Project of Quantitative Methods for Business

Group No: 18• 63-Ankita Nakum• 112-Dhara Vadodaria• 168-Heer Shah• 235-Khushi• 300-Meshwa Patel

 

Linear Programming Problem

Food Item                   Subject to        Constraints                                                                                                    

•                        ≥                                         

Linear Programming Problem• The main goal is to select a set of foods that meets certain daily 

nutritional requirements and preferences and additionally at minimum cost. Let´s consider the following list of foods with their nutritional profile to illustrate this application:

• We wish to propose a diet containing at least 2000 (Kcal), at least 55 grams of protein and at least 300g of carbohydrates. In addition, to provide some variety in the diet, some limits are set for the daily portions of food. This information is required to find the diet that has the lowest cost associated with meeting the above requirements.

Data RepresentationFood Portion

SizeCalories(Kcal)

Protein(grams)

Carbohy-drates

(grams)

Price (Rs/portion)

Limit(Portion/

day)Masala

Oats40g 629 27g 70g Rs.15 3

Eggs 2 eggs (100g)

140 12g 0g Rs.18 2

Milk(Amul Shakti)

500g 360 16g 24g Rs.25 4

Soya beans

250g 432.5 41.6g 24.8g Rs.25 2

CalculationsDecision variables:-• X1= Masala Oats• X2=Eggs• X3=Milk (Amul Shakti)• X4=Soya beans

Objective Function:-• Min Z=15X1+18X2+25X3+25X4

• Subject to Constraints:-• Minimum Calories(Kcal):629X1+140X2+360X3+432.5X4 ≥ 2000

• Minimum Protein(Grams): 27X1+12X2+16X3+41.6X4 ≥ 55

• Minimum Carbohydrates(Grams):70X1+0X2+24X3+24.8X4 ≥ 300

• Portion: X1≤ 3, X2 ≤ 2, X3≤ 4, X4≤ 2

• Non negativity: X1, X2, X3, X4 ≥ 0 

Calculations(cont…) 

Masala Oats Eggs

Milk(Amul Shakti) Soya beans

Food 15 18 25 25

RESOURCES USED

SIGN IN CONSTRAINTS

Resources Available

Minimum Calories 629 140 360 432.5 3358≥ 2000Minimum Protein 27 12 16 41.6

191.1333333≥ 55

Minimum Carbs 70 0 24 24.8 300≥ 300

Portion 1 1 0 0 0 3≤ 3

Portion 2 0 1 0 0 0≤ 2

Portion 3 0 0 1 01.68333333

3≤ 4

Portion 4 0 0 0 1 2≤ 2

total cost

Diet 3 01.68333333

3 2137.083333

3

Conclusion

• A Person can have a minimum cost on diet is Rs.145 and can have diet as shown in table above.   

Variables PortionMasala Oats 3Eggs 0Milk 2 (aprox)Soya Beans 2

Transportation Problem

Transportation Problem• Mr. Ankit Rathod is a merchant who is having shop in kalupur market. It is mainly

into business of selling of 3 types of products that is fruit(apples), vegetable(potatoes) and spices. He is having 3 shops for each of this products that is Ratnaraj:- fruit (apples), Harihar :-vegetable(potatoes) and Vasant masala:- Spices. This merchant is sending its goods for storage at four different warehouses/coldstorage.

• The four warehouses are at • Hapa Cold Storage - Naroada Gidc • Mahalaxmi Cold Storage - Changodar• Creata Cold Storage - Odhav • Vinod Cold Storage - Narmada• Transportation cost are given for sending goods from shop to warehouse in the

table below Mr. Ankit wants to know that where he can send goods more so that his transportation cost gets reduce. Get an optimum solution for the following case.

Data RepresentationShop/Warehouse

Hapa Cold Storage

Mahalaxmi Cold Storage

Creata Cold Storage

Vinod Cold Storage

Supply

Ratnaraj fruit shop

10 20 12 16 2000

Harihar vegetable shop

15 14 16 18 3000

Vasant masala shop

12 18 13 15 1500

Demand 4000 500 1000 1000 6500

Calculations: Initial Basic Feasible Sol. By VAM

WarehouseShop

Hapa Cold Storage

Mahalaxmi Cold Storage

Creata Cold Storage

Vinod Cold Storage

Supply D1 D2 D3

Ratnaraj fruit shop 

                                            10                 20                 12                 16

2000/0 2 2 -

Harihar vegetable shop 15

         14

        16

           18

3000/ 2500/ 500/0

1 1 1

Vasant masala shop

12 18 13 151500/0 1 1 1

Demand 4000/2000/500/1500/0

500/0 1000/0 1000/0 6500

D12 4 1 1

D22 - 1 1

D33 - 3 3

2000

500 10001000500

1500

Calculation: Check Optimality by Modi Method

Shop/Warehouse

Hapa Cold Storage

Mahalaxmi Cold Storage

Creata Cold Storage

Vinod Cold Storage

Supply  ui

Ratnaraj fruit Shop

                          

10

20

  

              12                  16 2000 10

Harihar vegetable  shop

 15

                         14 

16                     18 

       3000

                          

          15Vasant masala shop 

12                     18                    13                  15      1500

 

          12

Demand 4000 500 1000 1000    6500

Vj 0 -1 1 3

2000

500

1500

10001000500

+11 +1 +3

00+7

Conclusion• When dij= 0 , the current solution is basic 

feasible solution and it remains unaffected . • There may exist alternative solution.

Cost Allocations Total10 2000 2000015 500 750012 1500 1800014 500 700016 1000 1600018 1000 18000

Total Cost 86500

Assignment Problem

 Jobs                                                               Person

Assignment Problem

• A tuition class has three jobs to do. The three jobs are HW checking & sending messages, Accountant and paper checking. There are three persons available to do the job. Our goal is to assign one and only one job to each person in such a way that the total time of assignment is minimized. Following table shows the time(in hrs) taken by each person to do different jobs with some restrictions.

Data Representation

Job                Person

HW checking & messages

Accountant Paper checking

A(Kanubhai)

2

B(Shikha)

1.5 3 4.5

C(Kenny)

6 3.5 1

CalculationsJob                

Person

HW checking & messages

Accountant Paper checking

A K 2 K

B 1.5 3 4.5

C 6 3.5 1

CalculationJob                Person

HW checking & messages

Accountant Paper checking

A K 0 K

B 0 1.5 3

C 5 2.5 0

ConclusionJob Person Time(in 

hrs)

HW checking & messages

A 2

Accountant B 1.5

Paper checking C 1

                    Total   4.5

Replacement Problem

Replacement Problem

New mixing mill of the size which they havetoday cost of Rs. 2 lacs. Normally a mixing millused and handled carefully and with preventivemaintaince norms it can be used for a 10 yearswithout any major break down.

Data RepresentationYear Maintenance Resale 

Value(S)

1 30000 1500002 35000 1400003 40000 1300004 50000 1200005 55000 1100006 60000 1000007 70000 900008 80000 80000

Calculation

Year Maintenance

Maintenance  cost accumulate Ef(t)

Resale Value(S)

Total Cost C-S+Ef(t)

The Average Total Annual Cost C-S+Ef(t)/n

1 30000 30000 150000 80000 800002 35000 65000 140000 125000 625003 40000 105000 130000 175000 58333.334 50000 155000 120000 235000 587505 55000 210000 110000 300000 600006 60000 270000 100000 370000 61666.677 70000 340000 90000 450000 64285.718 80000 420000 80000 540000 67500

PERT CPM

PERT  CPMIntroduction to Case/Problem• A construction company made a winning bid of  Rs 35 crores to construct a new 

plan for a major manufacturer. The manufacturer needs the plant to go into operation within a year. Therefore the contract included following provisions

• A penalty of Rs 2 crore if the construction company did not complete the construction by the deadline of 47 weeks from the day of commencement.

• To provide additional incentive for speedy construction, a bonus of Rs 1.5 crore would be paid to the construction company if the plant is completed within 40 days.

• By using PERT/CPM method we will get valuable help for answering the following question:

• Develop the associated network for the project.• What is the total time taken by the construction company to complete the work 

if no delay occur?

Data RepresentationActivity Activity description Preceding activity Duration (in weeks)

ABCDEFGHIJKLMN

ExcavateLay the foundationPut up the rough wallPut up the roofInstall exterior plumbingInstall interior plumbingPut up exterior sidingDo exterior paintingDo electronic workPut up the wallboardInstall flooringDo interior paintingInstall exterior fixturesInstall interior fixtures

--ABCCED

E,GCF,IJJH

K,L

24

1064579784526

Calculations

1 2 3 4 136

9

5 78

10

12

11

Conclusion

• There is two dummy activities denoted as D1 and D2.

• Construction project will get complete at the end of the 44 weeks.