This document provides information about linear programming models, assumptions, and problem types. It discusses: 1) LP models make assumptions of proportionality, additivity, divisibility, and certainty. They can be iconic, analog, or mathematical representations of real problems. 2) Problem types discussed include product mix, blending, scheduling, inventory/portfolio problems, and multiple objective problems. 3) An example blending problem is provided to maximize profits by determining how to blend two petroleum components into regular and premium gasoline products based on specifications.

1. Models are used because 1) less expensive, time consuming and For a minimization problem hours of their shift (x6) and people coming into work to start their Let C2 = amount of money invested in certificates of deposit in year
risky, and more feasible. They are abstraction of a real object can - Reduced cost = positive or zero shift (x1). 2.
be (iconic (replica), analog, or mathematical) - Allowable increase = infinite Let Ri = amount of money invested in real estate at the beginning of
- Allowable decrease = reduced cost An Inventory Problem year i; i = 5; 6. And,
LP Assumptions: - If the RHS of a binding constraint changes within allowable limits The planning department of M-Kart, a producer of golf carts, must Let Ii = amount of money held idle and not invested at beginning of
Proportionality Assumption - Optimal solution changes determine aggregate production levels year i.
Contribution of a variable is proportional to its value. - OFV changes for their product over each of the next four quarters. The table below We’ve been told that our objective is to maximize the amount of
Additivity Assumptions - New OFV = old OFV + (change in RHS) x (shadow price) shows the relevant cost and demand cash at the end of the sixth year. Where
Contributions of variables are independent. - Shadow price for a maximization problem data. The inventory holding cost is per unit and is based on the end- does that cash come from? Well, we get it from the 2-year stock
Divisibility Assumption - Positive for ≤ type constraint of-quarter inventory. The cost of investment bought at the start of year 5,
Decision variables can take fractional values. - Negative for ≥ type constraint production varies over time. There are currently 10 units in the 3-year bond investment bought at the start of year 4, the real-
Certainty Assumption - Shadow price for a minimization problem inventory. The company would like to know estate investment purchases at the start
Each parameter is known with certainty. - Positive for ≥ type constraint how many units to produce and how many units to hold in inventory of year 6, and any idle cash we carried in year 6 as well. Thus our
Infeasible problem has every possible solution violate one or more - Negative for ≤ type constraint each quarter, so as to satisfy demand objective function is:
constraints. at minimum cost. MAX Z = 1:2S5 + 1:4B4 + 1:1R6 + I6
- If the RHS of a non-binding constraint changes within allowable
Unbounded problem is where Ob. function can increase indefinitely Production Inventory Holding We have zeros as coefficients on all of our other decision variables.
limits
without reaching a max value. Quarter: Demand Production(cost/unit) Invent Holding (cost/unit) That’s how we get money out of our
- Optimal solution is unchanged
Multiple optimal solutions can occur when the objective function 1 150 $200 $10 investments. How does the initial $1,000,000 get allocated? Well, at
- OFV is unchanged
is parallel to a constraint line. And when the feasible region is on a 2 150 $195 $10 the start of year 1, there are only three
- The shadow price for a non-binding constraint is always zero.
constraint. 3 120 $210 $10 places we can invest the money: stocks (S1), bonds (B1), and
- Non-binding constraint of ≤ type
Sensitivity range (Range of Optimality) is the range of coeff. 4 180 $200 $10 keeping it idle in some non-interest bearing
- Allowable Increase = infinity
values that will keep the current Opt. Sol optimal. account (I1). We can’t invest it in the certificate of deposit or the real
- Allowable decrease = RHS – final value
-To Find: equate the ratio of the current slope with the slope of the Decision Variables: estate because its too early. We have:
- Non-binding constraint of ≥ type
constraint in question on that axis, the difference between the Let Qi = Quantity of carts to produce in quarter i; i = 1; 2; 3; 4 and $1,000,000 = S1 + B1 + I1 as our first constraint.
- Allowable Increase = final value - RHS
current value of coeff and the calculated level is the range. Let Ii = Inventory of cards at the end of quarter i; i = 1; 2; 3; 4. If we invest S1 dollars into stocks in year one, we will have 1:2S1
- Allowable decrease = infinity
Shadow price is the amount of revenue associated with another Objective function: dollars to re-invest into other things
Problem Types:
unit of the constraint. The total amount of revenue that can be MIN Z = 200Q1 + 195Q2 + 210Q3 + 200Q4 + 10I1 + 10I2 + 10I3 + at the start of year three. Much like matter is neither created nor
1. Product Mix / Diet Problems / Marketing Mix: Xi = product i.
obtained at this marg. profit level, is the shadow price * allowable 10I4 (Production costs + Inventory destroyed, so it goes with money.
2. Make sure you understand how to articulate ‘ratio’ type
increase. (0 for non-binding constraint) carrying costs)
constraints!
Range of Feasibility is range of RHS of constraints that will 1) Constraints: the “innies” (sources of cash) must balance the “outties” (uses of
3. Blending problem: Xij = amount of component i in final product j.
keep the shadow price the same and 2) it’s the amount that a line 10 + Q1 = 150 + I1 Starting inventory (10) + quarterly production = cash) across the six years. Our constraint
4. Scheduling problems; draw picture: time vs. resources.
can go before it changes it’s binding/non-binding status. (the demand + ending inventory set can be pulled right off the diagram above. Here are the
5. Inventory / Portfolio: Inventory(t-1) + Production(t) = Demand(t) +
allowable increase/decrease). I1 + Q2 = 150 + I2 Second Quarter constraints in a nice tabular format:
Inventory(t). I.e. sources =
Binding Constraint: Remove and it changes the feasible region I2 + Q3 = 120 + I3 Third Quarter Start of Year Sources of $$ = Uses of $$
uses.
and Opt. Solution. I3 + Q4 = 180 + I4 Fourth Quarter One $1,000,000 = S1 + B1 + I1
up to 100%.
Non-Binding Constraint: Remove and it changes feasible region Qi; Ii _ 0; i = 1; 2; 3; 4 Non-negativity. Two I1 = S2 + B2 + C2 + I2
A Blending Problem
but not Opt. Sol. Shadow price is always 0. _ Helpful Hint: I like to use the “innies = outies” approach. Three 1:2S1 + I2 = S3 + B3 + I3
The Grand North Oil Company produces regular and premium
Redundant: Remove changes neither feasible region or Opt Sol. Whatever goes “in” (i.e. starting inventory Four 1:2S2 + 1:4B1 + I3 = S4 + B4 + I4
gasoline for independent service stations in
Standard Form: putting your constraints as equalities using + production for the quarter) must balance everything that goes out Five 1:2S3 + 1:4B2 + I4 = S5 + R5 + I5
Canada. The Grand North refinery manufactures the gasoline
slack/surplus var’s, (to meet demand or as inventory Six 1:2S4 + 1:4B3 + 1:1R5 + 1:8C2 + I5 = R6 + I6
products by blending 2 petroleum components.
Reduced Cost is the amount that an obj function coeff needs to at the end of the period). Of course you can re-arrange them so that you have all your DVs
The regular gasoline is sold for $1.00 per gallon and premium
improve by to make it part of the solution (not cheap/profitable on the LHS and a constant on the
gasoline for $1.08 per gallon. For the
enough to use resources on it). All used variables will have ‘0’ as Investment Problem (time value of money muraja) RHS to enter them into Excel. As mentioned previously, that’s just a
current production period, Grand North has obtained 5,000 gallons
reduced cost. An investment firm has $1 million to invest in stocks, bonds, mechanical exercise. The tough part
of component 1 at $0.50 per gallon and
It is -- for max, and + for min certificates of deposit, and real estate. The is thinking about the constraints and developing them in the first
10,000 gallons of component 2 at $0.60 per gallon.
Dual Cost is the amount the objective function will improve per unit firm wishes to determine the mix of investments that will maximize place.
The product specifications for the regular and premium gasolines
increase in constraint. (similar to Shadow price, but opposite sign the cash value at the end of 6 years. Recall:
restrict the amounts of each component
for min) Opportunities to invest in stocks and bonds will be available at the The total amount invested in stocks cannot exceed 30% of total
that can be used in each gasoline product. The percentage of
Non-Basic Variable: if reduced cost is 0, then an alternative beginning of each of the next 6 investments, and at least 25%
component 1 in the regular gasoline must
solution exists, Reduced cost will be ‘-‘ or ‘0’ for max prob, and will years. Each dollar invested in stocks will return $1.20 (a profit of of total investments must be in certificates of deposit.
be at most 30%, and the percentage of component 2 in the regular
be ‘+’ or ‘0’ for min problem. $0.20). 2 years later, the return can be S1 + S2 + S3 + S4 + S5/ S1 + S2 + S3 + S4 + S5 + B1 + B2 + B3 +
gasoline must be at least 40%. The
immediately reinvested in any alternative. Each dollar invested in B4 + C2 + R5 + R6 </30% (first constraint)
percentage of component 1 in the premium gasoline must be at
Transshippment Prob: Intermediate nodes to accept and ship bonds will return $1.40 three years later; C2 / S1 + S2 + S3 + S4 + S5 + B1 + B2 + B3 + B4 + C2 + R5 + R6
least 25%, and the percentage of component
goods. Must have constraint that has stuff going into to note = stuff the return can be reinvested immediately. </ 50% (2nd constraint)
2 in the premium gasoline must be at most 40%. Grand North wants
going out. Opportunities to invest in certificates of deposit will be available only
to determine how to blend the 2
Assignment Prob: Make sure that value of constraint<= 1, (only once, at the beginning of the Mixed Integer Example: Plant Closure
components into the 2 gasoline products and maximize profits.
can do 1 job). second year. Each dollar invested in certificates will return $1.80 A company is considering closing 1 or more of its plants, even
3.1 Formulation
four years later. Opportunities to invest though it may incur higher transportation
Decision Variables:
Multiple Objective Problems: 3 different approaches – weighted in real estate will be available at the beginning of the fifth and sixth costs to ship from the other plants to its distribution centres. The
Let xij = # of gallons of component i used in gasoline type j. i = 1; 2;
approach, absolute priorities, goal programming. years. Each dollar invested will return plant fixed and transportation variable
j = R; P.
$1.10 one year later. costs are given in the table below. Formulate a mixed integer
Objective function:
Integer Programming: 3 types – All integer, mixed integer, and To minimize risk, the firm has decided to diversify its investments. program to meet the regional demand at
MAX Z = $1:00(x1R + x2R) + 1:08(x1P + x2P ) ) 0:50(x1R + x1P )
binary (0/1) The total amount invested in stocks lowest total cost.
0:60(x2R + x2P )
- Project i conditional on project j: cannot exceed 30% of total investments, and at least 25% of total
Constraints:
xi - xj ≤ 0 or xj - xi ≥ 0 investments must be in certificates
x1R + x1P _ 5; 000 Availability of Component 1
- Project i co-requisite for project j: of deposit. The firm’s management wishes to determine the optimal
x2R + x2P _ 10; 000 Availability of Component 2
xj - xi = 0 mix of investments in the various
x1R _ 0:30(x1R + x2R) Regular gas specification
- Projects i & j mutually exclusive: alternative that will maximize the amount of cash at the end of the
x2R _ 0:40(x1R + x2R) Regular gas specification
xi + xj ≤ 1, or xi + xj = 1, if one must be done sixth year.
x1P _ 0:25(x1P + x2P ) Premium gas specification
Either Or Constraints: x2P _ 0:40(x1P + x2P ) Premium gas specification
Constraint written as 0< / x</ 1 xij _ 0 for all i and j Non-negativity.
If-then-constraints _Helpful Hint #1: If you’ve got i components and j products, you will
Simple if then: if f then g, yg > yf need i _ j decision variables.
If and only if Constraint _ Helpful Hint #2: If you put your components in rows and your
IF you invest in stocks 1 and 2, THEN you must invest in stock 3 products in columns, then adding up
y3 ≤ (y1 + y2)/2 and y3 ≥ y1 + y2 -1 the DVs in a row i will give you the total amount of component i
Methods to solve Int Prog Prob’s: consumed. Likewise at the bottom of the Decision Variables:
- Relaxation: round to adjacent int’s, can make infeasible solutions columns, adding up the DVs in column j will give you the total y(i) = 1 if plant i is open and = 0 if plant i is not open. i = 1; 2; 3; 4; 5
and less then optimal solutions. amount of product j produced. x(i; j) = amount to ship from plant i to warehouse j; j = A;B;C;D.
For max: Z(LP) ≥ Z(IP) A Scheduling Problem Objective Function:
For min: Z(LP) ≤ Z(IP) The personnel manager must schedule a security force in order to MIN Z = 2; 100; 000y(1) + 850; 000y(2) + 1; 800; 000y(3) + 1; 100;
- Enumeration: check all possible solutions satisfy staffing requirements shown below. 000y(4) + 900; 000y(5)
- Int Program Sensitivity Analysis is not meaningful since it Each worker has an eight hour shift and there are six such shifts +56x(1;A)+21x(1;B)+32x(1;C)+65x(1;D)+18x(2;A)+46x(2;B)+7x(2;C)
assumes continuous variables. each day. The starting and ending time for +35x(2;D)+12x(3;A)+
Multiple Changes - 100% Rule each of the 6 shifts is also given below. The personnel manager 71x(3;B)+41x(3;C)+52x(3;D)+30x(4;A)+24x(4;B)+61x(4;C)+28x(4;D)
Simultaneous changes in coefficients (Obj. Func. OR RHS) will not wants to determine how many people need +45x(5;A)+50x(5;B)+
change decision variables in optimal solution as long as sum of to work each shift in order to minimize the total number of officers 26x(5;C) + 31x(5;D)
percentages of change divided by corresponding maximum employed while satisfying the staffing Supply Constraints:
allowable change in range of optimality for each coefficient does not requirements. x(1;A) + x(1;B) + x(1;C) + x(1;D) _ 12; 000y(1)
exceed 100% Time: # Officers Req’d Shift No.: Shift Time This is a tougher problem than the investment problem from last x(2;A) + x(2;B) + x(2;C) + x(2;D) _ 18; 000y(2)
SENS. TREE If the change in objective function coefficient or RHS 12:00 AM – 4:00 AM 5 1 12:00 AM – 8:00 AM time (without time value) . Our decision variables will similar x(3;A) + x(3;B) + x(3;C) + x(3;D) _ 14; 000y(3)
of a constraint is outside allowable increase/decrease, the 4:00 AM – 8:00 AM 7 2 4:00 AM – NOON to before, e.g. Let X = amount of money invested in such-and-such x(4;A) + x(4;B) + x(4;C) + x(4;D) _ 10; 000y(4)
problem needs to be re-solved on the computer and we cannot 8:00 AM – NOON 15 3 8:00 AM – 4:00 PM investment. The key difference of x(5;A) + x(5;B) + x(5;C) + x(5;D) _ 16; 000y(5)
say anything about the objective function value (OFV) or the optimal NOON – 4:00 PM 7 4 NOON – 8:00 PM course is we are including the time value of money and the Demand Constraints:
solution (OS). 4:00 PM – 8:00 PM 12 5 4:00 PM – 12:00 AM investment runs for 6 years (vs. the previous x(1;A) + x(2;A) + x(3;A) + x(4;A) + x(5;A) = 6; 000
If the objective function coefficient of a basic variable (decision 8:00 PM – 12:00 AM 9 6 8:00 PM – 4:00 AM problem which was only looking at the initial allocation of funds). A x(1;B) + x(2;B) + x(3;B) + x(4;B) + x(5;B) = 14; 000
variable with non-zero soln) changes within allowable limits, picture may help clarify what is going x(1;C) + x(2;C) + x(3;C) + x(4;C) + x(5;C) = 8; 000
- Optimal solution is unchanged Decision Variables: on; draw a diagram showing the flow of investment during the 6 x(1;D) + x(2;D) + x(3;D) + x(4;D) + x(5;D) = 10; 000
- Objective function value changes Let xi = # of officers who start work on shift i; i = 1; 2; : : : ; 6. years. In many respects it is similar to the inventory problem Non-Negativity & Binary Constraints. . .
- The new OFV can be determined by substituting the OS into the Objective function: diagram we developed in the last class. Our x(i; j) _ 0 8 i; j and y(i) binary.
new objective function. MIN Z = x1 + x2 + x3 + x4 + x5 + x6 (Total # of officers) “inventory” in this case is money being held into future time periods. Either Or explanation
- Reduced cost is always zero for a basic variable Constraints: Our “inventory” isn’t static, it grows –If you have 2 constraints, but only 1 will be used. Ex:
- If the objective function coefficient of a non-basic variable x6 + x1 _ 5 12:00 AM – 4:00 AM coverage as it earns interest. The balancing equations from last time will hold making tables/chairs/desks, and can either make
(decision variable = 0) changes within allowable limits, x1 + x2 _ 7 4:00 AM – 8:00 AM coverage in this case as well — this is important them out of oak or pine (not a mix of both).
-Optimal solution is unchanged x2 + x3 _ 15 8:00 AM – NOON coverage to “link” the initial $1,000,000 investment with the final outcome at •Pine: 5xt+ 1xc+ 9xd≤ 1500
-Objective function value is unchanged x3 + x4 _ 7 NOON – 4:00 PM coverage the end of year six. But I’m getting
x4 + x5 _ 12 4:00 PM – 8:00 PM coverage ahead of myself. •Oak:2xt+ 3xc+ 4xd≤ 1000
Reduced cost is usually non-zero
x5 + x6 _ 9 8:00 PM – 12:00 AM coverage –So ADD a new constraint yp+ yo= 1 (so have to choose
-If reduced cost is zero, it indicates the existence of an alternate
xi _ 0; i = 1; : : : ; 6 Non-negativity. The decision variables will be the amount of money invested in each 1 of pine or oak)
optimal solution.
For a maximization problem, _ Helpful Hint: Set this problem up with a time-line running left to of the investment options at the
-Reduced cost is negative or zero right (columns) and the shifts start of the time period. Thus: –Then need to modify your pine/oak to be:
- Allowable increase = -reduced cost depicted as rows. Its easy then to figure out where the people are Let Si = amount of money invested in stocks at the beginning of •Pine:5xt+ 1xc+ 9xd≤ 1500 + M*(1-yp)
- Allowable decrease = infinite coming from for each of the required year i; i = 1; 2; 3; 4; 5. •Oak:2xt+ 3xc+ 4xd≤ 1000 + M*(yp)
(range of coeff of non basic var that will keep optimal the same) coverage periods (i.e. are the 12:00 AM – 4:00 AM time slot is Let Bi = amount of money invested in bonds at the beginning of the
covered by people finishing up the last 4 year i; i = 1; 2; 3; 4.

2. M is a random BIG number! If added, makes that
constraint useless..