00:02 Good day, my name is Afeez Adedokun

00:15 I'll be taking you for the math

00:17 and the topic is operations research.

00:19 At the end of this topic,

00:24 you should be able to differentiate the mooders

00:26 of operation research.

00:28 So, a practical example of problems

00:31 which involve the mooders.

00:33 Now, let's define what the operations research is.

00:40 This is an area of study which applies

00:42 mathematics and statistics to problem of management,

00:46 industry, business, commerce, organization, and production.

00:51 That is our operation research, what it's this with.

00:55 It's this with how we can solve problems

00:57 in some of these areas that is in management, industry,

01:01 business, commerce, and organization.

01:04 And it also deals with effective management of man

01:08 and man-shin system in any business enterprise,

01:11 or organization in order to minimize cost

01:15 and maximize profits.

01:17 What's the operation research also deals with

01:19 is how we can solve human problem

01:24 and how also the company, or a particular industry,

01:29 minimize cost and maximize profits.

01:32 Now, we have some steps in operations research.

01:38 The major steps involved in the application

01:41 of the operational research in solving management problems

01:44 are the first thing is to identify

01:46 and formulate the problems at hand.

01:48 That's the first thing.

01:49 We first of all know the problem.

01:51 What problem do we have in this company?

01:53 What problem do we have in this industry?

01:56 That's the first thing.

01:57 The second thing is to construct mathematical models

01:59 to solve their problems.

02:01 We'll construct mathematical problems.

02:03 That is, if we have identified the problem

02:06 that this is the problem we're having in our company,

02:08 then we use mathematical approach to do that.

02:12 Then the third one is to analyze or solve the model.

02:15 So having identified the problem,

02:18 having also analyzed the mathematical model for that,

02:22 then we analyze and solve the model.

02:25 Then we also draw conclusions from the solution

02:28 to the model.

02:29 Having solved the problem, the mathematical model,

02:33 then we can analyze, we can conclude that,

02:35 okay, this is the exact problem we have in our company.

02:38 Then the last one is implementation

02:42 of the findings of the results.

02:45 So having be able to conclude the problem we have,

02:49 then we can easily analyze,

02:50 and we can easily implement those things.

02:53 Those problems we have, we can implement it within our company.

02:57 Now, like we said, we have models in operational stage.

03:06 So we, these are the models.

03:09 The model from the context of operational stage

03:11 is an idealized representation of natural phenomena.

03:15 That is, the representation of the problem.

03:18 How will you represent our problem in mathematical

03:21 to solve a real life solution?

03:23 And that is what the operational stage

03:25 wants us to know.

03:27 Then the some models of operational stage

03:29 include, we have in linear programming,

03:32 models, we have a transportation models.

03:36 We also have an assignment models.

03:38 In all this model, the most popular method for solving,

03:45 the first one that we talked about,

03:46 and that is a linear programming models.

03:49 We have two methods of solving that.

03:51 We have a graphical method, and we have a simplest method.

03:55 Now, let's take a look to linear programming.

03:58 We want to see how we can maximize or minimize linear functions.

04:03 That is, we said earlier that the aim of the operational

04:05 stage is to minimize costs, I mean,

04:11 to minimize costs and maximize profits.

04:13 So we want to use this linear programming

04:16 to how we can solve problems in companies,

04:19 in industry, in commerce, and using a real life solution.

04:24 So we want to see how we can maximize linear functions.

04:28 When you have a linear function of z equals Ax plus BY.

04:35 So we represent this.

04:37 And over the region, we satisfy this

04:40 a given set of inequalities.

04:42 Linear function z equals Ax plus BY is called Objective Functions,

04:48 which the given sets of inequalities

04:51 are called constraints.

04:52 No mistake in our linear programming,

04:54 there are two things that we need.

04:55 We need what we call a linear function.

04:57 And we also need what we call the constraint.

05:02 And that is the set of inequalities

05:04 that we are given to solve the problems.

05:06 Having identified the problem in the company,

05:08 then using Mathematically, for example,

05:10 you have 2x plus 3y less than or 5.

05:14 This is a problem in the company.

05:17 How can we do that?

05:19 How can we use a linear programming

05:21 to solve this problem in mathematical model that we have?

05:28 So let's take a look at this example number one

05:30 to solve or to explain what the linear programming method is.

05:34 The number of units per kilogram of the two vitamin A

05:38 and B present in the two types of food x

05:41 and y are shown in the table below.

05:43 We have the table.

05:47 And that table is-- we have the first column

05:51 we have vitamin A. And the second column we have vitamin B.

05:55 We have in the first row we have x, y.

05:59 And that is 2, 1, 3, 4.

06:02 A patient P is allowed to take home

06:05 for a maximum of 19 units of vitamin A

06:08 and 17 units of vitamin B.

06:11 If the patient P consumes x kilogram of food x

06:17 and y kilogram of food y, write down the inequalities

06:21 involving x and y, if 15 nirah per kilogram is saved

06:26 in preparing food x, and 90 nirah per kilogram is used

06:32 to prepare food y, find how much each type of food the patient P

06:39 can take in order to save the greatest amounts.

06:42 And the solution to this--

06:49 don't forget that we said in our question

06:51 that a patient is allowed to take up

06:55 to a maximum number of 19 units of vitamin A.

06:58 We vitamin A for the food x is 2.

07:01 Vitamin A, vitamin B for food x is 1.

07:07 And that's for vitamin A for food y is 3.

07:10 And that of B is 4.

07:14 We have 2x plus 3y.

07:18 And we said 19 units of vitamin A. We need that less than or equal

07:22 to 19.

07:24 Then for x plus 4y, we add that it's going to be less than equals

07:29 to 17.

07:31 And we have x greater than equals to 0.

07:34 Y greater than equals to 0.

07:36 And since x and y cannot be negative,

07:40 our x cannot be negative, our y cannot be negative.

07:42 So that reason why it is x greater than equals to 0.

07:45 Y greater than equals to 0.

07:49 So from this, we have talked about inequalities earlier.

07:54 And we can easily draw these inequalities.

08:01 We can draw it in the form of a--

08:04 let's say we have something of this form.

08:07 And we have this.

08:13 This, the first one, the first straight line,

08:17 let's represent it to be 2x plus 3y less than or equal to 19.

08:21 And the second line, the second straight line,

08:24 let's say x plus 4y less than or equal to 17.

08:28 We add these two lines, means.

08:30 There's a point that we're going to note that that is 5, 3.

08:34 Then this is the origin of the two inequalities.

08:39 So these distributed portion A, if this is A,

08:46 we call this one A points.

08:47 Let's say this is C. There's a point A.

08:50 When your x is equal to 0, we have a number A.

08:54 When your x is-- when your y is equal to 0,

08:56 we have a particular number A is 2 as well.

08:58 So the points, the points, the distributed parts

09:05 and close by the polygon satisfy the four inequalities.

09:09 These four inequalities that we have, this,

09:12 this point that we have, the points when your A, when your A,

09:19 when it equals 0, 0, that is, when your x is 0, when your y is 0.

09:27 Your B is when you have your x to 0 and your y is 17 over 4.

09:34 And your C is when your x is 5 and your y is 3,

09:40 that is, the point where the first inequality

09:43 and second equality means.

09:45 Then D, we have 19 over 2 and we have your y to be 0.

09:53 Now, we put the value of--

09:57 you know, we said our x is 0, when your y is 0.

10:00 So we put the value of each of these x, y

10:05 in the formula 50, x plus 90 y.

10:10 And that is, you know, let's there be the amounts saved

10:15 in preparing S, K, S kilogram of food x

10:18 and y kilogram of food y.

10:20 Then we have this.

10:21 So when your x is equal to 0, 50, 0, 90, 0.

10:28 We have this, our z is going to be 0.

10:30 Then when B, as B, x is equal to 0, y is equal to 17 over 4.

10:40 So we put the value in that delinear programming.

10:44 Then we have 50 times 0 plus 90 times 10 over 4.

10:47 Then we have 3, it's 2.5.

10:50 Then as C, when our x is equal to 5 and y is equal to 3,

10:55 we put the value in the delinear programming

10:57 that is z is equal to 50 x plus 90 y.

11:00 Then our x is 5 times 5 times 5.

11:05 Then plus 90 times 3, we have 520.

11:07 Then as 90-- when your x is 19 over 2 and your y is 0,

11:12 we have that, and that is 4.75.

11:15 Now from this calculation, you see that the greatest amount

11:18 saved is 520 Naira.

11:21 You can see that the greatest amount we have here is 520 Naira.

11:24 And in our operation usage, it's due

11:30 with how we can maximize profits and minimize cost.

11:36 Now the second model is what we call the transportation

11:38 model, and that has to do with the class of linear programming

11:42 problems in operation stage.

11:45 And it is the transportation problem.

11:49 It is a type of allocation problem.

11:51 The basic problem of the type of linear programming

11:54 problem has to do with a distributor who

11:56 has a storage deposit.

11:58 And it's just like you want to move in a particular product

12:03 from one point to another.

12:05 So you have to move it, how can you save your time?

12:09 Number one, that is number one.

12:11 That is the cost of the production.

12:12 I mean, the cost of the production

12:14 and the cost of the transportation.

12:15 If you want to move your product A to a particular place,

12:22 how do you want to cost cost?

12:25 And that is what the transportation model

12:26 wants to teach us.

12:28 Then we have the cost of delivery depends

12:29 on the distance, that is where you want to take it to.

12:33 Then the time-taking, how long will it take you

12:36 to move your product A to a particular place?

12:39 Then the model of the mode of transportation

12:42 used, is it by road?

12:46 Probably you want to use a trailer to carry it

12:49 from the particular place.

12:50 You want to use the car.

12:51 And that is-- and the basic assumption

12:53 are the cost of delivery.

12:55 And second one is available work, supplies.

12:57 That is, what we need.

13:00 If a particular company said we need a particular amount

13:03 of goods, so you need to supply them

13:06 that's a amount of food that they need.

13:08 So in transportation model, we have the method of solving

13:12 that we have the least cost method

13:13 and we have the Northwest corner rule.

13:17 So we want to use this example to identify the least cost

13:21 method.

13:22 A plastic manufacturing company has

13:24 a production plant in Lagos, if I don't have an area.

13:26 The available stocks in the three plants are 50 tones,

13:30 40 tones and 30 tones respectively.

13:32 So we use this table.

13:34 OK, the company has a warehouse in the Shobu and Enugong,

13:37 which can accommodate 65 tones, 55 tones respectively.

13:41 Because of transportation, each tones of plastic

13:44 from each plant to each warehouse in Android of NERA

13:48 are given in the table below.

13:50 We have the table.

13:51 So if I'm moving the product from Lagos to Shobu,

13:55 I will have this unit of--

13:58 I will have seven.

13:59 If I'm moving from Lagos to Enugong, I will have 15,

14:02 then if I'm moving from advantage to Shobu, I will have 5,

14:04 if I'm going to Enugong, I will have it.

14:06 I will have 12, if I'm going to Enugong, I will have 6.

14:09 Now, we want to determine how much should be transported

14:12 from each point to each warehouse

14:15 as to minimize transportation cost.

14:18 And we want to find-- we want to use the least cost method

14:21 to find the transportation cost.

14:25 Now, we have the next table that we have in the solution

14:28 is we want to expand the table above to include

14:31 the demands and supplies--

14:33 or accompanying demands for your particular products.

14:35 And how many of that products are

14:38 used to applying to the company.

14:39 So the table in the previous slide, you know,

14:42 talks about what we want to send.

14:46 And we want to solve-- we want to use a least cost method.

14:50 In our second step is to include the supply and demand

14:55 to the table.

14:58 And that is we have the supply yet.

15:00 We have the first column, we have the second column yet,

15:03 and we have the first through second row and third row.

15:05 So we have 55 tombs, we have 55 tombs,

15:08 and we have 120 for the total amount of products

15:12 to be supplied at its 120.

15:14 And we proceed to the first stage using the table, table 1.

15:19 So we have this as well.

15:22 Now, this table, we proceed to the second stage

15:28 using the table below.

15:31 Now, the first thing is that we first of all identify

15:36 the shortest routes that we can take to transport our product

15:42 from the particular, from warehouse to the company.

15:46 So if I want to transport the vehicle at this table,

15:49 we have-- this is the least.

15:53 This is the least.

15:54 If you look at this table from here,

15:56 you see that this is the least.

15:57 So it means that the short test, the least cost method

16:02 is illustrated in the table, the preverse table

16:05 that we have, and explain as follows.

16:07 From the preverse table, the least value is five.

16:10 Like I said earlier, and the cheapest route

16:12 is the balance of showbook.

16:14 So we will now allocate 40 units.

16:17 We allocate 40 units of products to these

16:21 because it's short.

16:23 So the reason why we allocate 40 is because it's short,

16:27 and we want to make sure that we minimize the cost.

16:30 So we will make sure that the amount of products

16:34 be sent to that would be much.

16:36 So we now allocate 40.

16:38 So when we allocate 40, we write 40 beside that five.

16:42 Then also, the second least is six.

16:44 So we allocate 32 units.

16:46 And the supply of-- the supply from Ibadam to Ibadam

16:53 to Ushubu will not be zero because this is what we want to supply.

16:58 And we are ready to apply 40.

16:59 So it will move from 40 to zero.

17:02 And we now write 40 slash zero.

17:05 Then also, we allocate 38.

17:08 So our supply will also become zero.

17:11 Then also, we also allocate the next one is seven.

17:16 We allocate 25.

17:18 That's legal to Ushubu, 25.

17:20 legal to any one, 25.

17:21 So that will make out the supplies to be zero as well.

17:24 Then the least cost method is illustrated like I have explained.

17:29 And the next piece, I have explained this.

17:31 So we observe that how the units of the product from Ibadam

17:39 to Ibadam have been transported.

17:41 It remains only 25 units to legal splans.

17:44 We now allocate these 25 units remaining

17:48 through the legal to Ibadam.

17:49 If you look at this, you see that we legal to Enugun,

17:53 we also take another 25.

17:55 Legos to Ushubu with allocate 25.

17:57 Legos to Enugun will allocate 25 as well.

18:01 And that makes it to supply to be zero from 50 to 25 to 0.

18:06 And the transportation plan is now completed as shown in the table.

18:12 So we now have this.

18:14 We can see Legos to Ushubu is 25 now.

18:16 Legos to Enugun is 25.

18:18 The total supply is 50.

18:21 Ibadam to Ushubu is 40.

18:22 You can see we don't have anything here again 40.

18:25 How about to Ushubu?

18:25 We don't have anything 30, 30.

18:27 Then if you look at the amount of product that we

18:31 transported is 25 plus 465, 25 plus 30 is 55.

18:35 And the supplier, the supply product is 50, 40, 30,

18:39 and at the zone 20.

18:41 And this table shows the initial feasible transportation plan.

18:46 It shows the number of units of items allocated through each root.

18:52 So we now calculate--

18:54 this is where the list method comes in.

18:57 We calculate the list to the transportation cost

19:00 by multiplying each unit caused by the number of unit

19:05 allocated through that truth.

19:07 So the list to the cost equals 7 times 25 plus 15 times 25

19:21 plus 5 times 40 plus 6 times 30 whole multiplied by 100.

19:30 So we have the value for 7 times 25.

19:34 We have 175 plus 15 times 25.

19:39 We have 375.

19:42 Then 5 times 40.

19:44 We have 200.

19:47 6 times 30.

19:47 We have 180.

19:49 So adding all this to the data, they multiplied by 100.

19:53 We give us 93,000.

19:57 So this is the list to the cost to transport the product

20:02 from a particular from A to B.

20:10 Now, we also have an assignments model.

20:13 We have an assignments model.

20:17 And the assignment model is like the transportation model.

20:23 It's a type of allocation model.

20:26 In transportation model, some jobs

20:28 require more than one resource.

20:31 Or a resource can be used for more than one job.

20:35 In assignment model, the same number of jobs

20:38 are associated with numbers of resources.

20:40 The objective of the assignment model

20:42 is to assign each resource or facility to one and one

20:46 and only one job, so as to maximize the total cost

20:50 or maximize total profit.

20:55 Example of this is a state of universal basic education

20:59 board.

20:59 So the site to build four blocks of six classroom

21:02 at different location A, B, C, D of the state

21:06 true for contracts.

21:08 The contractors are W, X, Y, Z.

21:11 Each contractor is to be given only one block of six

21:14 classroom.

21:15 So we denote this in our table that the cost of building

21:20 each block of six classroom is determined by the beads.

21:24 Detenda by each contract, as shown in the table below.

21:28 You can see that this is a--

21:31 we have the contractors, and we have the location A, B, C.

21:38 And we want to assign the different locations

21:40 to contractors that will minimize the total cost.

21:46 And in doing this, how do we minimize the cost?

21:51 There are several techniques of solving

21:53 this type of problem.

21:57 One of the methods is we use what we call Hungarian method.

22:02 Or we call it a matrix reduced method.

22:05 We observe that the table we have above--

22:08 you see that this is--

22:10 we have the row, the first row, second row, the fourth row.

22:12 You can see this is the four by four matrix.

22:14 You can see.

22:15 So another reason why--

22:17 and how do we reduce-- how do we do this to minimize

22:21 the cost of the transportation?

22:23 And we observe that the table above is a rectangular array

22:25 of numbers, which consists of four rows and four columns.

22:29 Such a rectangular or square array of numbers

22:33 is called matrices.

22:35 And we observe that the row has the number--

22:38 it has 10, 8, 9, 12.

22:41 The row B has four--

22:43 the row B has four entries, which are 6, 7, 10, 8.

22:46 And row C has 8, 7, 5, 6.

22:48 Row D has four entries as 8, 10, 12, 9, 11.

22:53 So how do we solve-- how the first step is,

22:56 we stop trot the minimum entry of each row from all the entries.

23:01 If we have in our first-- in our fourths, this, in our fourths,

23:18 We have WXYZ, we have A, B, C, D, and we have 10 hits, 9/12, and we have 6/7, 10 hits, 8/7, 5/6, 10/12, 9/11.

23:44 This is a, we have, so the first step is, we subtract the minimum entry of each row,

23:54 the each we have the row, the first row, 8, each row from all the entries of the row,

24:00 and check if there's at least one zero in each row and column, and if you look at the first row we have

24:09 here, we, we, the list of the number we have here is 8, many that we deduct, we minus 8 from each

24:17 of the element we have in, in the column A, so we subtract, we subtract 8 from this,

24:24 we subtract 8 from this, we subtract 8 from this, we subtract 8 from this, the list number

24:32 we have here is 6, so we subtract 6 from here, we subtract 6 from here, we subtract 6 from here,

24:38 we subtract 6 from here, the list number we have here is 5, we subtract 5 from here,

24:42 we subtract 5 from here, we subtract 5 from here, we subtract 5, the list number we have in this

24:48 row 2 is 9, so we subtract 9, we subtract 9, we subtract 9, we subtract 9, and you see that this

24:57 will give us 2, so we have the table here, it give us 2, 0, 1, 4, this one 2 will give us 0, 1, 4, 2,

25:07 3, 2, 0, 1, 1, 3, 0, 2, and if we look at this, you see that the first row has at least 1, 0 entry,

25:16 the second row also has at least 1, 0 entry, the third row also has at least 1, 0 entry,

25:23 and the fourth row also has at least 1, 0 entry, now let's look at the column as well,

25:28 the column has at least 1, 0 entry, the second column also has at least 1, 0 entry,

25:36 at least 1, 0 entry, it can't be 2, but the last column does not have 0, so that means we have to

25:42 proceed by deducting the matches as well, so many data, we also deduct the list value from each

25:52 of the boxes that we have here, this is 0, if we deduct 0 from here to give us 2, 0 from here to give

25:59 us 2, 0 from here, give us 0, 0 from here we give you 0, 0 from here we give you 1, and if we look at this

26:06 also, if we deduct 0, this is 0, this is the list number, 0 minus 0, we give out 0, 1 minus 0, 1,

26:13 to give out the same number, if deduct 0 is the list number as well, and the list number here is 1,

26:18 so we minus 1 from this, 4 minus 1, that will give us 3, 2 minus 1, that will give us 1, 1 minus 1,

26:26 give us 0, 2 minus 1, we give us 1, you can see that the last column 2 has 0 entry,

26:31 and that is the table we have here, you can see 2, 0, 1, 3, the previous table was reduced by deducting 1

26:38 from this, so we have 3, 1, 0, 4, you can see that we have achieved that each row and each column,

26:44 we have at least 1, 0 entry, and the top step is we check each row of the table above to find an

26:55 on mark 0 entry, put a box, you can see we put a box, a kind of bracket, we put it where we have 0,

27:02 that is the first row half is at least 1, we have this, we have this as well,

27:06 and the first step is you check each column also to find an on mark 0 and place the mark of,

27:16 you place that also around the on mark 0 and cross the order 0 in the same row, we observe from

27:22 the last table above that there are assignments in each of the row and in each column, there is assignments

27:29 of model in each of these, where we have 0, 0, and you know, and it's just to minimize the cost,

27:36 so you can see that our row a column x has 0, also row b column w has 0 to row c,

27:46 column z has 0 row d column y has 0, and hence the optimal assignment plan is the location a,

27:55 you know, our look a, b, c are the locations, y, x, y, z, x, w, w, x, y, z are the contractors,

28:04 so we have location a, contractor x, location b, contractor y, location c, contractor z,

28:10 locations d contractor y, the row a column x, row b column w, row c column z, row d column y,

28:20 and hence the ultimate assignment plan is the location a has contractor w,

28:26 location b has contractor x, location c has contractor y, location d has contractor z,

28:35 and to find a total minimum cost, we combine the tables with all the tables that we have in

28:42 our previous explanation, we combine them to play that and each assignment is given in numeric

28:48 value of 1, meaning that where we have an assignment plan, each we are located 1 for where there is

28:57 an assignment plan, so you can see that where there's an assignment plan we have 1 yet in brackets,

29:04 where there's an assignment plan we have 1 yet in brackets, where there's an assignment plan here,

29:09 we have 1, that is where our zero is, and where an assignment plan here we have 1 as well,

29:15 so we had all these to find total minimum cost, that is 8 times 1 plus 6 times 1 plus 6 times 1,

29:24 and that is what we have, the total minimum cost is going to be 8 times 1 plus 6 times 1 plus 6 times 1

29:32 plus 9 times 1, then we multiply it by 1 million, and that is 8 times 1 we give us 8 times 6 we give us 6,

29:42 and 6 times 1 we give us 6, and 9 times 1 we give us 9, so having this together,

29:46 we have that is 8 plus 6 plus 6 plus 9, and that will give us 14, 20, 29, 29 times 1 million

29:58 we give us 29 million, and that is the total minimum cost, so having this cost, the three

30:04 models in solving in an operational research, how we can and the aim of the objective is to

30:13 how we can minimize the cost and maximize the profits in any content that we have, and we have

30:21 this evolution show graphically the region how which satisfies the set of inequalities,

30:28 so the evolution so far now is we are to show graphically the region how which satisfy the set

30:37 of inequalities, we have the two x plus 3 y less than or equals to 26, x plus 2 y less than or

30:43 equals to 16, x greater than or equals to 0, y greater than or equal to 0, our x and y cannot be

30:51 negative, so that is the reason why we have x greater than or equals to 0, you can see that the

30:56 first inequality can be less than, so it can be negative, but our x and y cannot be negative, so

31:02 thanks for watching, my name is Afiz, till next time.