Guideline Answers for May 2013 CA Final Advanced Management Accountig Guideline Answers to May 2013 FinalExam Advanced Management Accounting Question No.1 is compulsory (4 × 5 = 20 Marks). Answer any 5 questions from the remaining 6 questions (16 × 5 = 80 Marks). [Any 4 out of 5 in Q.7] 1(a): Relevant Cost Analysis – Basic Decision Making M 13 (5 Marks) A Process Industry Unit manufactures three Joint Products, A, B and C. C has no realizable value unless it undergoes further processing after the point of separation. The cost details of C are as follows – Particulars Per Unit Upto point of separation Marginal Cost 30 Fixed Cost 20 After point of separation Marginal Cost 15 Fixed Cost 5 Total Cost 70 C can be sold at ` 37 per unit and no more. 1. Would you recommend production of C? 2. Would your recommendation be different if A, B and C are not Joint Products? Solution: Case 1: If C is a Joint Product – Net Benefit if C is processed and sold (37 – 15) = ` 22 Conclusion: Hence, Production of C can be recommended. Note: 1. Cost upto the point of separation is sunk / irrelevant, if C is a Joint Product. 2. Fixed Cost p.u after separation, being apportioned, is irrelevant. Case 2: If C is not a Joint Product (i.e. Separate Product) – Net Benefit or (Loss) (37 – 30 – 15) = Loss ` 8 Conclusion: Hence, Production of C is not worthwhile. Note: Fixed Cost p.u (before and after separation), being apportionment, is irrelevant. 1(b): Pricing Decision – Price Reduction to increase Sales Quantity M 13 (5 Marks) HTM Ltd by using 12,00,000 units of a Material M produces jointly 2,00,000 units of H and 4,00,000 units of T. The Costs and Sales details are – Particulars Amount Direct Material M at ` 5 per unit 60,00,000 Other Variable Costs 42,00,000 Total Fixed Costs 18,00,000 Selling Price of H per unit 25 Selling Price of T per unit 20 The Company receives an additional order for 40,000 units of T at the rate of ` 15 per unit. If this order has been accepted, the existing Price of T will not be affected. However, the present Price of H should be reduced evenly on the entire Sale of H to market the additional units to be produced.Find the Minimum Average Unit Price to be charged on H to sustain the increased Sales. Solution: Similar to Page 3.15, Illustration 3 (N 99) 1. Computation of Additional Raw Material required Particulars Raw Material M Product H Product T Given relationship 12,00,000 units 2,00,000 units 4,00,000 units To produce extra qtty of T (pro–rata) 1,20,000 units (pro–rata) 20,000 units (Given) 40,000 units Downloaded From cacracker, visit: cacracker for more updates & files... Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.2 2. Computation of Revised Selling Price of H (a) Additional Raw Material Costs = 1,20,000 units × ` 5 ` 6,00,000 (b) Additional Variable Processing Costs = 1,20,000 units × (` 42,00,000 ÷ 12,00,000 units) `4,20,000 (c) Total Additional Costs (a + b) `10,20,000 (d) Additional Sale Revenue obtained from Product T = 40,000 units × ` 15 `6,00,000 (e) Net Recovery required from Product H (c – d) For 20,000 units `4,20,000 (f) Present Sale Revenue from Product H = 2,00,000 units × ` 25 per unit For 2,00,000 units ` 50,00,000 (g) Total Desired Sale Revenue from 2,20,000 units of Product H (e + f) For 2,20,000 units ` 54,20,000 (h) Desired Selling Price per kg of Product H = ` 54,20,000 ÷ 2,20,000 units ` 24.64 per unit 1(c): Assignment – Steps (Theory) M 13 (5 Marks) Prescribe the steps to be followed to solve an assignment problem. Solution: Refer Page 16.1, Q.No. 1, Point 4 1(d): Marginal Costing – Indifference Point Interpretation and Decision Making M 13 (5 Marks) X Ltd wants to replace one of its old Machines. Three alternative Machines namely M1, M2 and M3 are under its consideration. The Costs associated with these Machines are – Particulars M1 M2 M3 Direct Material Cost p.u. 50 100 150 Direct Labour Cost p.u. 40 70 200 Variable Overhead p.u. 10 30 50 Fixed Cost p.a. 2,50,000 1,50,000 70,000 You are required to compute the Cost Indifference Points for these alternatives. Based on these points suggest a most economical alternative Machine to replace the old one when the expected level of Annual Production is 1,200 units. Solution: Similar to Page 2.30, Illustration 2.1 (N 00, M 02) Particulars M1 M2 M3 Fixed Costs ` 2,50,000 ` 1,50,000 ` 70,000 Variable Costs per report ` 100 ` 200 ` 400 Indifference Point = Difference in Variable Cost per Unit Difference in Fixed Costs , Between M1 and M2 = 200 100 2,50,000 1,50,000 ` ` ` ` − − = 1,000 units. Between M2 and M3 = 400 200 1,50,000 70,000 ` ` ` ` − − = 400 units. Between M3 and M1 = 400 100 2,50,000 70,000 ` ` ` ` − − = 600 units. 2. Interpretation of Indifference Points and Decisions: (the numbers indicate the number of reports / cases handled) No. of Cases Nil 400 600 1000 Choice M3 M2 or M3 M2 M1 or M3 M2 M1 or M2 M1 Number of Cases / Reports Choice of Method Reason Less than 400 units M3 Due to Lower Fixed Cost. Exactly 400 units Either M2 or M3 Indifference Point. Above 400 but less than 1,000 units M2 Next Range of Lower Fixed Costs. Exactly 1,000 units Either M1 or M2 Indifference Point. Above 1,000 units M1 Lower Variable Costs per unit. Expected Production = 1,200 units. This is in the range – above 1,000 units, and hence M1 can be purchased. Note: Indifference Point between M1 & M3 (600 units) is not relevant for decision–making since M2 is profitable in the range 400 units to 1,000 units. This Indifference Point will be relevant only if the choice lies between M1 and M3. Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.3 3. Cost of various options at 600units: Particulars M1 M2 M3 Fixed Costs ` 2,50,000 ` 1,50,000 ` 70,000 Variable Costs ` 100×600 = ` 60,000 ` 200×600 = ` 1,20,000 ` 400×600 = ` 2,40,000 Total Costs ` 3,10,000 ` 2,70,000 ` 3,10,000 At this level, M2 is more profitable than M1 and M3. Hence, this Indifference Point becomes irrelevant in the overall analysis among M1, M2 and M3. 2(a): ABC in Service Sector – Cost of Rendering Services M 13 (8 Marks) DEF Bank operated for years under the assumption that profitability can be increased by increasing Rupee volumes. But that has not been the case. Cost Analysis has revealed the following – Activity Activity Cost (`) Activity Driver Activity Capacity Providing ATM Service 1,00,000 No. of transactions 2,00,000 Computer Processing 10,00,000 No. of transactions 25,00,000 Issuing Statements 8,00,000 No. of statements 5,00,000 Customer Inquiries 3,60,000 Telephone minutes 6,00,000 The following annual information on three products was also made available – Checking Accounts Personal Loans Gold Visa Units of Product 30,000 5,000 10,000 ATM Transactions 1,80,000 0 20,000 Computer Transactions 20,00,000 2,00,000 3,00,000 Number of Statements 3,00,000 50,000 1,50,000 Telephone Minutes 3,50,000 90,000 1,60,000 Required: 1. Calculate rates for each activity. 2. Using the rates computed in requirement (1) above, calculate the Cost of each Product. Solution: Similar to Page 8.26, Illustration 12 (N 09) Activity Cost Driver Rates Checking Accounts Personal Loans Gold Visa Providing ATM Service 2,00,000 ` 1,00,000 = 0.50 per transaction 1,80,000 × 0.50 = 90,000 – 20,000 × 0.50 = 10,000 Computer Processing 25,00,000 ` 10,00,000 = 0.40 per transaction 20,00,000 × 0.40 = 8,00,000 2,00,000 × 0.40 = 80,000 3,00,000 × 0.40 = 1,20,000 Issuing Statements 5,00,000 ` 8,00,000 = 1.60 per statement 3,00,000 × 1.60 = 4,80,000 50,000 × 1.60 = 80,000 1,50,000 × 1.60 = 2,40,000 Customer Services 6,00,000 ` 3,60,000 = 0.60 per telephone minute 3,50,000 × 0.60 = 2,10,000 90,000 × 0.6 = 54,000 1,60,000 × 0.60 = 96,000 Total Cost 15,80,000 2,14,000 4,66,000 Units of Product 30,000 5,000 10,000 Cost per unit 52.67 42.80 46.60 2(b): Production and Material Usage Budget M 13 (8 Marks) KG Ltd is engaged in the production of two Products K and G. One unit of Product K requires two units of Material A and four units of Material B. Each unit of Product G needs four units of Material A, two units of Material B and four units of Material C. Material C is locally produced in the factory of the Company, by using two units of Material B for each unit of C. Materials A and B are purchased in the open market. Production of Products K, G and C is carried out evenly throughout the year. At present the Company has purchased its 3 months requirements of A and B in one purchase. That is four purchases per annum. The other particulars provided by the Company are – Downloaded From cacracker, visit: cacracker for more updates & files... Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.4 Particulars K (In Units) G (In Units) Budgeted Sales for the next year 40,000 75,000 Desired Stock at the end of the year 5,000 10,000 Expected Stock at the beginning of the year 15,000 25,000 Particulars A B Purchase Price p.u (`) 15 25 Ordering Cost per Order (`) 1,000 1,000 Carrying Cost p.a. 10% 10% You are required to: 1. Prepare a Production Budget and a Material Requirement Budget for the next year. 2. Calculate the number of Material Purchases to be made, if the Company wants to Purchase Materials in optimal quantity. Solution: Similar to Page 7.10, Illustration 2 (M 06, M 10) 1. Computation of Budgeted Production Quantities Particulars Product K Product G Budgeted Sales 40,000 75,000 Add: Stock at the end of the year 5,000 10,000 Sub – Total 45,000 85,000 Less: Stock at the beginning of the year 15,000 25,000 Budgeted Production Quantity 30,000 60,000 2. Computation of Budgeted Component Requirements and EOQ Particulars Material A Material B Material C Product K: 30,000 units 2 × 30,000 = 60,000 4 × 30,000 = 1,20,000 NA Product G: 60,000 units 4 × 60,000 = 2,40,000 2 × 60,000 = 1,20,000 +(for making C) 2,40,000 × 2=4,80,000 4 × 60,000 = 2,40,000 (to be made internally) Total Requirement p.a. 3,00,000 units 7,20,000 units (already considered in B) Buying Cost per Order ` 1,000 ` 1,000 Carrying Cost p.u. p.a. 10% of ` 15 = ` 1.50 10% of `25 = `2.5 EOQ = C 2AB 20,000 units 24,000 units No. of Purchases p.a. 3,00,000 ÷ 20,000 = 15 times 7,20,000 ÷ 24,000 = 30 times 3(a): Linear Programming – Maximisation under Simplex Method M 13 (8 Marks) A Company manufactures two Products A and B, involving three Departments – Machining, Fabrication and Assembly. The process time, profit/unit and total capacity of each department is given in the following table – Particulars Machining (Hrs) Fabrication(Hrs) Assembly(Hrs) Profit(`) A 1 5 3 80 B 2 4 1 100 Capacity 720 1,800 900 Required – 1. Set up Linear Programming Problem to maximize Profits. 2. What will be the Product–Mix at Maximum Profit Level? 3. What will be the Profit at that Level? Solution: Similar to Page 18.9, Illustration 4 [M 05 (Mod.), M 11] Let x, y be the number of units of the two products A and B respectively. Basic Data: A B Profitp.u. `80 `100 Hours conditions: Available Machining 1 hr 2 hrs 720 hrs Fabrication 5 hrs 4 hrs 1,800 hrs Assembly 3 hrs 1 hr 900 hrs The LPP is as under – Maximise Z = 80 x + 100 y Subject to: x + 2y ≤ 720 5x + 4y ≤ 1,800 3x + y ≤ 900 x, y ≥ 0 (non–negativity) Introducing Slack Variables, we have Maximise Z = 80x + 100y + 0S1+0S2+0S3 Subject to: x + 2y + S1= 720 5x + 4y + S2 = 1,800 3x + y + S3 = 900 x, y, S1, S2, S3 ≥ 0 S1, S2 and S3are the slacks for the 3 Constraints above. Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.5 First Simplex Table: Fixed Ratio Program Profit Quantity x y S1 S2 S3 Repl. Ratio NA S1 0 720 1 2 1 0 0 360Min non–ve 2 S2 0 1800 5 4 0 1 0 450 1/2 S3 0 900 3 1 0 0 1 900 Decision: In = Key Column = y Out= Key Row = S1 Z (Objective Value) 80 100 0 0 0 C (Computed Value) 0 0 0 0 0 Net Evaluation Row 80 100Max +ve 0 0 0 Note:For Non–Key Rows, A = (Previous Table Corresponding Row Element) LessB = (Key Row Element × Fixed Ratio) Computation for S2 Row Computation for S3 Row A 1800 5 4 0 1 0 A 900 3 1 0 0 1 – B 1440 2 4 2 0 0 – B 360 1/2 1 1/2 0 0 A–B 360 3 0 –2 1 0 A–B 540 5/2 0 –1/2 0 1 The above A–B values are carried over to the Second Simplex Table in S2 and S3 Rows (being Non–Key Rows of 1st Table). Second Simplex Table: Fixed Ratio Program Profit Quantity x y S1 S2 S3 Repl. Ratio 1/6 Y 100 360 1/2 1 1/2 0 0 720 NA S2 0 360 3 0 –2 1 0 120Min non–ve 5/6 S3 0 540 5/2 0 –1/2 0 1 216 Decision: In = Key Column = x Out= Key Row = S2 Z (Objective Value) 80 100 0 0 0 C (Computed Value) 50 100 50 0 0 Net Evaluation Row 30Max +ve 0 –50 0 0 Note:For Non–Key Rows, A = (Previous Table Corresponding Row Element) LessB = (Key Row Element × Fixed Ratio) Computation for y Row Computation for S3 Row A 360 1/2 1 1/2 0 0 A 540 5/2 0 –1/2 0 1 – B 60 1/2 0 –1/3 1/6 0 – B 300 5/2 0 –5/3 5/6 0 A–B 300 0 1 5/6 –1/6 0 A–B 240 0 0 7/6 –5/6 1 The above A–B values are carried over to the Third Simplex Table in yand S3 Rows (being Non–Key Rows of 2nd Table). Third Simplex Table: Fixed Ratio Program Profit Quantity x y S1 S2 S3 Repl. Ratio Y 100 300 0 1 5/6 –1/6 0 x 80 120 1 0 –2/3 1/3 0 S3 0 240 0 0 7/6 –5/6 1 Decision:Since all NER ≤ 0 for max. objective, the Third Table is optimal. Z (Objective Value) 80 100 0 0 0 C (Computed Value) 80 100 30 10 0 Net Evaluation Row 0 0 –30 –10 0 Answer: The Company should produce 120 units of Product A and 300 units of Product B. Maximum Contribution = 80 x + 100 y = (80 × 120) + (100 × 300) = 9,600 + 30,000 = ` 39,600. 3(b): Reverse Working – Overhead Variance Only M 13 (8 Marks) The following are the information regarding Overheads of a Company – 1. Overheads Cost Variance –` 2,800 (A) 2. Overheads Volume Variance –` 2,000 (A) 3. Budgeted Overheads –` 12,000 4. Actual Overhead Recovery Rate –` 8 per hour 5. Budgeted Hours for the period – 2,400 hours You are required to compute the following: 1. Overheads Expenditure Variance. 2. Actual Incurred Overheads. 3. Actual Hours for Actual Production. 4. Overheads Capacity Variance. 5. Overheads Efficiency Variance. 6. Standard Hours for Actual Production. Downloaded From cacracker, visit: cacracker for more updates & files... Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.6 Solution: 1. Expenditure Variance = Cost Variance 2,800 A (–) Volume Variance 2,000A = ` 800A 2. Actual OH = Budgeted Overhead 12,000 + Expenditure Variance 800A = ` 12,800 3. Actual Hours = Actual Re cov ery Rate Actual OH = 8 12800 = 1,600 hours 4. Capacity Variance = (AH – BH) × SR p.h = (1,600 – 2,400) × 5 = ` 4,000 A 5. Efficiency Variance = Volume Variance 2,000A – Capacity Variance 4,000A = ` 2,000 F 6. (SH – AH) × SR = Efficiency Variance 2,000 F So, (SH – 1,600) × 5 = 2000 So, SH = 2,000 hours 4(a): Transportation Problem – Optimality Test M 13 (8 Marks) XYZ Company has three Plants and four Warehouses. The Supply and Demand in units and the corresponding Transportation Costs are given. The table below shows the details taken from the solution procedure of the Transportation Problem – Warehouse I II III IV Supply A 5 10 10 4 5 10 B 20 6 8 7 5 2 25 C 5 4 10 2 5 5 7 20 Demand 25 10 15 5 55 Answer the following questions. Give brief reasons – 1. Is this solution feasible? 2. Is this solution degenerate? 3. Is this solution optimum? Solution: 1. Feasible Solution: (a) The Initial Basic Feasible Solution is given in the question. From the allocation of units it can be interpreted that the solution is determined as per Vogel’s Approximation Method. (See Note Below) (b) Since the above IBFS satisfies all the Row and Column Totals, the solution can be said to be a feasible solution. Note: Least Cost Cell is not used since the Least cost of 5 (A–I, A–IV) remains unallocated. North West Corner Rule is not used since the North West Corner remains unallocated (A–I). 2. Degeneracy: If the Number of Allocations < (m + n – 1), the solution is said to be degenerate. Since in this solution the Number of allocations (6) = m + n – 1 (3 + 4 – 1 = 6) the solution is not degenerate, and can be tested for optimality. 3. Optimality Test: Table 1 = U, V for allocated cells computed as below: U & V 0 + 4 = 4 0 + 2 = 2 0 + 5 = 5 2 – 2 = 0 4 – 5 = –1 5 10 10 4 5 6 – 4 = 2 20 6 8 7 5 2 Base = 0 5 4 10 2 5 5 7 Table 2 = U + V for Unallocated Cells Table 3 = Net Evaluation Table (NET) = Table 1 – Table 2 for Unallocated Cells 4 – 1 = 3 2 – 1 = 1 0 – 1 = –1 5 – 3 = 2 10 – 1 = 9 5 – (–1) = 6 2 + 2 = 4 5 + 2 = 7 8 – 4 = 4 7 – 7 = 0 0 + 0 = 0 7 – 0 = 7 Conclusion: Since all the elements in NET are ≥ 0, the IBFS is optimal but not unique. Since there is a “0” in the NET, there is an Alternate Solution. Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.7 4(b): TOC – Bottleneck Identification using TA Ratio and Resource Allocation M 13 (8 Marks) Gupta Ltd produces 4 Products P, Q, R and S by using three different Machines X, Y and Z. Each Machine Capacity is limited to 6,000 hours per month. The details given below are for July – Particulars P Q R S Selling Price p.u. (`) 10,000 8,000 6,000 4,000 Variable Cost p.u (`) 7,000 5,600 4,000 2,800 Machine Hours required p.u. Machine X Machine Y Machine Z 20 20 20 12 18 6 4 6 2 2 3 1 Expected Demand (units) 200 200 200 200 Required: 1. Find out the Bottleneck Activity. 2. Allocate the Machine Hours on the basis of the Bottleneck. 3. Ascertain the Profit expected in the month if the monthly Fixed Cost amounts to ` 9,50,000. 4. Calculate the unused Spare Hours of each Machine. Solution: Similar to Page 12.2, Illustration 1 (M 09) 1. Identification of Bottleneck Activity Machine Time required for Products (Demand × M/c Hr p.u.) Total Time reqd (Hrs) Time Available (Hrs) Machine P Q R S Utilization (a) (b) (c) (d) (e) = (a+b+c+d) (f) = given (g) = (e ÷ f) X 4,000 2,400 800 400 7,600 6000 126.67% Y 4,000 3,600 1,200 600 9,400 6000 156.67% Z 4,000 1,200 400 200 5,800 6000 96.67% Since Machine Y has the highest Machine Utilization (i.e. TA Ratio), it represents the bottleneck activity. Hence product, ranking & resource allocation should be based on Contribution per Machine Hour of Machine Y. 2. Allocation of Resources and overall Profit Particulars P Q R S Total (a) Contribution per unit ` ` 3,000 ` 2,400 ` 2,000 ` 1,200 (b) Time required in Machine Y 20 hours 18 hours 6 hours 3 hours (c) Contribution per Machine–hour ` 150 ` 133.33 ` 333.33 ` 400 (d) Rank based on (c) above III IV II I (e) Allocation of Machine Y time 200×20 = 4000 (bal. fig.)200 200×6 = 1200 200×3 = 600 6,000.00 (f) Production Quantity (e÷b) 200 units 11.11 units 200 units 200 units (g) Allocation of Machine X time 200 × 20 = 4,000 hours 11.11 ×12 = 133.32 hours 200 × 4 = 800 hours 200 × 2 = 400 hours 5333.32 (h) Allocation of Machine Z time 200 × 20 = 4,000 hours 11.11×6 = 66.66hours 200 × 2 = 400 hours 200 × 1 = 200 hours 4666.66 (i) Contribution based on allocation 200 × 3,000 = 6,00,000 11.11 × 2,400 = 26,664 200 × 2,000 = 4,00,000 200 × 1,200 = 2,40,000 12,66,664 (j) Fixed Cost for the month – – – – (9,50,000) (k) Profit for the Month – – – – 3,16,664 Note: Spare Capacity – 1. Machine X: (6000 – 5333.32) = 666.68 Hours 2. Machine Z: (6000 – 4666.66) = 1333.34 Hours 5(a): Profitability Analysis – Evaluation of Alternatives M 13 (12 Marks) Better and Best Ltd manufactures only one Product. Production is regular throughout the year and the capacity of the Factory is 1,50,000 units per annum. The summarized Profit and Loss Account for the year ended 31st December is being reviewed by the Board of Directors. Downloaded From cacracker, visit: cacracker for more updates & files... Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.8 Particulars ` Sales at ` 10 per unit 10,00,000 Cost of Sales – Direct Materials Direct Labour 2,50,000 1,50,000 Production Overheads – Variable Fixed 30,000 2,30,000 Administrative Overheads (Fixed) 1,00,000 Selling and Distribution Overhead – Variable Fixed 50,000 1,50,000 1. The Production Director proposed to reduce Selling Price to ` 9 in order to utilize Full Capacity. 2. The Sales Director proposed to increase Selling Price by 20%. By spending ` 2,25,000 on Advertisement, Sales will be increased to 1,20,000 units per annum. 3. The Personnel Director pleaded for a change in the method of Wage payment. For the present Piece Rate of ` 1.50 per unit, a Bonus Scheme (for each 2% increase in Production over the target, there would be an increase of 1% in the Basic Wage of each employee) will be implemented. A target of 2,000 units per week for the Company will be set for 50 week year. Selling Price increase by 10%. With an additional Advertisement Cost of ` 1,60,000, 20% increase in present Sales will be achieved. 4. The Chairman felt that the packaging of the product required improvement. He wanted to know the Sales required to earn a target Profit of 10% on Turnover with the introduction of an improved packing at an additional cost of 20 paise per unit (no change in Selling Price). You are required to evaluate individually the proposals of each of the Board Member and give your recommendation. Solution: 1. Profitability Statement (Present) Particulars Per Unit Per Unit Total Total Sales (10,00,000 ÷ 10 per unit = 1,00,000 units) 10 Less: Variable Costs Direct Material (2,50,000 ÷ 1,00,000 = 2.50 p.u.) Direct Labour (1,50,000 ÷ 1,00,000 = 1.50 p.u.) Variable POH (30,000 ÷ 1,00,000 = 0.30 p.u.) Variable SOH (50,000 ÷ 1,00,000 = 0.50 p.u.) 2.50 1.50 0.30 0.50 4.80 Contribution (5.20 p.u. × 1,00,000 units) 5.20 5,20,000 Less: Fixed Costs (only for Total) POH AOH SOH 2,30,000 1,00,000 1,50,000 4,80,000 Profit 40,000 2. Proposal I (Reduction in Selling Price) Particulars Total Contribution [(9.00 – 4.80) × 1,50,000 units] 6,30,000 Less: Fixed Costs 4,80,000 Profit 1,50,000 Note: Selling Price is reduced to 9.00 p.u. and also quantity is increased to full capacity 1,50,000 units. 3. Proposal II (Increase in Selling Price and Advertisement Cost) Particulars Total Contribution [(12.00 – 4.80) × 1,20,000 units] 8,64,000 Less: Fixed Costs (Existing Fixed Costs + Additional Advertising Costs) (4,80,000 + 2,25,000) 7,05,000 Profit 1,61,000 Note: Increase in Selling Price = 10 + 20% = 12 p.u. The No. of units is increased to 1,20,000 (given). Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.9 4. Proposal III (Increase in Sales, Advertisement Costs and Wage Revision) Alternative 1: It is assumed that 20% increase pertains to “Sales Value”. Alternative 2: It is assumed that 20% increase pertains to “Sales Quantity”. Alternative 1 Alternative 2 Particulars Per Unit Total Per Unit Total Sales 11 11 Less: Variable Costs Direct Material Direct Labour (See Note Below) Variable POH Variable SOH 2.50 1.57 0.30 0.50 2.50 1.65 0.30 0.50 Contribution (6.13 p.u. × 1,09,091 units) (6.05 p.u. × 1,20,000 units) 6.13 6,68,728 6.05 7,26,000 Less: Fixed Costs (only for Total) Fixed POH Fixed AOH Fixed SOH Additional Advertisement Cost 2,30,000 1,00,000 1,50,000 1,60,000 2,30,000 1,00,000 1,50,000 1,60,000 Profit 28,728 86,000 Note: Particulars Alternative 1 Alternative 2 Revised Selling Price 10 + 10% = 11 p.u. 10 + 10% = 11 p.u. Present Production 2,000 units per week × 50 weeks = 1,00,000 units 2,000 units per week × 50 weeks = 1,00,000 units Expected Production (a) Increase in overall Sales = 10,00,000 + 20% = 12,00,000 (b) Therefore Expected Production = 12,00,000 ÷ 11 p.u. = 1,09,091 Units 1,00,000 + 20% = 1,20,000 Units Increase in Production (Units) 1,09,091 – 1,00,000 = 9,091 units 1,20,000 – 1,00,000 = 20,000 Increase in Production (%) 9,091 ÷ 1,00,000 = 9.09% 20% (Given) Increase in Wage Rate due to increase in Production 9.09 ÷ 2 = 4.55% 20.00 ÷ 2 = 10.00%. Increase in Wage Rate 1.50 p.u + 4.55% = 1.57 1.50 p.u + 10.00% = 1.65 Note: For every 2% increase in production, 1% Bonus is given. Therefore the ratio is ½. 5. Proposal IV (Sales Quantity to achieve the required Profit) Let required Quantity be “x”. Particulars Total Sales (X units × 10 p.u.) 10x Less: Variable Costs (Existing + Additional Packing Costs) (4.80 + 0.20) 5x Contribution 5x Less: Fixed Costs 4,80,000 Profit (Given) 10% × 10x Number of Units to be sold and overall Profit: 5x – 4,80,000 = 1x (10% × 10x) So, 5x – x = 4,80,000 4x = 4,80,000 X = 1,20,000 Therefore, Profit = 10% (1,20,000 × 10 p.u.) = ` 1,20,000 Recommendation: Particulars Present Proposal I Proposal II Proposal III (Alt 1) Proposal III (Alt 2) Proposal IV Units Produced 1,00,000 1,50,000 1,20,000 1,09,091 1,20,000 1,20,000 Selling Price 10 9 12 11 11 10 Contribution 5,20,000 6,30,000 8,64,000 6,68,728 7,26,000 6,00,000 Profit 40,000 1,50,000 1,61,000 28,728 86,000 1,20,000 Based on the criteria of Profit, the Proposal II made by the Sales Director is more beneficial to the Company. Downloaded From cacracker, visit: cacracker for more updates & files... Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.10 5(b): Activity Based Costing – Direct Product Profitability (Thoery) M 13 (4 Marks) What do you mean by DPP? What are its benefits? Solution: Refer Page 8.9, Q.No 22 6(a): Network Analysis – Project Crashing M 13 (7 Marks) The Noida Nirman Authority intends to install a 4–road traffic regulating signal in a heavy traffic prone area. The total installation work has been broken down into six activities. The Normal Duration, Crash Duration and Crashing Cost of the activities as expected are given in the following table: Activity Normal Duration (Days) Crash Duration (Days) Crashing Cost per day 1–2 9 6 30,000 1–3 8 5 40,000 1–4 15 10 45,000 2–4 5 3 15,000 3–4 10 6 20,000 4–5 2 1 60,000 You are required to: 1. Draw the Network and find the Normal and Minimum Duration of the work. 2. Compute the Additional Cost involved if the authority wants to complete the work in the shortest duration. Solution: Similar to Page 19.28, Illustration 20 (M 02) Note: Slope = Crashing Cost per day =given in the question. 1. Paths Table 2. Network Diagram Path Normal Duration (Normal Days) Min. Duration (Crash Days) Duration after Stage I II III IV V Path X: 1–2–4–5 9+5+2 = 16 6+3+1 = 10 16 16–1 = 15 15 15–2 = 13 13–1 = 12 Path Y: 1–4–5 15+2 = 17 10+1 = 11 17 17–1 = 16 16–1 = 15 15–2 = 13 13–1 = 12 Path Z: 1–3–4–5 8+10+2 = 20 (Initial CP) 5+6+1 = 12 (Min. Duration) 20 – 3 = 17 17–1 = 16 16–1 = 15 15–2 = 13 13–1 = 12 Note: Since Minimum Duration is only 12 days on Path Z, the Project can be crashed and brought to min. 12 days only. 3. Crashing Process Stage Decision on Crashing Crash Costs Stage I Initial CP is Path Z, and Activity with least slope is 3 – 4. This can be crashed for 10 – 6 = 4 days maximum, but the time gap between Paths Z & Y (i.e. next longest path) is only 3 days. So, Activity 3 – 4 is crashed for the permissible period of 3 days.(least) ` 30,000 per day for 3 days = ` 90,000 Stage II Paths Y and Z are the CPs (17 days). Activities available for crashing are – (1) Common Activity 4 – 5 (with Slope 60,000) or (2) Separate Activities (with higher slope). So, Common Activity 4 – 5 is crashed for 1 day (being maximum time reduction possible). ` 60,000 per day for 1 day = ` 60,000 Stage III Paths Y and Z are the CPs (16 days). Activity Combinations available for crashing are – (1) 3–4 & 1–4 (with Slope 20,000 + 45,000 = 65,000) and (2) 1–3 & 1–4 (with Slope 40,000 + 45,000 = 85,000). The first combination is chosen due to lower slope, and crashed for 1 day. ` 65,000 per day for 1 day = ` 65,000 Stage IV All Paths are critical. Since Common Activity is fully crashed in Stage II, the only possible crashing is to crash 2–4, 1–4 & 1–3 by 2 days (being the least of the maximum time reduction possible for 2–4). Slope is 15,000 + 45,000 + 40,000 = 1,00,000 ` 1,00,000 per day for 2 days = ` 2,00,000 Stage V All Paths are critical. The only activities available for crashing are 1–2 (3), 1–4 (2) and 1–3 (1) (with maximum time reduction possible in brackets. Each activity is reduced by 1 day (being minimum of the three). Slope is 30,000 + 45,000 + 40,000 = 1,15,000 ` 1,15,000 per day for 1 day = ` 1,15,000 After this, no further Crashing is possible since Minimum Duration of the Project is 12 days as per Note in WN 2 above. 1 2 3 4 5 9 5 15 8 10 2 Gurukripa’s Guideline Answers for May 2013 CA Final Advanced Management Accounting May 2013.11 4. Costs Table (on a per day basis) Stage Duration Activities Crashed Crash Costs Nature of Costs Initial 20 days – NIL Normal Duration Cost I 19 days (3 – 4) ` 30,000 18 days (3 – 4) ` 60,000 17 days (3 – 4) ` 90,000 II 16 days (4 – 5) ` 1,50,000 III 15 days (3–4 & 1–4) ` 2,15,000 Optimal Duration Cost IV 14 days (2–4, 1–4 & 1–3) ` 3,15,000 13 days (2–4, 1–4 & 1–3) ` 4,15,000 V 12 days (1–2, 1–4 & 1–3) ` 5,30,000 Minimum Duration Cost 6(b): Learning Curve (Theory) M 13 (4 Marks) Bring out the main applications of Learning Curve. Solution: Refer Page 20.1, Q.No. 4 (RTP, M 03, M 04, M 06, M 07, N 07) 6(c): Inter–Firm Comparison (Theory) M 13 (5 Marks) State the advantages available in Inter–Firm Comparison. Solution: Refer Page 15.4, Q.No. 10. 7: Theory – Various Topics – Any 4 out of 5 M 13 (4 × 4 = 16 Marks) Question Answer Reference (a) What are the focuses of Theory of Constraints? How it differs with regard to Cost Behaviour? Page No.12.1, 12.2 Q.No.3,5,6 (RTP, N 03, M 06, M 08) (b) Brief the reasons for using Simulation Technique to solve problems. Page No.21.1, Q.No.4 (N 96, M 93, M 04, M 09) (c) List out the qualities required for a Good Pricing Policy. Page No.3.6, Q.No.15 (N 08) (d) Under what circumstance PERT is more relevant? How? Page No.19.7, Q.No.19 (RTP, N 87, N 98, N 00, M 11) (e) Enumerate the expected disadvantages in taking Divisions as Profit Centers. Page No.5.7, Q.No.18 (RTP,N 01, N 08)
Posted on: Sat, 15 Jun 2013 05:42:05 +0000
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