{"id":16551,"date":"2024-03-13T11:46:16","date_gmt":"2024-03-13T06:16:16","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=16551"},"modified":"2024-03-14T16:26:50","modified_gmt":"2024-03-14T10:56:50","slug":"samacheer-kalvi-12th-business-maths-solutions-chapter-10-miscellaneous-problems","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-business-maths-solutions-chapter-10-miscellaneous-problems\/","title":{"rendered":"Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems"},"content":{"rendered":"

Students can download 12th Business Maths Chapter 10 Operations Research Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems<\/h2>\n

Question 1.
\nThe following table summarizes the supply, demand and cost information for four factors S1<\/sub>, S2<\/sub>, S3<\/sub>, S4<\/sub> shipping goods to three warehouses D1<\/sub>, D2<\/sub>, D3<\/sub>.
\n\"Samacheer
\nFind an initial solution by using the north-west corner rule. What is the total cost of this solution?
\nSolution:
\nLet ‘ai<\/sub>‘ denote the supply and ‘bj<\/sub>‘ denote the demand.
\nThen total supply = 5 + 8 + 7 + 14 = 34 and Total demand = 7 + 9 + 18 = 34
\n\u03a3ai<\/sub> = \u03a3bj<\/sub>. So the problem is a balanced transportation problem and we can find a basic feasible solution, by North-west comer rule.
\nFirst allocation:
\n\"Samacheer
\nSecond allocation:
\n\"Samacheer
\nThird allocation:
\n\"Samacheer
\nFourth allocation:
\n\"Samacheer
\nFifth allocation:
\n\"Samacheer
\nWe first allot 4 units to cell (S3<\/sub>, D3<\/sub>) and then the balance 14 units to cell (S4<\/sub>, D3<\/sub>).
\nThus we get the following allocations:
\n\"Samacheer
\nThe transportation schedule:
\nS1<\/sub> \u2192 D1<\/sub>, S2<\/sub> \u2192 D1<\/sub>, S2<\/sub> \u2192 D2<\/sub>, S3<\/sub> \u2192 D2<\/sub>, S3<\/sub> \u2192 D3<\/sub>, S4<\/sub> \u2192 D3<\/sub>
\n(i.e) x11<\/sub> = 5, x21<\/sub> = 2, x22<\/sub> = 6, x32<\/sub> = 3, x33<\/sub> = 4, x43<\/sub> = 14
\nTotal cost = (5 \u00d7 2) + (2 \u00d7 3) + (6 \u00d7 3) + (3 \u00d7 4) + (4 \u00d7 7) + (14 \u00d7 2)
\n= 10 + 6+ 18 + 12 + 28 + 28
\n= 102
\nThus the initial basic solution is got by NWC method and minimum cost is Rs. 102.<\/p>\n

Question 2.
\nConsider the following transportation problem
\n\"Samacheer
\nDetermine an initial basic feasible solution using
\n(a) Least cost method
\n(b) Vogel’s approximation method
\nSolution:
\nLet ‘ai<\/sub>‘ denote the availability and ‘bj<\/sub>‘ denote the requirement. Then,
\n\u03a3ai<\/sub> = 30 + 50 + 20 = 100 and \u03a3bj<\/sub> = 30 + 40 + 20 + 10 = 100
\nSince \u03a3ai<\/sub> = \u03a3bj<\/sub>. the given problem is a balanced transportation problem and we can get an initial basic feasible solution.
\n(a) Least Cost Method (LCM)
\nFirst allocation:
\n\"Samacheer
\nSecond allocation:
\n\"Samacheer
\nThird allocation:
\n\"Samacheer
\nFourth allocation:
\n\"Samacheer
\nFifth allocation:
\n\"Samacheer
\nWe first allocate 10 units to cell (O2<\/sub>, D2<\/sub>). Since it has a minimum cost. Then we allocate the balance 10 units to cell (O1<\/sub>, D2<\/sub>). Thus we get the final allocation table as given below.
\n\"Samacheer
\nTransportation schedule:
\nO1<\/sub> \u2192 D2<\/sub>, O1<\/sub> \u2192 D3<\/sub>, O2<\/sub> \u2192 D1<\/sub>, O2<\/sub> \u2192 D2<\/sub>, O2<\/sub> \u2192 D4<\/sub>, O3<\/sub> \u2192 D2<\/sub>
\n(i.e) x12<\/sub> = 10, x13<\/sub> = 20, x21<\/sub> = 30, x22<\/sub> = 10, x24<\/sub> = 10, x32<\/sub> = 20
\nThe total cost is = (10 \u00d7 8) + (20 \u00d7 3) + (30 \u00d7 4) + (10 \u00d7 5) + (10 \u00d7 4) + (20 \u00d7 2)
\n= 80 + 60+ 120 + 50 + 40 + 40
\n= 390
\nThus the LCM, we get the minimum cost for the transportation problem as Rs. 390.
\n(b) Vogel’s approximation method:
\nFirst allocation:
\n\"Samacheer
\nThe largest penalty = 3. So allocate 20 units to the cell (O3<\/sub>, D2<\/sub>) which has the least cost in column D2<\/sub>
\nSecond allocation:
\n\"Samacheer
\nLargest penalty = 4. So allocate min (20, 30) units to the cell (O1<\/sub>, D3<\/sub>) which has the least cost in column D3<\/sub>
\nThird allocation:
\n\"Samacheer
\nThe largest penalty is 3. So allocate min (20, 50) to the cell (O2<\/sub>, D2<\/sub>) which has the least cost in column D2<\/sub>.
\nFourth allocation:
\n\"Samacheer
\nLargest penalty = 2. So allocate min (10, 30) to cell (O2<\/sub>, D4<\/sub>) which has the least cost in column D4<\/sub>
\nFifth allocation:
\n\"Samacheer
\nLargest penalty = 1. Allocate min (30, 20) to cell (O2<\/sub>, D1<\/sub>) which has the least cost in column D1<\/sub>. Finally allot the balance 10 units to cell (O1<\/sub>, D1<\/sub>)
\nWe get the final allocation table as follows.
\n\"Samacheer
\nTransportation schedule:
\nO1<\/sub> \u2192 D1<\/sub>, O1<\/sub> \u2192 D3<\/sub>, O2<\/sub> \u2192 D1<\/sub>, O2<\/sub> \u2192 D2<\/sub>, O2<\/sub> \u2192 D4<\/sub>, O3<\/sub> \u2192 D2<\/sub>
\n(i.e) x11<\/sub> = 10, x13<\/sub> = 20, x21<\/sub> = 20, x22<\/sub> = 20, x24<\/sub> = 10, x32<\/sub> = 20
\nTotal cost is given by = (10 \u00d7 5) + (20 \u00d7 3) + (20 \u00d7 4) + (20 \u00d7 5) + (10 \u00d7 4) + (20 \u00d7 2)
\n= 50 + 60 + 80+ 100 + 40 + 40
\n= 370
\nHence the minimum cost by YAM is Rs. 370.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nDetermine an initial basic feasible solution to the following transportation problem by using (i) North-West Corner rule (ii) least-cost method.
\n\"Samacheer
\nSolution:
\nTotal supply = 25 + 35 + 40 = 100 = \u03a3ai<\/sub>
\nTotal requirement = 30 + 25 + 45 = 100 = \u03a3bj<\/sub>
\nSince \u03a3ai<\/sub> = \u03a3bj<\/sub> the given transportation problem is balanced and we can find an initial basic feasible solution.
\n(i) North West Corner Rule
\nFirst allocation:
\n\"Samacheer
\nSecond allocation:
\n\"Samacheer
\nThird allocation:
\n\"Samacheer
\nFourth allocation:
\n\"Samacheer
\nWe first allot 5 units to cell (S2<\/sub>, D3<\/sub>). Then balance 40 units we allot to cell (S3<\/sub>, D3<\/sub>).
\nThe final allotment is given as follows
\n\"Samacheer
\nTransportation schedule:
\nS1<\/sub> \u2192 D1<\/sub>, S2<\/sub> \u2192 D1<\/sub>, S2<\/sub> \u2192 D2<\/sub>, S2<\/sub> \u2192 D3<\/sub>, S3<\/sub> \u2192 D3<\/sub>
\n(i.e) x11<\/sub> = 25, x21<\/sub> = 5, x22<\/sub> = 25, x23<\/sub> = 5, x33<\/sub> = 40
\nTotal cost is = (25 \u00d7 9) + (5 \u00d7 6) + (25 \u00d7 8) + (5 \u00d7 4) + (40 \u00d7 9)
\n= 225 + 30 + 200 + 20 + 360
\n= 835
\nHence the minimum cost is Rs. 835 by NWC method.<\/p>\n

(ii) Least cost method (LCM)
\nFirst allocation:
\n\"Samacheer
\nSecond allocation:
\n\"Samacheer
\nThird allocation:
\n\"Samacheer
\nFourth allocation:
\n\"Samacheer
\nWe first allot 15 units to cell (S3<\/sub>, D1<\/sub>) since it has the least cost. Then we allot the balance 15 units to cell (S1<\/sub>, D1<\/sub>).
\nThe final allotment is given as follows
\n\"Samacheer
\nTransportation schedule:
\nS1<\/sub> \u2192 D1<\/sub>, S1<\/sub> \u2192 D3<\/sub>, S2<\/sub> \u2192 D3<\/sub>, S3<\/sub> \u2192 D1<\/sub>, S3<\/sub> \u2192 D2<\/sub>
\n(i.e) x11<\/sub> = 15, x13<\/sub> = 10, x23<\/sub> = 35, x31<\/sub> = 15, x32<\/sub> = 25
\nTotal cost is = (15 \u00d7 9) + (10 \u00d7 5) + (35 \u00d7 4) + (15 \u00d7 7) + (25 \u00d7 6)
\n= 136 + 50 + 140 + 105 + 150
\n= 580
\nThe optimal cost by LCM is Rs. 580.<\/p>\n

Question 4.
\nExplain Vogel’s approximation method by obtaining an initial basic feasible solution to the following transportation problem.
\n\"Samacheer
\nSolution:
\nTotal supply (ai<\/sub>) = 6+1 + 10=17
\nTotal demand (bj<\/sub>) = 7 + 5 + 3 + 2 = 17
\n\u03a3ai<\/sub> = \u03a3bj<\/sub>. So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.
\nFirst allocation:
\n\"Samacheer
\nThe largest penalty is 6. So allot min (2, 1) to the cell (O2<\/sub>, D4<\/sub>) which has the least cost in column D4<\/sub>.
\nSecond allocation:
\n\"Samacheer
\nThe largest penalty is 5. So allot min (5, 6) to the cell (O1<\/sub>, D2<\/sub>) which has the least cost in column D2<\/sub>.
\nThird allocation:
\n\"Samacheer
\nThe largest penalty is 5. So allot min (7, 1) to the cell (O1<\/sub>, D1<\/sub>) which has the least cost in row O1<\/sub>.
\nFourth allocation:
\n\"Samacheer
\nLargest penalty = 4. So allocate min (6, 10) to the cell (O3<\/sub>, D1<\/sub>) which has the least cost in row O3<\/sub>
\nFifth allocation:
\n\"Samacheer
\nThe largest penalty is 6. So allot min (1, 4) to the cell (O3<\/sub>, D4<\/sub>) which has the least cost in row O3<\/sub>. The balance 3 units is allotted to the cell (O3<\/sub>, D3<\/sub>). We get the final allocation as given below.
\n\"Samacheer
\nTransportation schedule:
\nO1<\/sub> \u2192 D1<\/sub>, O1<\/sub> \u2192 D2<\/sub>, O2<\/sub> \u2192 D4<\/sub>, O3<\/sub> \u2192 D1<\/sub>, O3<\/sub> \u2192 D3<\/sub>, O3<\/sub> \u2192 D4<\/sub>
\n(i.e) x11<\/sub> = 1, x12<\/sub> = 5, x24<\/sub> = 1, x31<\/sub> = 6, x33<\/sub> = 3, x34<\/sub> = 1
\nTotal cost is given by = (1 \u00d7 2) + (5 \u00d7 3) + (1 \u00d7 1) + (6 \u00d7 5) + (3 \u00d7 15) + (1 \u00d7 9)
\n= 2 + 15 + 1 + 30 + 45 + 9
\n= 102
\nThus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is Rs. 102.<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nA car hire company has one car at each of five depots a, b, c, d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given in the following distance matrix.
\n\"Samacheer
\nHow should the cars be assigned to the customers so as to minimize the distance travelled?
\nSolution:
\nIn the given assignment problem, the number of rows equals the number of columns. So the problem is balanced and we can get an optimal solution.
\nStep 1:
\n\"Samacheer
\nWe have subtracted the minimum distance of each row from all elements of that row.
\nStep 2:
\n\"Samacheer
\nWe have subtracted the minimum distance of each column from all elements of that column.
\nStep 3: (Assignment)
\n\"Samacheer
\nWe are not able to assign depots for C and E. So we proceed as below.
\n\"Samacheer
\nWe have drawn a minimum number of lines to cover all zeros. Now 15 is the smallest element from all uncovered elements. Subtract this from all uncovered elements and add to the elements which lie at the intersection of two lines. Thus we obtain the following reduced matrix for fresh assignment.
\n\"Samacheer
\nAssignment Schedule
\n\"Samacheer
\nThus the minimum distance is 570 miles.<\/p>\n

Question 6.
\nA natural truck-rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometres) between the cities with a surplus and the cities with a deficit are displayed below:
\n\"Samacheer
\nHow should the truck be dispersed so as to minimize the total distance travelled?
\nSolution:
\nThe number of rows equals the number of columns. So the problem is a balanced assignment problem and we can get an optimal solution.
\nStep 1: Subtract the minimum element of each row from all elements of that row.
\n\"Samacheer
\nStep 2: Subtract the least element of each column from all elements of that column.
\n\"Samacheer
\nStep 3: (Assignment)
\n\"Samacheer
\nWe are not able to assign city for starting place 4. We proceed as follows.
\nDraw a minimum number of lines to cover all the zeros of the matrix. Subtract the smallest element from all uncovered elements and add to the elements which lie at the intersection of two lines.
\n\"Samacheer
\nThen we obtain a new reduced matrix for fresh assignment.
\n\"Samacheer
\nFinal dispersal
\n\"Samacheer
\nThus the minimum total distance the trucks should travel is 125 km.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nA person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay-off matrix based on three potential economic conditions is given in the following table:
\n\"Samacheer
\nDetermine the best investment plan using each of the following criteria (i) Maximin (ii) Minimax
\nSolution:
\n(i) Maximin rule
\n\"Samacheer
\nMax (3000, 1000, 6000) = 6000. Since the maximum value is 6000, he must invest in debentures by the maximin criteria.
\n(ii) Minimax rule
\n\"Samacheer
\nMin (10000, 8000, 6000) = 6000. Since the minimum value is 6000, he must invest in debentures by the minimax criteria.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download 12th Business Maths Chapter 10 Operations Research Miscellaneous Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems Question …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16551"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=16551"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16551\/revisions"}],"predecessor-version":[{"id":60776,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/16551\/revisions\/60776"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=16551"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=16551"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=16551"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}