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Supply Chain Lab Summary
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HC 1How does spreadsheet modeling help? Models generates insight for better decisions, improves thinking skills (Break problems down
in components and makes assumptions explicit). Modeling improves quantitative skills and sharpens intuition. Can lead to significant
time and cost savings in logistics and operations management.
Formulating LP Models
1. Understand the problem
2. Describe objective and each constraint in words
3. Define the decision variables
4. Express the objective and each constraint in terms of the decision variables
→ Find ‘constraint words’: at most, at least, no more/no less than, up to, exactly, required, limited, restricted, must, cannot, budget
Three categories of constraints; – Limited resources: “at most”
– Minimum performance: “at least”
– Conservation: “exactly”
Solving LP’s; Graphical Method
1. Draw the constraint boundary line for each constraint. Use origin to determine which side of the line is permitted by the constraint.
2. Find the feasible region by determining where all constraints are satisfied simultaneously.
3. Select value for objective function and determine slope of one objective function line. All other function lines have the same slope.
4. Shift objective function in the direction of improving values of the objective function. Stop at the last point in the feasible region.
This point (or this set of points) is the optimal solution.
maximize Z = 5X1 + 7X2
s.t. X1 ≤ 6
2X1 + 3X2 ≤ 19
X1 + X2 ≤ 8
X1 , X2 ≥ 0
Limitations;
Linearity of Objective Function and Constraints; - Proportionality
- Implies; Constant returns to scale and No “fixed charges” (no constants)
Additivity: – F ( X1, X2,…, Xn) = f(X1) + f(X2) +…f(Xn)
– No interactive effects among Xi terms
– Assumes that individual segments of the problem operate as well independently as together
Non-negativity; – Xi ≥ 0
– No fundamental difficulties except in particular situations
Divisibility; Decision variables can have fractional values
Certainty;All parameters of the model are known
Linear objective function => Optimum will be on an edge of the feasible region
LP Solver; File – Options – Add-ins - Solver
Write down the expression for the objective function using the cells containing the objective function coefficients and those that
will contain the values of the decision variables.
Use Simplex LP as solver
The feasible region for a two-variable LP problem can be nonexistent, a single point, a line, a polygon, or an unbounded area.
Any linear program falls in one of 3 categories: – is infeasible; no point satisfies all the constraints
– has unique or alternate optimal solution; single point or line segment is optimal
– has an objective function that can be increased or decreased without bound.
VB. Infeasible solution
maximize 2X1 + 6X2
subject to (s.t.) 4X1 + 3X2 ≤ 12
2X1 + X2 ≥ 8
X1, X2 ≥ 0
There are no points that satisfy both constraints, hence this problem has no feasible region, and no optimal solution.
VB. Unbounded problem
maximize 3X1 + 4X2
subject to (s.t.) X 1 + X2 ≥ 5
3X1 + X2 ≥ 8
X1, X2 ≥ 0
The feasible region is unbounded and the objective
function line can be moved parallel to itself without
bound so that z can be increased infinitely.
,HC 2
VB Digital Imaging (DI) produces photo printers with profit of $42 for each DI-910 and $87 for each DI-950. Plants are automated
and uses two lines. Line 1 performs assembly operation with times of three minutes per DI-910 printer and six minutes per DI-950
printer. Line 2 performs both testing and packaging operations with four minutes per DI-910 printer and two minutes per DI-950
printer. Both manufacturing lines operate eight-hour shift a day. Determine how many units of each printer to produce.
D1= number of units of the DI-910 produced, D2= number of units of the DI-950 produced
Maximize 42D1 + 87D2
s.t. 3D1 + 6D2 <= 480 (Line 1 Capacity)
4D1 + 2D2 <= 480 (Line 2 Capacity) Note; line capacity is 8 hours×60 minutes/hour = 480 min per day
D1, D2 >= 0
Management wants to produce at least as many DI-910 as DI-950 printers → NEW CONSTRAINT
D1 >= D2
New constraints limit difference between T1 and T2 to 30 min or less. T1 = 3D1 + 6D2 and T2 = 4D1 + 2D2
T1= total time spent on Line 1 and T2= total time spent on line 2
Whatever value of T2, T1 <= T2 + 30
T1 >= T2 - 30
Leads to: 3D1 + 6D2 <= 4D1 + 2D2 + 30 → -D1 + 4D2 <= 30
3D1 + 6D2 >= 4D1 + 2D2 - 30 → -D1+ 4D2 >= -30 = D1 - 4D2 <= 30
Sensitivity analysis; used to determine how the optimal solution is affected by changes, within specified ranges in the objective
function coefficients and the right-hand side (RHS) values. → Sensitivity usually focuses on changing a single coefficient at a time
Brute force: resolving model with updated data from scratch
Classical sensitivity analysis: based on relationships between initial tableau and optimum tableau to quickly update the optimum
solution for changes in the original tableau
Computer-based ranging: specify how much RHS variables and coefficients of the objective function can change without
fundamentally changing the solution.
Simultaneous changes; 100% rule or Cost coefficients and RHS.
How can a change in RHS affect the feasible region and cause a change in optimal solution?
- The shadow price associated with a constraint is change in objective function value per unit increase in RHS of constraint.
- As RHS increases (or decreases), other constraints can become binding and limit the change in the value of the objective function.
- The shadow price for a non binding constraint is 0.
Range of feasibility; range over which the shadow price is applicable → determined by finding the values of a RHS coefficient such
that the same two lines that determined the original optimal solution continue to determine the optimal solution for the problem.
Range of optimality for each objective function coefficient provides range of values over which current solution will remain optimal
→ Found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.
Reduced costs indicate how much objective function coefficient of each decision variable would have to improve (change) before
variable gets a positive value (increase) in the optional solution. Reduced cost for a decision variable with a positive value is 0.
Wyndor Glass Co. EXAMPLE
, Hoorcollege 3
Truck Leasing Strategy BOOK
Ministry estimates it takes 4 months with 10,12,14 and 8 trucks in month 1 till 4 for a project. Goal; minimize costs. Availability of
current long term leased trucks; 1 truck in month 1. 2 trucks in month 2, 3 trucks in month 3 and 1 truck in month 4. Long term
leasing contract has monthly cost of €600,- per truck (already paid) and truck driver for €20,- an hour, fuel is approximately €100 per
day. Each truck used on project will be operating 8 hours a day, 5 days a week for approximately 4 weeks/month. Due to no layoff
policy, current long term trucks need to be used. Short term leasing cost(includes driver&truck costs).
Decision Variables;
Xij; # trucks from short term lease signed in month i for a period of j months
Yi; # trucks from long term lease used in month i
MIN 6000 (x11 + x21 + x31 + x41) + 11400 (x12 + x22 + x32) +
15675 (x13 + x 23) + 20160 (x14) + 2000 (y1 + y2 + y3 + y4)
s.t. x11 + x12 + x13 + x14 + y1 =10 demand month 1
x12 + x13 + x14 + x21 + x22 + x23 + y2 =12 demand month 2
x13 + x14 + x22 + x23 + x31 + x32 + y3 =14 demand month 3
x14 + x23 + x32 + x41 + y4 =8 demand month 4
y1 <= 1 # long term lease trucks available month 1
y2 <= 2 # long term lease trucks available month 2 Becomes = with no lay off policy
y3 <= 3 # long term lease trucks available month 3
y4 <= 1 # long term lease trucks available month 4
Xij, Yi >= 0 (non-negativity) i= 1,2,3,4 and j=1,2,3,4
What is optimal leasing plan and what are costs associated? Total cost associated with the leasing plan is €203.660,-
What are cost for ministry to maintain no layoffs policy? Replacing Yi with €5200(include driver costs of €3200).
To use all the available trucks from the long-term lease, we add the following constraints; y1 =1, y2 =2, y3 =3, y4 =1
With the constraint “=“ ; the minimum leasing costs are €226,060
With the constraint “<=“ ; the minimum leasing costs are €224,620
Maintains no layoff policy, extra costs are €226,060 - €224,620 = $1,440 (obligatory of long term lease)
Case Airplane Loading
A cargo plane has three compartments for storing cargo: front, center and back. These compartments have capacity limits on both
weight and space. Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that
compartment’s weight capacity to maintain the balance of the airplane.
Four cargoes are offered for shipment. Any portion of these cargoes can be accepted. How much of each cargo should be accepted
and how to distribute each among the compartments to maximize total profit for the flight?
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