:D37, enter
the formula =TRANSPOSE(ShippedOut2_Col) and press
Ctrl-Shift-Enter. This allows us to compare a row with a
row when we specify the equation in the Solver dialog box.
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Model Formulation
Facility Location and Logistics Planning at Huntco
• Upper limit on amount shipped out of each warehouse: For each
warehouse we need a constraint of the form
Total shipped out of warehouse  UpperBound * Fixed-cost
variable for warehouse
UpperBound is an upper bound on the most that could possibly be
shipped out of any warehouse. We’ll use the smaller of the total
demand for all customers and the total capacity for all plants.
If a warehouse’s fixed-cost variable is 0, then warehouse cannot be
used
If the fixed-cost variable is 1, then this inequality is easily satisfied.
We already have the total shipped out of warehouse
Warehouse upper bound: enter formula
=E21*MIN(SUM(Capacities),(SUM(Demands)) in cell H42 and
copy
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Model Formulation
Facility Location and Logistics Planning at Huntco
• Amount received by each customer: enter the
formula =SUM(B42:B44) in cell B45 and copy
• Shipping costs: Enter formulas for total costs of
shipping from plants to warehouses and from
warehouses to customers
=SUMPRODUCT(UnitCosts1,Shipped1) in cell B50
=SUMPRODUCT(UnitCosts2,Shipped2) in cell B51
• Fixed costs: Calculate annual fixed costs for operating
plants and warehouses
=SUMPRODUCT(FCosts1,UsePlants) in cell B52
=SUMPRODUCT(FCosts2,UseWhses) in cell B53
• Total cost: enter formula
=SUM(ShipCosts,FixedCosts) in cell B54
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Set Solver Settings
Facility Location and Logistics Planning at Huntco
• The completed Solver dialog box is shown here.
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Set Solver Settings
Facility Location and Logistics Planning at Huntco
• The following is the explanation of the setup of the
previous dialog box.
– Objective: Minimize total annual cost
– Changing cells: There are four sets of changing cells – two
sets for amounts to ship and two sets of binary variables for
which plants and warehouses to use
– Plant upper bounds: The constraint
ShippedOut1<=UpBounds1 operationalizes the 1st inequality.
– Warehouse upper bounds: The constraint
ShippedOut2_Col<=UpBounds2 operationalizes the 2nd
inequality.
– Warehouse balance: The constraint
ShippedIn1=ShippedOut2_Row operationalizes the equality.
– Demand constraints: The constraint ShippedIn2>=Demands
ensures that each customer received the required amount.
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Optimal Solution Results
Facility Location and Logistics Planning at Huntco
• Huntco should use plants 2, 3, and 5 and warehouses 2 and 3
(notice Shipped1 & Shipped2 ranges)
• Total annual cost of $700,500 (check that tolerance setting is
set to 0)
• At this point, you might want to review the inputs for this
problem and see whether the optimal solution appears
reasonable from an economic point of view
• For example, although plant 1 has a relatively small fixed cost,
it has relatively large unit shipping costs (so it is not used)
• However, the situation is not so obvious for plant 4 or
warehouse 1. We think you will agree that on logistics
problems such as this – and this is not even a large problem –
more than intuition is necessary!
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Sensitivity Analysis
Facility Location and Logistics Planning at Huntco
• We will not report any specific sensitivity analyses for
this model, but many are possible.
– For example, we might check whether adding larger
capacities at plants 1 and 4 would induce Huntco to
open them.
– Or we might see what would happen if all the fixed
costs increases by some percentage.
– Or we might see what would happen if all customer
demands increased by some percentage.
• SolverTable, after some slight model modifications,
can easily analyze any of these situations.
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Models with Either-Or Constraints
Manufacturing at Dorian Auto
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Background Information
Manufacturing at Dorian Auto
• Dorian Auto is considering manufacturing three types of
cars – compact, midsize, and large
• Each car type uses steel and labor resources and earns
a profit
• At present, 6000 tons of steel and 60,000 hours of labor
are available
• If any cars of a given type are produced, production of
that type of car will be economically feasible only if at
least 1000 cars of that type are produced
• Dorian wants to find a production schedule that
maximizes its profit
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Model Formulation
Manufacturing at Dorian Auto
• In addition to tracking the number of each car type produced,
labor hours, steel used, and profit, we must ensure that Dorian
produces either 0 or at least 1000 cars of each type – this is an
either-or constraint
• To develop the model, follow these steps:
• Inputs: Enter the input data into the shaded ranges
• Numbers of cars produced: Enter trial values for the number
of cars of each type produced in the UnitsProduced range
• Binary variables for minimum production: Enter any trial 0-1
values in the ProduceMin range (1=Dorian must produce at
least the minimum number of the corresponding car type, 0=do
not produce any of that type car)
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Model Formulation
Manufacturing at Dorian Auto
• The either-or constraints are implemented with the binary variables in
row 13 and the inequalities in rows 15 and 19
• Lower limits on production: enter the formula =B7*B13 in cell B15
and copy (if the binary variable in row 13 = 1, then produce at least
the minimum number of that car type, if the binary variable = 0, then
the lower bound is 0 and production must be nonnegative)
• Upper limits on production: enter the formula
=B13*MIN(SteelAvail/B5,LaborAvail/B6) in cell B19 and copy
- the MIN term in this formula is the max number of type cars Dorian
could make if it devoted all of its resources to that type car
- If the binary variable in row 13 is 1, this upper limit is essentially
redundant – we could never produce more than this many cars
- If the binary variable is 0, then the upper limit is 0, which prevents
Dorian from making any cars of this type
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Model Formulation
Manufacturing at Dorian Auto
• Summarizing the lower & upper limits, if the binary variable is 1,
the production limits become
Min production required  Production  Maxproduction
possible
• If the binary variable is 0, the limits become 0  Production 0
• Exactly one of these cases must hold for each car type, so they
successfully implement the either-or constraints
• Steel / labor used: enter the formula
=SUMPRODUCT(B5:D5,UnitsProduced) in cell B22 and copy
to calculate the tons of steel and number of labor hours used
• Profit: Calculate the profit in the Profit cell with the formula
=SUMPRODUCT(UnitProfits,UnitsProduced).
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Set Solver Settings
Manufacturing at Dorian Auto
• The completed Solver dialog box is shown here
•The objective is to maximize profit, the changing cells
are the production limits and resource availabilities
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Optimal Solution
Manufacturing at Dorian Auto
• The optimal solution shown in the earlier figure
indicates, by the 0 values in row 13, that Dorian
should not produce any compact or large cars
• The value of 1 in cell C13, however, indicates that
Dorian must produce at least the minimum number,
1000, of midsize cars
• Actually, midsize cars are quite profitable, so Dorian
produces as many as possible, 2000, before running
into the steel availability constraint
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Sensitivity Analysis
Manufacturing at Dorian Auto
• What type of incentive might cause the company to produce
more than one car type?
– One possible answer is that the minimum production levels for
each type, all currently 1000, are perhaps too high
– Use SolverTable to see the effect of decreasing each of these
minimum production levels by the same factor (the model must be
modified slightly: go to the worksheet labeled Sensitivity)
– Input the original minimum production levels in range F5:H5, and
enter the formula =DecrFactor*H5 in cell B7 and copy it across
row 7
– Invoke SolverTable with the DecrFactor cell as the single input
cell, varied from 0.2 to 1 in increments of 0.2, and keep track of
profit and all changing cell values
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Sensitivity Analysis
Manufacturing at Dorian Auto
• SolverTable results indicate that when the minimum production
levels are reduced to 200 or 400, Dorian produces both compact
and midsize cars above the minimum level
• When the minimum production level is 600, Dorian still produces
both types, but it does not produce any more compacts than
necessary
• Finally, when the minimum production level is 800 or 1000, Dorian
produces only midsize cars – as many as steel availability allows
• Does this sound correct? We checked it, and it is correct, but here is
a test of your economic reasoning
• The results seems to imply that compacts extract less profit from the
resources than midsize cars. But if this is the case, why doesn’t
Dorian produce the minimum number of compacts when the
minimum production is 200 or 400? (for example, when it is 400,
why doesn’t Dorian produce 400 compacts and 1800 midsize cars?)
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Goal Programming
Determining an Advertising Schedule
at Leon Burnit
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Background Information
Determining an Advertising Schedule at Leon Burnit
• The Leon Burnit Ad Agency is trying to determine a TV advertising
schedule for a client
• The client has three goals (listed in descending order of importance)
– Goal 1: at least 65 million high-income men (HIM)
– Goal 2: at least 72 million high-income women (HIW)
– Goal 3: at least 70 million low-income people (LP)
• Burnit can purchase ads to air on different program types (live sports
shows, game shows, news shows, sitcoms, dramas, and soap
operas), each with different ad costs & audience makeup
• At most $700,000 total can be spent on ads
• The client requires that at least two ads be placed on sports shows,
news shows, and dramas, and that no more than ten ads be placed
on any single type of show
• Burnit wants to find the advertising plan that best meets its client’s
goals
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Solution Approach
Determining an Advertising Schedule at Leon Burnit
• First, we build a spreadsheet model to see whether all
of the goals can be met simultaneously.
• In the spreadsheet model we must keep track of the
following:
– The number of ads placed for each show type
– The cost of the ads
– The number of exposures to each group (HIM, HIW, LIP)
– The deviation from the exposure goal of each group
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Model Formulation
Determining an Advertising Schedule at Leon Burnit
• Inputs: Enter all inputs in the shaded ranges
• Number of ads: Enter any trial values for the numbers of ads in the
Ads range
• Total cost: Enter the formula =SUMPRODUCT(UnitCosts,Ads) in the
TotCost cell
• Exposures obtained: enter the formula =SUMPRODUCT(B7:G7,Ads)
in cell B26 and copy to
calculate exposures
for the three groups
• The completed Solver
dialog box is shown
here (notice there is
no target cell – we are
checking for feasibility
only)
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Solution Results
Determining an Advertising Schedule at Leon Burnit
• When we click on Solve, we get the message that there
is no feasible solution
• It is impossible to meet all of the client’s goals and stay
within this budget
• To see how large the budget must be, we ran
SolverTable with the Budget cell as the single input cell,
varied from 700 to 850, and any cell as the output cell
• They show that unless the budget is greater then
$775,000, it is impossible to meet all of the client’s goals
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Goal Programming
Determining an Advertising Schedule at Leon Burnit
• Since a $700,000 budget is not sufficient to meet all of the
client’s goals, use goal programming to see how close Burnit
can come to their goals
• The upper and lower limits on the ads of each type and the
budget constraints are considered hard constraints in this
model (they cannot be violated under any circumstances)
• The goals on exposures, on the other hand, are considered soft
constraints (the client wants to satisfy these goals, but it is
willing to come up somewhat short – in fact, it must because of
the limited budget)
• In goal programming models the soft constraints are prioritized
– We first try to satisfy the goals with the highest priority
– If there is still any room to maneuver, we then try to satisfy the
goals with the next highest priority
– If there is still room to maneuver, we move on to the goals with the
third highest priority, and so on
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Goal Programming Model
Determining an Advertising Schedule at Leon Burnit
• In general, goal programming requires several consecutive
Solver runs, one for each priority level
• However, it is possible to set up the model so that we can
make these consecutive runs with only minor changes from
one run to the next (use the worksheet Goal Programming)
• New changing cells: We add additional changing cells in the
DevUnder and DevOver ranges (Dev is short for “deviations”)
to indicate how much or over each goal we are In the solution
• At least one of these two types of deviations will always be 0
for each goal – we will either be below the goal or above the
goal, but not both
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Goal Programming Model
Determining an Advertising Schedule at Leon Burnit
• Balance equations: enter the formula =B26+C26-D26 in
cell E26 and copy to tie the deviation cells to the rest of
the model
• The balance equation for each group specifies that the
actual number of exposures plus the number under the
goal minus the number over the goal (Col E) must be
equal the goal (Col G)
• Constraints on deviations under: enter the formula
=C26 in cell B32 and copy to measure the “under”
deviations
• The client is concerned only with too few exposures, not
with too many
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Goal Programming Model
Determining an Advertising Schedule at Leon Burnit
• Highest priority goal: First attempt to Minimize the Dev1 Cell to try to
achieve the highest priority goal
• The constraints include the hard constraints, the balance constraint,
and the DevUnder1 <= Obtained constraint
• We have entered the goals themselves in the Obtained range
(therefore, the DevUnder1 <= Obtained constraint at this point is
redundant – the “under”
deviations cannot
possibly be greater than
the actual goals)
• We include it because
it will become important
in later Solver runs, which
will then require only
minimal modifications
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Goal Programming Model
Determining an Advertising Schedule at Leon Burnit
• The solution shows that Burnit can satisfy the HIM goal
completely (the other two goals are not satisfied because their
“under” deviations are positive)
• Second highest priority goal: Now we come to the key aspect
of goal programming
– Once a high priority goal is satisfied as fully as possible, we move
on to the next highest priority goal
– Constrain the “under” deviation for the first goal to be no greater
than what we already achieved (we achieved a deviation of 0 for
HIM, so enter 0 in cell D32)
– Run the Solver again, changing only the target cell in to Dev2 (we
are minimizing the “under” deviation for the HIW group)
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Goal Programming Model
Determining an Advertising Schedule at Leon Burnit
• The solution from this second Solver run shows
that the HIM goal has not suffered at all, but we are
now a little closer to the HIW goal than before (it
was under 11.75 before, and now it is under by only 11)
• The lowest priority goal get worse, moving from under
by 11.25 to under by 18 (it could either improve or get
worse)
• Lowest priority goal: Minimize the Dev3 cell, the
deviation for the LIP group, while ensuring that the two
higher priority goals are achieved (make sure cell
D32=0 and cell D33=11)
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Goal Programming Solution
Determining an Advertising Schedule at Leon Burnit
• When you run Solver this time, the solution remains
exactly the same as shown (this occurs frequently in
goal programming models)
• After satisfying the first goal or two as fully as possible,
there is often no room to improve later goals.
• To summarize Burnit’s situation, the budget of
$700,000 allows it to satisfy the client’s HIM goal, miss
the HIW goal by 11 million, and miss the LIP goal by 18
million
• Given the priorities on these three goals, this is the best
possible solution
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Sensitivity Analysis
Determining an Advertising Schedule at Leon Burnit
• There is no quick way to do sensitivity analysis on a
goal programming model
• SolverTable works on only a single objective,
whereas goal programming requires a sequence of
objectives
• If we want to see how the solution to Burnit’s model
changes with different budget levels, we would need
to go through the above steps several times and
keep track of the results manually
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Effect of changing priorities
Determining an Advertising Schedule at Leon Burnit
• With three goals, there are six possible orderings of the
goals
• The goal programming solutions corresponding to these
orderings are listed in the table shown below.
• The solution can change dramatically if the priorities of the
goals change (for example, when we give the HIW goal
the highest priority, none of the goals are achieved
completely)
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Pareto Optimality and
Trade-off Curves
Maximizing Profit and Minimizing
Pollution at Chemcon
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Background Information
Max Profit and Min Pollution at Chemcon
• Chemcon plans to produce eight products, each with
different requirements for labor, raw materials, min
and max production levels, and profit margins
• Currently 1300 labor hours and 1000 units of raw
material are available
• Chemcon’s two objectives are to maximize profit and
minimize pollution produced
• Chemcon wants to graph the trade-off curve for this
problem
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• We want the product mix that stays within the
lower and upper production limits, uses no more
labor or raw material than are available, keeps
pollution low, and keeps profit high
• To develop a trade-off curve, we let profit be
objective 1 and pollution be objective 2
• To obtain one endpoint of the curve, we
maximize profit and ignore pollution (maximize
the Profit cell and delete the Pollution constraint)
• The optimal solution had profit $20,089 and
pollution level 8980
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• At the other extreme, we can minimize the pollution in
cell B26 and ignore any constraint on profit
• You can check that this solution has pollution level
3560 and profit $8360
• Profit can get as high as $20,089 by ignoring
pollution or as low as $8360 by focusing entirely on
pollution, and pollution can get as low as 3560 by
ignoring profit or as high as 8980 by focusing entirely
on profit
• These establish the extremes - now we can search
for points in between
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• To develop the trade-off curve, we will add the
pollution constraint back into the model and use
SolverTable
• We need to constrain pollution to various
degrees and see how large profit can be
• This is indicated in the model, where the
objective is to maximize profit with an upper limit
on pollution
• The only upper limits on pollution we need to
consider are those between the extremes, 3560
and 8980
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• Use SolverTable with the
setup shown
• We have used the option to
enter non-equally-spaced
inputs: 3560, 4000, 4500 and
so on, ending with 8980
• Alternatively, equally-spaced
inputs could be used
• All we require is a
representative set of values
between the extremes
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• These results show that as we allow more pollution,
profit increases (the product mix also shifts
considerably)
• Profit 8, a low polluter with a low profit margin,
eventually leaves the mix when pollution is allowed to
increase, which makes sense
• It is less clear why the level of product 6 increases so
dramatically (it is only a moderate polluter and has a
moderate profit margin, so the key is evidently that it
requires low levels of labor and raw materials.)
• The trade-off curve is created as an X_Y chart directly
from columns J and K of the table
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Developing a Trade-Off Curve
Max Profit and Min Pollution at Chemcon
• The chart indicates that profit indeed increases as
Chemcon allows more pollution, but at a decreasing rate
• For example, when pollution is allowed to increase from
4000 to 4500, Chemcon can make an extra $3187 in
profit
• However, when pollution is allowed to increase from
8000 to 8500, the extra profit is only $532
• All points below the curve are dominated – for a given
level of pollution, the company can achieve a larger
profit – and all points above the curve are unattainable
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