Chapter 5

LP formulations

LP formulations of four basic problem

Resource allocation problem Transportation problem Feed mix problem Joint products problem

We will examine:

Basic Structure Formulation Example application Answer interpretation

 

Resource Allocation Problem

The classical LP problem involves the allocation of an endowment of scarce resources among a number of competing products so as to maximize profits.

– Objective: Maximize Profits– Competing products index is j; scarce resources index is i

– Major decision variable Xj is the number of units of the jth product made

– Non negative production (Xj ≥ 0)– Resource usage across all production possibilities is less than

or equal to the resource endowment

Algebraic Set UP

cj: profit per unit of the jth product aij: number of units of the ith resource used

when producing one unit of the jth product bi: the endowment of the ith resource

ij jj

j

i

jj

j

s.t. a X b for a

Ma

X 0 for al

ll

x c

i

l j

X

Resource Allocation Problem: E-Z Chair

Objective: find the number of two types of chairs to produce that will maximize profits.

Chair Types: Functional and Fancy Resources: Large & Small Lathe, Chair

Bottom Carver, and Labor Profit Contributions: (revenue – material

cost - cost increase due to lathe shifts)

Information for Problem

OBJ Functional $82 - $15 = $67Fancy $105 - $25 = $80

Resource Requirements When Using The Normal Pattern

Hours of Use per Chair Type

Functional Fancy

Small Lathe 0.8 1.2

Large Lathe 0.5 0.7

Chair Bottom Carver 0.4 1.0

Labor 1.0 0.8

Resource Requirements and Increased Costs for Alternative Methods of Production in Hours of Use per Chair and Dollars

Maximum Use of Small Lathe Maximum Use of Large Lathe

Functional Fancy Functional Fancy

Small Lathe 1.30 1.70 0.20 0.50

Large Lathe 0.20 0.30 1.30 1.50

Chair Bottom Carver 0.40 1.00 0.40 1.00

Labor 1.05 0.82 1.10 0.84

Cost Increase $1.00 $1.50 $0.70 $1.60

Alternative Production Method

Resource Limits

Small lathe: 140 hours Large lathe: 90 hours Chair bottom carver: 120 hours Labor: 125 hours

Production alternatives and profits

Functional, regular method (X1) : $67 (c1) Functional, max small lathe (X2): $66 (c2) Functional, max lg lathe (X3): $66.30 (c3) Fancy , regular method (X4): $80 (c4) Fancy, max small lathe (X5): $78.50 (c5) Fancy, max lg lathe (X6): $78.40 (c6)

Empirical Set-UP

Max 67X1 + 66X2 + 66.3X3 + 80X4 + 78.5X5 + 78.4X6

s.t. 0.8X1 + 1.3X2 + 0.2X3 + 1.2X

4

+ 1.7X5 + 0.5X6 ≤ 140

0.5X1 + 0.2X2 + 1.3X3 + 0.7X

4

+ 0.3X5 + 1.5X6 ≤ 90

0.4X1 + 0.4X2 + 0.4X3 + X4 + X5 + X6 ≤ 120

X1 + 1.05X2 + 1.1X3 + 0.8X

4

+ 0.82X5 + 0.84X6 ≤ 125

X1 , X2 , X3 , X4 , X5 , X6 ≤ 0

X1 X2 X3 X4 X5 X6 RHS Used S.P.Obj 67 66 66.3 80 78.5 78.4 10417.29Small Lathe 0.8 1.3 0.2 1.2 1.7 0.5 le 140 140.00 33.33Large lathe 0.5 0.2 1.3 0.7 0.3 1.5 le 90 90.00 25.79Chair bottom carver 0.4 0.4 0.4 1 1 1 le 120 103.09 0.00Labor 1 1.05 1.1 0.8 0.82 0.84 le 125 125.00 27.44answers 62.23 0.00 0.00 73.02 0.00 5.18reduced cost 0.00 -11.30 -4.08 0.00 -8.40 0.00

Solution from Excel Solver

McCarl provides GAMS code and solution from GAMS (same as this solution).

Interpretation

Produce 62 functional chairs using the traditional method, 73 fancy chairs using the traditional method, and 5 fancy chairs using the maximum large lathe method to earn profits of $10,417

Producing functional chairs by the max small lathe method reduces profits by $11.30 per chair made, producing functional chairs by max large lathe reduces profits by $4.08/chair, and producing fancy chairs by max small lathe method reduces profits by $8.40/chair.

An hour more small lathe time would increase profits by $33.33; an hour more large lathe time would increase profits by $25.79; an hour more labor would increase profits by 27.44.

Dual of the problem (general)

i allfor 0 U

j allfor caUs.t.

bUMin

i

ji

iji

iii

Empirical Dual for E-Z Chair

Min 140U1 + 90U2 + 120U3 + 125U4    

s.t. 0.8U1 + 0.5U2 + 0.4U3 + U4 ≥ 67

  1.3U1 + 0.2U2 + 0.4U3 + 1.05U4 ≥ 66

  0.2U1 + 1.3U2 + 0.4U3 + 1.1U4 ≥ 66.3

  1.2U1 + 0.7U2 + U3 + 0.8U4 ≥ 80

  1.7U1 + 0.3U2 + U3 + 0.82U4 ≥ 78.5

  0.5U1 + 1.5U2 + U3 + 0.84U4 ≥ 78.4

  U1 , U2 , U3 , U4 ≥ 0

Dual Solution

u1 u2 u3 u4 cj Used S.P.OBJ 140.00 90.00 120.00 125.00 10417.29x1 0.80 0.50 0.40 1.00 ge 67.00 67.00 62.23x2 1.30 0.20 0.40 1.05 ge 66.00 77.30 0.00x3 0.20 1.30 0.40 1.10 ge 66.30 70.38 0.00x4 1.20 0.70 1.00 0.80 ge 80.00 80.00 73.02x5 1.70 0.30 1.00 0.82 ge 78.50 86.90 0.00x6 0.50 1.50 1.00 0.84 ge 78.40 78.40 5.18answers 33.33 25.79 0.00 27.44red c 0.00 0.00 16.91 0.00

= slack on constraint three in primal

Shadow prices in primal

answers toprimal

Transportation Problem

This problem involves the shipment of a homogeneous product from a number of supply locations to a number of demand locations.

– Objective: Minimize cost– Variables: Quantity of goods shipped from each supply

point to each demand point– Restrictions: Non negative shipments – Supply availability at supply point – Demand need at a demand point

Diagram of Problem

Supply Locations Demand Locations

1

2...m

A

B...n

Formulate the problem

Supply locations as supplyi Demand locations as demandj Decision variable = Movesupplyi,demandj

– Decision to move a quantity of supply from location i to demand location j

costsupplyi,demandj = the cost of moving one unit of product from location i to demand location j

Constraints

Supply availability: limiting shipments from each supply point so that the sum of outgoing shipments from point supplyi to all possible destinations (demandj) doesn't exceed supplyi

Minimum demand: requiring shipments into the demandjth demand point be greater than or equal to demand at that point. Incoming shipments include shipments from all possible supply points to the demandjth demand point.

Nonnegative shipments:

Algebra

Minimize s du empp andjlyi costsupplyi,demandjMovesupplyi,demandj

demandjsupplyi supply, iM supplyovedemandj

demans dup jplyisupp

demandjl

, yi

dMo emandve

,suppl demand yi jMove 0

Transportation Problem Example: Shipping Goods

Three plants: New York, Chicago, Los Angeles

Four demand markets: Miami, Houston, Minneapolis, Portland

Minimize the cost of shipping product from the three plants to the four demand markets.

Quantities

Supply Available Demand Required New York 100 Miami 30

Chicago 75 Houston 75

Los Angeles 90 Minneapolis 90

Portland 50

Distances

Distances: Miami Houston Minneapolis Portland

New York 30 7 6 23

Chicago 9 11 3 13

Los Angeles 17 6 13 7

Transportation costs = 5 +5*Distance

Miami Houston Minneapolis Portland

New York 20 40 35 120

Chicago 50 60 20 70

Los Angeles 90 35 70 40

NY-MI

NY-H

NY-MN

NY-P C-MI C-H C-MN C-P LA-MI

LA-H LA-MN

LA-PUsed SP

20.00 40.00 35.00 120.00 50.00 60.00 20.00 70.00 90.00 35.00 70.00 40.00 Min 7425NY 1.00 1.00 1.00 1.00 LE 100 80 0Chicago 1.00 1.00 1.00 1.00 LE 75 75 -15LA 1.00 1.00 1.00 1.00 LE 90 90 -5Miami 1.00 1.00 1.00 GE 30 30 20Houston 1.00 1.00 1.00 GE 75 75 40Minn 1.00 1.00 1.00 GE 90 90 35Portland 1.00 1.00 1.00 GE 50 50 45answers 30 35 15 0 0 0 75 0 0 40 0 50 r.c. 0 0 0 75 45 35 0 40 75 0 40 0

Minimization problem. Note the first three inequalities.

Interpretation

shadow price represents marginal values of the resources i.e.marginal value of additional units in Chicago = $15

reduced cost represents marginal costs of forcing

non-basic variable into the solution i.e. shipments from New York to Portland increase costs by $75.

twenty units are left in New York

U1 U2 U3 U4 U5 U6 U7 RHS USED S.P.obj -100 -75 -90 30 75 90 50 7425

r1 -1.00 0.00 0.00 1.00 0.00 0.00 0.00 le 20.00 20 30

r2 -1.00 0.00 0.00 0.00 1.00 0.00 0.00 le 40.00 40 35

r3 -1.00 0.00 0.00 0.00 0.00 1.00 0.00 le 35.00 35 15

r4 -1.00 0.00 0.00 0.00 0.00 0.00 1.00 le 120.00 45 0

r5 0.00 -1.00 0.00 1.00 0.00 0.00 0.00 le 50.00 5 0

r6 0.00 -1.00 0.00 0.00 1.00 0.00 0.00 le 60.00 25 0

r7 0.00 -1.00 0.00 0.00 0.00 1.00 0.00 le 20.00 20 75

r8 0.00 -1.00 0.00 0.00 0.00 0.00 1.00 le 70.00 30 0

r9 0.00 0.00 -1.00 1.00 0.00 0.00 0.00 le 90.00 15 0

r10 0.00 0.00 -1.00 0.00 1.00 0.00 0.00 le 35.00 35 40

r11 0.00 0.00 -1.00 0.00 0.00 1.00 0.00 le 70.00 30 0

r12 0.00 0.00 -1.00 0.00 0.00 0.00 1.00 le 40.00 40 50answers 0 15 5 20 40 35 45red cost 20 0 0 0 0 0 0

Dual: Note the signs on U1, U2, U3 (the constraints were LE in primal)

The dual is a maximization

Feeding Problem

Objective: Minimize total diet costs Variables: how much of each feedstuff is

used in the diet Restrictions: Non negative feedstuff Minimum requirements by nutrient Maximum requirements by nutrient Total volume of the diet 

Indices needed

Ingredients: How many possible ingredients can be used in the ration? (corn, soybeans, molasses, hay, etc.)

Nutrients: What are the essential nutrients to consider? {protein, calories, vitamin A, etc.}

Constraints

restricting the sum of the nutrients generated from each feedstuff to meet the dietary minimum

restricting the sum of the nutrients generated from each feedstuff not to exceed the dietary maximum

the ingredients in the diet equal the required weight of the diet.

nonnegative feedstuff

Algebraic Representation

ingredientj ingredientjnutrient nu, trienta Feed minimumingredientj

ingredientj ingredientjnutrient nu, trienta Feed maximumingredientj

ingredientingredient

jj

Feed = 1

ingredientjFeed 0

Minimize ingredientj costingredientj Feedingredientj which is the per unit

Example: Cattle Feeding

Seven nutritional characteristics: energy, digestible protein, fat, vitamin A, calcium, salt, phosphorus

Seven feed ingredient availability: corn, hay, soybeans, urea, dical phosphate, salt, vitamin A

New product: potato slurry

Information for feeding problem

Ingredient Costs for Diet Problem Example per kg

Corn $0.133 Dical $0.498 Alfalfa hay $0.077 Salt $0.110 Soybeans $0.300 Vitamin A $0.286 Urea $0.332

New ingredient: Potato Slurry, initially set cost to $.01/kg

More information for problem

Required Nutrient Characteristics per Kilogram Nutrient Unit Minimum

Amount

Maximum

Amount Net energy Mega calories 1.34351 -- Digestible protein Kilograms 0.071 0.13 Fat Kilograms -- 0.05 Vitamin A International Units 2200 -- Salt Kilograms 0.015 0.02 Calcium Kilograms 0.0025 0.01 Phosphorus Kilograms 0.0035 0.012 Weight Kilograms 1 1

Nutrient Content per kilogram

Characteristic

Corn

Hay

Soybean

Urea

Dical

Phosphate

Salt

Vitamin A

Concentrate

Potato

Slurry

Net energy 1.48 0.49 1.29 1.39

Digestible protein 0.075 0.127 0.438 2.62 0.032

Fat 0.0357 0.022 0.013 0.009

Vitamin A 600 50880 80 2204600

Salt 1

Calcium 0.0002 0.0125 0.0036 0.2313 0.002

Phosphorus 0.0035 0.0023 0.0075 0.68 0.1865 0.0024

Corn Hay Soybean Urea Dical Salt Vitamin A Slurry RHS Used0.133 0.077 0.3 0.332 0.498 0.11 0.286 0.01 0.021

Protein 0.075 0.127 0.438 2.62 0 0 0 0.032 le 0.13 0.071 0Fat 0.0357 0.022 0.013 0 0 0 0 0.009 le 0.05 0.008776 0Salt 0 0 0 0 0 1 0 0 le 0.02 0.015 0Calcium 0.0002 0.0125 0.036 0 0.2313 0 0 0.002 le 0.01 0.002857 0Phospor 0.0035 0.0023 0.0075 0.68 0.1865 0 0 0.0024 le 0.012 0.012 -2.206765Energy 1.48 0.49 1.29 0 0 0 0 1.39 ge 1.34351 1.34351 0.065432Protein 0.075 0.127 0.438 2.62 0 0 0 0.032 ge 0.071 0.071 0.740747Vit A 600 50880 80 0 0 0 2204600 0 ge 2200 2200 0.00Salt 0 0 0 0 0 1 0 0 ge 0.015 0.015 0.218158Calcium 0.0002 0.0125 0.0036 0 0.2313 0 0 0.002 ge 0.0025 0.0025 4.399996Phospor 0.0035 0.0023 0.0075 0.68 0.1865 0 0 0.0024 ge 0.0035 0.012 0volume 1 1 1 1 1 1 1 1 eq 1 1 -0.1082answers 0 0.00134 0.01102 0.0135 0.0023 0.015 0.000966 0.9558

0.095 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Minimization problem. Note the LE constraints, and the equality.

u1 u2 u3 u4 u5 u6 u7 u8 u9 u10 u11 u12p u12n-0.13 -0.05 -0.02 -0.01 -0.012 1.34351 0.071 2200 0.015 0.0025 0.0035 1 -1 le-0.075 -0.0357 0 -0.0002 -0.0035 1.48 0.075 600 0 0.0002 0.0035 1 -1 le 0.133-0.127 -0.022 0 -0.0125 -0.0023 0.49 0.127 50880 0 0.0125 0.0023 1 -1 le 0.077-0.438 -0.013 0 -0.036 -0.0075 1.29 0.438 80 0 0.0036 0.0075 1 -1 le 0.3-2.62 0 0 0 -0.68 0 2.62 0 0 0 0.68 1 -1 le 0.332

0 0 0 -0.2313 -0.1865 0 0 0 0 0.2313 0.1865 1 -1 le 0.4980 0 -1 0 0 0 0 0 1 0 0 1 -1 le 0.110 0 0 0 0 0 0 2204600 0 0 0 1 -1 le 0.286

-0.032 -0.009 0 -0.002 -0.0024 1.39 0.032 0 0 0.002 0.0024 1 -1 le 0.01

u1-u5 associated with LE constraints in the min.

U12 is associated with an equality constraint. So u12 is unrestricted in sign, which here we work as two parts, u12p and u12n.

The dual problem

Tracing out the derived demand for slurry

Start at $0.01 for slurry (which we've solved) and production is 96% slurry.

Sensitivity report – tells you the price at which the answers would change. Look at the "allowable increase" for this price. It is 0.09630319. So at slurry prices greater than .01 + 0.09630319, the optimal solution will change.

Plug in .1163, and the slurry falls to 87% of the ration. In the new sensitivity report, the next allowable increase is

0.005930872. At a price of .1222, slurry use falls to 64%. At prices close to .13, use of slurry falls to 0.

Joint Products Problem

One production process yields multiple products, such as lambs and wool, or wheat grain and straw

Maximize profits when each production possibility yields multiple products, uses some inputs with a fixed market price, and uses some resources that are available in fixed quantity.

Variables in model

the amount of each product produced for sale

the production process chosen to produce the products

the amount of market inputs to purchase

Technical consideration

How much of each output will a given production process generate?

How much of each market input must be bought for that production process?

How much of each fixed resource is used in the production process?

Transfer row

"Transfer row" is the name given to a certaintype of row in an LP problem. These rowstake the production generated by a processand distribute it to sales activities. The RHSof a transfer row is generally zero.

Example: Wheat-Straw Problem

7 possible processes to produce wheat and straw with different yields.

Wheat sells for $4 per bushel, straw for $.50 per small (square) bale.

Fertilizer and seed can be purchased and are needed in different amounts for the 7 practices.

Fertilizer costs $2/lb and seed costs $.20/lb. Land is limited to 500 acres.

Number of decision variables

2 sales variables (wheat and straw) 7 production variables (the 7 possible

processes) 2 purchase variables (seed and fertilizer)

Number of constraints

Transfer row for wheat sales (from processes to sales activity)

Transfer row for straw sales (from processes to sales activity)

Transfer row for fertilizer purchases (distribute purchased fertilizer to production processes)

Transfer row for seed purchases (distribute purchased seed to processes)

Limit on land

Other information: Yields from Processes

Process Wheat Straw1 30 102 50 173 65 224 75 265 80 296 80 317 75 32

Max 4Sale1 + .5Sale2 - 5Y1 - 5Y2 - 5Y3 - 5Y4 - 5Y5 - 5Y6 - 5Y7 - 2Z1 - .2Z2

s.t. Sale1 - 30Y1 - 50Y2 - 65Y3 - 75Y4 - 80Y5 - 80Y6 - 75Y7 ≤ 0

Sale2 - 10Y1 - 17Y2 - 22Y3 - 26Y4 - 29Y5 - 31Y6 - 32Y7 ≤ 0

+ 5Y2 + 10Y3 + 15Y4 + 20Y5 + 25Y6 + 30Y7 - Z1 ≤ 0

10Y1 + 10Y2 + 10Y3 + 10Y4 + 10Y5 + 10Y6 + 10Y7 - Z2 ≤ 0

Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 ≤ 500

Sale1 , Sale2 , Y1 , Y2 , Y3 , Y4 , Y5 , Y6 , Y7 , Z1 , Z2 ≥ 0

Joint Product Problem: Wheat-Straw

Note the transfer rows have zero on RHS. It is important to get the inequalityin the right direction. Here Sales must be LE production. Use of inputmust be LE purchases.

Excel Solver Solution

Sale1 Sale2 Y1 Y2 Y3 Y4 Y5 Y6 Y7 Z1 Z2 MAX RHS Used S.P.

Obj 4 0.5 -5 -5 -5 -5 -5 -5 -5 -2 -0.2 143750Grain 1 0 -30 -50 -65 -75 -80 -80 -75 0 0 le 0 0 4Straw 0 1 -10 -17 -22 -26 -29 -31 -32 0 0 le 0 0 0.5fert 0 0 0 5 10 15 20 25 30 -1 0 le 0 0 2seed 0 0 10 10 10 10 10 10 10 0 -1 le 0 0 0.2land 0 0 1 1 1 1 1 1 1 0 0 le 500 500 287.5answers 40000 14500 0 0 0 0 500 0 0 10000 5000

0 0 -169 -96 -43.5 -11.5 0 -9 -38.5 0 0

Dual

Grain Straw fert seed land RHSOBJ 0 0 0 0 500 MAX 143750Sale1 1 0 0 0 0 GE 4 4Sale2 0 1 0 0 0 GE 0.5 0.5Y1 -30 -10 0 10 1 GE -5 164.5Y2 -50 -17 5 10 1 GE -5 91Y3 -65 -22 10 10 1 GE -5 38.5Y4 -75 -26 15 10 1 GE -5 6.5Y5 -80 -29 20 10 1 GE -5 -5Y6 -80 -31 25 10 1 GE -5 4Y7 -75 -32 30 10 1 GE -5 33.5Z1 0 0 -1 0 0 GE -2 -2Z2 0 0 0 -1 0 GE -0.2 -0.2answers 4 0.5 2 0.2 287.5