quantitative methods for business project ppt
TRANSCRIPT
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.