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3QA3/2DA3

Management Science for Business

Lecture 03

C01: June 28, 2021

C02: June 29, 2021

Instructors:

Seyyed Hossein Alavi Zdravko Dimitrov

Practice Question P2-13?

 Let us look at Handout 1.1

LP Models in the Business World

Develop a written LP model on paper before

attempting to implement it on Excel!

 We deal with problems:

 with more than 2 variables

 require more complicated modelling techniques

✓ Understand the logic behind a model before implementing it on the

computer!

✓ If the model is incorrect or incomplete from a logical perspective

(even if it is correct from a mathematical perspective), Excel cannot

recognize the logical error!

1- Blending Problem

Determining the optimal (least costly) mix of ingredients

 Oil Companies  Mix of different crude oils and other chemicals for gasoline

 Lawn Care Companies  Mix of chemicals and other products to make fertilizers

 Agriculture Industry (Feed mix problem)  Mix of grain, corn, and minerals to make different feeds for various farm

animals

 Diet Problem  Select a set of foods that will satisfy a set of daily nutritional requirement

1- Blending Problem

 Handout 1.2.

 An agriculture company received an order for 8,000 pounds of feed

 At least 20% corn, 15% grain, 15% minerals

 What should the company do to minimize the cost?

Nutrient Percent of Nutrition in

Feed 1 Feed 2 Feed 3 Feed 4

Corn 30% 5% 20% 10%

Grain 10% 30% 15% 10%

Minerals 20% 20% 20% 30%

Cost/pound $0.25 $0.30 $0.32 $0.15

1- Blending Problem

 Decision Variables

 𝑥1 = amount of feed 1 in pounds to use in the blend

 𝑥2 = amount of feed 2 in pounds to use in the blend

 𝑥3 = amount of feed 3 in pounds to use in the blend

 𝑥4 = amount of feed 4 in pounds to use in the blend

 Objective Function

Minimize 0.25𝑥1 + 0.30𝑥2 + 0.32 𝑥3 + 0.15𝑥4

1- Blending Problem

 Constraints

 Pounds of feed required

𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 8,000

 Min % of corn required

0.30𝑥1 + 0.05𝑥2 + 0.20𝑥3 + 0.10𝑥4 ≥ 0.20(8,000)

 Min % of grain required

0.10𝑥1 + 0.30𝑥2 + 0.15𝑥3 + 0.10𝑥4 ≥ 0.15(8,000)

 Min % of minerals required

0.20𝑥1 + 0.20𝑥2 + 0.20𝑥3 + 0.30𝑥4 ≥ 0.15(8,000)

 Non-negativity constraints

𝑥1, 𝑥2, 𝑥3, 𝑥4 ≥ 0 8

2- Investment/Portfolio Problem

 Finance Problems

 Maximizing the expected return of an investment

 Selecting from a variety of alternatives (companies)

 Satisfying certain cash flow requirements

 Satisfying certain risk constraints

 Satisfying certain policy requirements

2- Investment/Portfolio Problem

 Investing $750,000 in below companies

 No more than 25% of money in any company

 At least half of the money be invested in long-term bonds that mature in 10 or more years

 No more than 35% in low rating companies (3 and 4)

Company Rate of return Years to Maturity Rating

1 8.65% 11 1-Excellent

2 9.50% 10 3-Good

3 10.00% 6 4-Fair

4 8.75% 10 1-Excellent

5 9.25% 7 3-Good

6 9.00% 13 2-Very Good

2- Investment/Portfolio Problem

 Decision Variables

 𝑥1 = amount of money to invest in Company 1

 𝑥2 = amount of money to invest in Company 2

 𝑥3 = amount of money to invest in Company 3

 𝑥4 = amount of money to invest in Company 4

 𝑥5 = amount of money to invest in Company 5

 𝑥6 = amount of money to invest in Company 6

 Objective Function

Maximize .0865𝑥1 + .095𝑥2 + .10𝑥3 + .0875𝑥4 + .0925𝑥5 + .09𝑥6

2- Investment/Portfolio Problem

 Constraints

 Total investment

𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 = 750,000

 Max amount in one bond

𝑥1, 𝑥2, 𝑥3, 𝑥4 , 𝑥5, 𝑥6 ≤ 0.25 750,000

 Min amount invested in long-term bond

𝑥1 + 𝑥2 + 𝑥4 + 𝑥6 ≥ 0.5 750,000

 Max amount invested in low rating bond

𝑥2 + 𝑥3 + 𝑥5 ≤ 0.35 750,000

 Non-negativity constraints

𝑥1,𝑥2, 𝑥3,𝑥4 ,𝑥5,𝑥6 ≥ 0 12

3- Transportation Problem

Distribution of goods from several supply points to a number of

demand points

 Each supply point has a capacity for production.

 Each demand point has a certain demand.

 Minimizing the distance/cost

 Can also be defined as maximization of profit

3- Transportation Problem

Sources Destinations

Supply A B C

D $5 $4 $3 200

E $6 $7 $5 300

F $3 $8 $5 400

Demand 350 150 400 900=900

3- Transportation Problem

Sources Destinations

200 350

150

400

300

400

3- Transportation Problem

 Decision Variables

𝑥𝑖𝑗 = number of goods shipped from source 𝑖 to destination 𝑗

where 𝑖 = D, E or F, and 𝑗 = A, B or C

 Objective Function

Minimize 5𝑥DA + 4𝑥DB + 3𝑥DC

+ 6𝑥EA + 7𝑥EB + 5𝑥EC

+ 3𝑥𝐹𝐴 + 8𝑥FB + 5𝑥FC 16

3- Transportation Problem

 Constraints (Flow in = Flow out)

 Net flow at supply node D

200 = 𝑥DA + 𝑥DB +𝑥DC

 Net flow at supply node E

300 = 𝑥EA + 𝑥EB +𝑥EC

 Net flow at supply node F

400 = 𝑥FA + 𝑥FB +𝑥FC

 Net flow at demand node A

𝑥DA + 𝑥EA +𝑥FA = 350

 Net flow at demand node B

𝑥DB + 𝑥EB +𝑥FB = 150

 Net flow at demand node C 𝑥DC + 𝑥EC +𝑥FC = 400

 Non-negativity constraints

𝑥𝑖𝑗 ≥ 017

3- Transportation Problem

 Supply > Demand

 Supply equality must change to inequality

 𝑆𝑢𝑝𝑝𝑙𝑦 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒(𝐹𝑙𝑜𝑤 𝑖𝑛) ≥ 𝐹𝑙𝑜𝑤 𝑜𝑢𝑡

e.g. 300 ≥ 𝑥DA + 𝑥DB +𝑥DC

Sources Destinations

Supply A B C

D $5 $4 $3 300

E $6 $7 $5 300

F $3 $8 $5 400

Demand 350 150 400 900/1000

3- Transportation Problem

 Supply < Demand

 Demand equality must change to inequality

 𝐹𝑙𝑜𝑤 𝑖𝑛 ≤ 𝐷𝑒𝑚𝑎𝑛𝑑 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑(𝐹𝑙𝑜𝑤 𝑜𝑢𝑡)

e.g. 𝑥DA + 𝑥EA +𝑥FA ≤ 350

Sources Destinations

Supply A B C

D $5 $4 $3 200

E $6 $7 $5 300

F $3 $8 $5 400

Demand 350 150 500 1000/900

4- Crew/Labor Scheduling

 Determining the optimal number of staff for a specific planning

horizon (month, week, shift, flight, …)

 Staffing needs could be different during different time periods of

the planning horizon

4- Crew/Labor Scheduling  A bank requires between 10 and 18 tellers

 Can be full-time (at most 12) and part-time

 Part-time salary: $28/day full-time: $90/day

 Part-time work is 4 hours, can start from 9 am to1:00 pm

 Full-time tellers work from 9 am to 5 pm with 1 hour for lunch

 50% have lunch at 11:00 am, others at noon

 Part-time hours cannot exceed 50% of the total required

Time Period # Required

9 a.m.–10 a.m. 10

10 a.m.–11 a.m. 12

11 a.m.–Noon 14

Noon–1 p.m. 16

1 p.m.–2 p.m. 18

2 p.m.–3 p.m. 17

3 p.m.–4 p.m. 15

4 p.m.–5 p.m. 10

4- Crew/Labor Scheduling

 Decision Variables

 𝐹 = number of full−time tellers to use (starting from 9 am to 5 pm)

 𝑃1 = number of part-time tellers starting from 9 am to 1 pm

 𝑃2 = number of part-time tellers starting from 10 am to 2 pm

 𝑃3 = number of part-time tellers starting from 11 am to 3 pm

 𝑃4 = number of part-time tellers starting from 12 pm to 4 pm

 𝑃5 = number of part-time tellers starting from 1 pm to 5 pm

 Objective Function

Minimize 90𝐹 + 28(𝑃1 + 𝑃2 + 𝑃3 + 𝑃4 + 𝑃5) 22

4- Crew/Labor Scheduling

 Constraints

 𝐹+𝑃1 ≥ 10 (9 am-10 am requirement)

 𝐹+𝑃1+𝑃2 ≥ 12 (10 am-11am requirement)

 0.5𝐹+𝑃1+𝑃2+𝑃3 ≥ 14 (11 am-12 pm requirement)

 0.5𝐹+𝑃1+𝑃2+𝑃3+𝑃4 ≥ 16 (12 pm-1 pm requirement)

 𝐹+𝑃2+𝑃3+𝑃4+𝑃5 ≥ 18 (1 pm-2 pm requirement)

 𝐹+𝑃3+𝑃4+𝑃5 ≥ 17 (2 pm-3 pm requirement)

 𝐹+𝑃4+𝑃5 ≥ 15 (3 pm-4 pm requirement)

 𝐹+𝑃5 ≥ 10 (4 pm-5 pm requirement)

 𝐹 ≤ 12 (Max # of full-timers)

 4(𝑃1+𝑃2+𝑃3+𝑃4+𝑃5) ≤ 0.5(10+12+14+16+18+17+15+10) = 56 (Max hours of part-timers)

 𝐹,𝑃1,𝑃2,𝑃3,𝑃4,𝑃5 ≥ 0 (Non-negativity constraints) 23

5- Multi-period Applications

 Determine the optimal decisions for several periods

 It is challenging because the decisions are not independent!

 The decision of period 𝑡 depends on the decision of periods 𝑡 − 1, 𝑡 − 2, …

 Applications in:

 Production

 Finance

 etc.

5- Multi-period Production Planning

 A company is trying to plan its production, with the objective of minimizing the total production and total inventory carrying cost.

 Inventory Capacity: 6,000 (Current inventory: 2,000)

 Safety stock: 1,500

 Produce no less than half of max capacity

 Cost of carrying inventory: 1.5% of unit cost

 Inventory carrying cost is accounted based on the ending inventory each month (after demand is met)

 What is the optimal quantity to produce for each month?

Month

1 2 3

Unit cost $240 $250 $265

Demand 1,000 4,500 4,000

Max Prod.

Capacity 4,000 3,500 4,000

5- Multi-period Production Planning

 Decision Variables

• 𝑃1 = number of units to produce in month 1

• 𝑃2 = number of untis to produce in month 2

• 𝑃3 = number of units to produce in month 3

• 𝐼1 = ending inventory in month 1

• 𝐼2 = ending inventory in month 2

• 𝐼3 = ending inventory in month 3

 Objective Function

Minimize 240𝑃1 + 250𝑃2 + 265𝑃3 +3.6𝐼1 + 3.75𝐼2 + 3.98𝐼3

5- Multi-period Production Planning

 Constraints

 Production constraints:

 2,000 ≤ 𝑃1≤ 4,000 (production level for month 1)

 1,750 ≤ 𝑃2≤ 3,500 (production level for month 2)

 2,000 ≤ 𝑃3≤ 4,000 (production level for month 3)

5- Multi-period Production Planning

 Constraints

 Balance Constraints:

Ending Inventory = Beginning Inventory + Units Produced − Demand

 𝐼1 = 𝐼0 + 𝑃1 − 1,000 (balance constraint for month 1)

 𝐼2 = 𝐼1 + 𝑃2 − 4,500 (balance constraint for month 2)

 𝐼3 = 𝐼2 + 𝑃3 − 4,000 (balance constraint for month 3)

 Warehouse Capacity & Safety Stock

 1,500 ≤ 𝐼1≤ 6,000 (ending inventory for month 1)

 1,500 ≤ 𝐼2≤ 6,000 (ending inventory for month 2)

 1,500 ≤ 𝐼3 ≤ 6,000 (ending inventory for month 3)28

6- Vehicle Loading Problem

 Deciding on items to load on a vehicle

 Maximizing the value of the load shipped

 Constraints

 Weight and volume limit of vehicle

 Minimum level of certain items

6- Vehicle Loading problem

 The company has 2 vehicles with

 Weight capacity 10,000 pounds

 Volume capacity 900 cubic feet

 The weight of both vehicles must be equal.

 The objective is to maximize the value of loaded items.

ITEM VALUE WEIGHT

(pounds) Value per pound

VOLUME

(cu. ft. per pound)

1 $15,500 5,000 3.10 0.125

2 $14,400 4,500 3.20 0.064

3 $10,350 3,000 3.45 0.144

4 $14,525 3,500 4.15 0.448

5 $13,000 4,000 3.25 0.048

6 $9,625 3,500 2.75 0.018

Practice Problems

 CHAPTER 3

 Problems 3, 7, 9, 12, 13, 17, 21

 CHAPTER 5

 Problem 17