Costing/or solution paper
Question 1
a) Statement Showing “Cost Reduction Programme”
X (D) Y (D) (i) Current Period Selling Price per unit 10,000 12,000 (ii) Variable Cost for X (75%) & Y (70%) of (i) 7,500 8,400 (iii) Next Period Selling Price per unit 8,000 9,000 (iv) Variable Cost for X (75%) & Y (70%) of (iii) 6,000 6,300 (v) Cost Reduction in Variable Cost per unit (ii)- (iv) 1,500 2,100 (vi) Next Period Contribution per unit (iii)- (iv) 2,000 2,700
(vii) Desired BEP in units 400 units 400 units
(viii) Fixed Cost for next period (vi × vii) 8,00,000 10,80,000 (ix) Current Period’s Fixed Cost 10,00,000 15,00,000 (x) Cost Reduction in Fixed Cost 2,00,000 4,20,000
b) Working Notes
(1) Calculation of Actual Sales at Budgeted Prices
(D)
Actual Sales at Actual Price 11,07,000
Less: Sales Price Variance (F) 17,000
Actual Sales at Budgeted Prices 10,90,000
Activity Level = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑆𝑎𝑙𝑒𝑠 𝑎𝑡 𝐵𝑢𝑑𝑔𝑒𝑡𝑒𝑑 𝑃𝑟𝑖𝑐𝑒𝑠
𝐵𝑢𝑑𝑔𝑒𝑡𝑒𝑑 𝑆𝑎𝑙𝑒𝑠 𝑎𝑡 𝐹𝑢𝑙𝑙 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 x 100
= 𝑅𝑠.10,90,000
𝑅𝑠.13,50,000 x 100
= 80.74%
(2) Segregation of Fixed &Variable Cost Element from Semi- Variable Overheads
Variable Overhead = 𝑂𝑣𝑒𝑟ℎ𝑒𝑎𝑑 𝑎𝑡 𝐹𝑢𝑙𝑙 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦−𝑂𝑣𝑒𝑟ℎ𝑒𝑎𝑑 𝑎𝑡 75% 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝐴𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝐿𝑒𝑣𝑒𝑙
= 𝑅𝑠.3,65,000−𝑅𝑠.3,23,750
25
= D 1,650
Fixed Overhead = Total SV Overheads at 100% Level – Variable Overheads at 100% level
= D 3,65,000– (D 1,650 × 100)
= D 2,00,000
Flexible Budget at 80.74% Activity Level
(Amount in D)
Sales 10,90,000
Less:
Direct Material (D 4,25,000× 80.74 %) 3,43,148
Direct Labour (D 1,85,000× 80.74 %) 1,49,370
Variable Overheads (D 2,15,000× 80.74%) 1,73,593 Semi-Variable Overheads
Variable Cost (D 1,650 × 80.74.) [W.N.-2] 1,33,222
Fixed Cost [W.N.-2] 2,00,000
Profit 90,667
c) Dummy machine (X5) is inserted to make it a balanced cost matrix and assume its installation cost to be zero. Cost of install at cell X3 (P) and X2 (R) is very high marked as M.
P Q R S T
X1 9 11 15 10 11
X2 12 9 M 10 9
X3 M 11 14 11 7
X4 14 8 12 7 8
X5 (Dummy) 0 0 0 0 0
Step 1
Subtract the minimum element of each row from each element of that row-
P Q R S T
X1 0 2 6 1 2
X2 3 0 M 1 0
X3 M 4 7 4 0
X4 7 1 5 0 1
X5 (Dummy) 0 0 0 0 0
Step 2
Subtract the minimum element of each column from each element of that column-
P Q R S T
X1 0 2 6 1 2
X2 3 0 M 1 0
X3 M 4 7 4 0
X4 7 1 5 0 1
0
0
0
0
0
X5 (Dummy) 0 0 0 0 0
Step 3
Draw lines to connect the zeros as under-
P Q R S T
X1 0 2 6 1 2
X2 3 0 M 1 0
X3 M 4 7 4 0
X4 7 1 5 0 1
X5 (Dummy) 0 0 0 0 0
There are five lines which are equal to the order of the matrix. Hence the solution is optimal. We may proceed to make the assignment as under-
P Q R S T
X1 0 2 6 1 2
X2 3 0 M 1 0
X3 M 4 7 4 0
X4 7 1 5 0 1
X5 (Dummy) 0 0 0 0 0
The following is the assignment which keeps the total cost at minimum-
Machines Location Costs (Rs)
X1 P 9
X2 Q 9
X3 T 7
X4 S 7
X5 (Dummy) R 0
Total 32
d) Let
u = Investment in “Mutual Fund ‘XY’”
v = Investment in “Mutual Fund ‘HN’”
w = Investment in “Money Market Fund”
x = Investment in “Government Bonds”
y = Investment in “Share ‘P’“
z = Investment in “Share ‘Q’“
Objective function:
Maximize
Z = 0.15u + 0.09v + 0.08w + 0.0875x + 0.17y + 0.18z Condition-1:
D 15,00,000 to be invested - u + v + w + x + y + z ≤ 15,00,000 Condition-2:
At least 40% of investment in Government Bonds- x ≥ (u + v + w + x + y + z) × 0.40 Or 2u +2v + 2w − 3x + 2y + 2z ≤ 0
Condition-3:
Combined Investment in two shares not to exceed Rs. 2,60,000-y + z ≤ 2,60,000
Condition-4:
At least 25% of the investment in the money market fund- W ≥ (u + v + w + x + y + z) × 0.25
Or u + v − 3w + x + y + z ≤ 0
Condition-5:
Amount of money invested in shares should not exceed the amount invested in mutual funds-
y + z ≤ u + v
Or − u − v + y + z ≤ 0
Condition-6:
Amount invested in mutual fund ‘XY’ should be not be more than the amount invested in mutual fund ‘HN’-
u ≤ v
Or u – v ≤ 0
Maximize
Z= 0.15u + 0.09v + 0.08w + 0.0875x + 0.17y + 0.18z
Subject to the Constraints:
u + v + w + x + y + z ≤ 15,00,000 2u +2v + 2w − 3x + 2y + 2z ≤ 0
y + z ≤ 2,60,000
u + v − 3w + x + y + z ≤ 0
− u − v + y + z ≤ 0
u – v ≤ 0
u, v, w, x, y, z ≤ 0
Question 2 a)
(i) Assumed Quotation Price ‘P’, Quantity ‘Q’
The Marginal Cost of a ‘Wagon’ is Rs.3,60,000 (Rs.2,20,000 × 4 CA snub Bogies + Rs.4,80,000) Demand Function for a ‘Wagon’
P = Rs.17,10,000 – (Rs. 50,000 / 2) × Q
Revenue (R) = Q × [17,10,000 – 25,000 × Q]
= 17,10,000 Q – 25,000 Q2 Marginal Revenue (MR) = 17,10,000 – 50,000 Q Marginal Cost (MC) = 13,60,000
Profit is Maximum where Marginal Revenue (MR) equals to Marginal Cost (MC)
17,10,000 – 50,000 Q = 13,60,000
Q = 7.00 units
By putting the value of ‘Q’ in Demand Function, value of ‘P’ is obtained.
P = 17,10,000 – (50,000/ 2) × Q
= 17,10,000 – 25,000 × 7.00
= D 15,35,000
At Rs.15,35,000 unit Quotation Price of a Wagon the Eastern Company Ltd.’s Profit will be Maximum.
(ii) At CBD the Divisional Manager would ensure that Divisional Marginal Revenue should be equal to Division’s Marginal Cost so that Profit can be Maximum.
MR of a CA snub Bogies = MC of Manufacturing a CA snub Bogies 3,20,000 – 2 (10,000/ 30) × Q = 2,20,000
Q = 150 units
Selling Price of a CA snub Bogie ‘P’ is
P = 3,20,000 – (10,000/ 30) × 150
= D.2,70,000
CBD will earn Maximum Profit when it will Quote Rs.2,70,000 to the Outside Market. Since, Outside Market Quotation is Transfer Price as well, so Transfer Price to WD will be D. 2,70,000 and it forms part of WD’s Marginal Cost.
At WD, Division Manager would ensure that Divisional Marginal Revenue should be equal to Division’s Marginal Cost so that Profit can be
Maximum.
MR of a Wagon = MC of Manufacturing a Wagon
17,10,000 – 50,000 × Q = (D 2,70,000 × 4 CA snub Bogies) + D 4,80,000
Q = 3.00 units
Quotation Price of a Wagon ‘P’ should be:
P = D 17,10,000 – 25,000 × 3.00
= D 16,35,000
The unit Quotation Price of Wagon that emerges as a result of Market Based Transfer Pricing is D16,35,000.
b)
If Plant is Continued If Plant is Shutdown
Sales 7,60,000 ---
Less: Variable Cost 5,70,000 ---
Contribution 1,90,000 ---
Less: Fixed Cost 3,50,000 1,30,000
Additional Cost --- 15,000
Operating Loss 1,60,000 1,45,000
A comparison of loss figures indicated as above points out that loss is reduced by Rs. 15,000 (D16,000 – D14,500) if plant is shut down.
Shut Down Point = 3,50,000− 1,45,000 8−6
= 1,02,500 units Capacity Level of Shut Down Point
At 100% Level Production = 1,18,750 (90,000 𝑢𝑛𝑖𝑡𝑠 0.80 ) Capacity Level at Shut Down = 86.32% (1,02,500 𝑢𝑛𝑖𝑡𝑠
1,18,750 𝑢𝑛𝑖𝑡𝑠)
Question 3
a) Working Notes
Particulars P Q
Production / Sales Quantity (units) 1,00,000 50,000
Batch Size (units) 1,000 500
No. of Batches …(a ÷ b) 100 100
Setup Time per Batch (hours) 30 36
Total Setup Hours (hours) …(c × d) 3,000 3,600 Machine Setup Cost D 4,62,000
Cost Driver per Machine Setup Hour = 6,600 ℎ𝑜𝑢𝑟𝑠4,62,000 = D 70
Testing Time per Unit (hours) 5 9
Total Testing Time (hours) …(a × h) 5,00,000 4,50,000 Testing Cost D 23,75,000
Cost Driver per Testing Hour = 𝑅𝑠.23,75,000
9,50,000 ℎ𝑜𝑢𝑟𝑠 = Rs. 2.50
(i) Statement Showing “Cost per unit- Activity Based Costing”
Particulars of Costs
Basis P Q
Direct Material Direct 42,00,000 30,00,000
Direct Labour Direct 15,00,000 10,00,000
Direct Machine Cost
Direct 7,00,000 5,50,000
Machine Setup Cost
3,000 hrs. @ D70 2,10,000 ----
3,600 hrs. @ D70 ---- 2,52,000
Testing Cost 5,00,000 hrs. @ D 2.50 12,50,000 ---- 4,50,000 hrs. @ D 2.50 ---- 11,25,000 Engineering Cost Allocated 8,40,000 14,10,000
Total Cost (D) 87,00,000 73,37,000
Cost per unit (D) 87.00 146.74 (ii) Statement Showing “Mark-up (full cost basis)- Product P”
Particulars Per unit
Selling Price 100.05
Less: Full Cost 87.00
Markup 13.05
Percentage of Markup on Full Cost [13.0587.00 x 100] 15%
(iii) Statement Showing “Target Cost of Product P” (After New Design is Implemented)
Particulars (D)
Target Price (given) 86.25
Mark-up [115.0086.25 x 15] 11.25
Target Cost per unit 75.00
(iv) Statement Showing “Cost of P (New Design)”
Particulars of Costs Basis of Costs Rate* Total Cost Direct Material Decrease by D 5 p.u. 37.00 37,00,000 Direct Labour Decrease by D 2 p.u. 13.00 13,00,000 Direct Machining Cost No Change as Machine is
Dedicated
7.00 7,00,000 Machine Setup Cost 100 Setup × 28 hrs. × D 70 1.96 1,96,000 Testing Cost 1,00,000 units × D 2.50 × 4
hrs.
10.00 10,00,000
Engineering Cost No Change 8.40 8,40,000
Total Cost 77.36 77,36,00
0 Rate per unit
The target cost is D 75 p.u. and estimated cost (new design) is D 77.36 p.u. The new design does not achieve the target cost set by NEC Ltd.
Hence the target mark up shall not be achieved.
Possible Management Action
- Value engineering and value analysis to reduce the direct material costs.
- Time and motion study in order to redefine the direct labour time and related costs.
- Exploring possibility of cost reduction in direct machining cost by using appropriate techniques.
- Identification of non-value added activities and eliminating them in order to reduce overheads.
- The expected selling price based on estimated cost of D 77.36 per unit is (D 77.36 +15%) D 88.96. Introduce sensitivity analysis after
implementation of new design to study the sales quantity changes in the price range of D 86.25 to D 88.96.
b)
(i) Statement of Comparative Cost of Repairs and Maintenance to ascertain the optimum amount of maintenance each week
Breakdown Hours (per week)
0 5 10 15 20 25*
(D) (D) (D) (D) (D) (D) Maintenance Costs 23,00
0
13,00 0
6,50 0
3,00 0
1,50 0
750
Repairs Cost 0 2,50
0
3,00 0
5,00 0
6,50 0
7,50
Total 23,00 0
0
15,50 0
9,50 0
8,00 0
8,00 0
8,25 Cost of Idle Time 0
(Breakdown hrs - 10 hrs) x D 60
-- -
-- -
-- -
300 600 900
Total Cost 23,00
0
15,50 0
9,50 0
8,30 0
8,60 0
9,15
* At present 0
It is seen from the above table that the optimum amount of maintenance is for break-down of 15 hours a week.
(ii) Additional Revenue that will Result from the Optimal Level;
Compared with Present Situation
(D) Value of Additional Output (10 hours saved × 6 tonnes × D 70) 4,200
Less: Wages (10 hours × D 60) 600
Less: Material (10 hours saved × 6 tonnes × D 35) 2,100 Add: Saving in Maintenance and Repairs Cost (D 9,150 – D
8,300)
850
Total Additional Revenue 2,350
Question 4 a.
(i) The Initial basic solution worked out by the shipping clerk is as follows-
Warehouse Market Supply
I II III IV IV
A
5
12 2
1 4
9
3 22
B 15
4 8 1 6 15
C 7
4 6
1
7 5 8
Req 7 12 17 9 45
The initial solution is tested for optimality. The total number of independent allocations is 6 which is equal to the desired (m +n – 1) allocations. We introduce ui’s (i = 1, 2, 3) and vj’s (j = 1, 2, 3, 4). Let us assume u1 = 0, remaining ui’s and vj’s are calculated as below-
(ui + vj) Matrix for Allocated / Unallocated Cells Ui
1 2 4 3 0
-2 -1 1 0 -3
4 5 7 6 3
y 1 2 4 3
Now we calculate ∆ij = Cij – (ui +vj) for non basic cells which are given in the table below-
∆ij Matrix
4
6 9 6
1 -1
Since one of the Δij’s is negative, the schedule worked out by the clerk is not the optimal solution.
(ii) Introduce in the cell with negative Δij [R3C4], an assignment. The reallocation is done as follows-
12 1
+1
9
-1 15
7 1
-1 +1
Revised Allocation Table
12 2 8 15
7 1
Now we test the above improved initial solution for optimality- (ui + vj) Matrix for Allocated / Unallocated Cells
Ui
2 2 4 3 0
-1 -1 1 0 -3
4 4 6 5 2
y 2 3
Now we calculate ∆ij = Cij – (ui +vj) for non basic cells which are given in the table below
∆ij Matrix
3
5 9 6
2 1
Since all Δij for non basic cells are positive, the solution as calculated in the above table is the optimal solution. The supply of units from each warehouse to markets, along with the transportation cost is given below-
Warehouse Market Units Cost per unit (Rs)
Total Cost (Rs)
A II 12 2 24
A III 2 4 8
A IV 8 3 24
B III 15 1 15
C I 7 4 28
C IV 1 5 5
Minimum Total Shipping Cost 104
(iii) If the clerk wants to consider the carrier of route C to II only, instead of 7 units to I and 1 unit to IV, it will involve shifting of 7 units from (A, II) to (A, I) and 1 unit to (A, IV) which results in the following table-
Warehouse Market
I II III IV IV
A 7 5
4 2
2 4
9
3 22
B 4 8 15
1 6 15
C
4
8
6 7 5 8
Req 7 12 17 9 45
The transportation cost will become-
Warehouse Market Units Cost per unit (Rs) Total Cost (Rs)
A I 7 5 35
A II 4 2 8
A III 2 4 8
A IV 9 3 27
B III 15 1 15
C II 8 6 48
Minimum Total Shipping Cost 141
The total shipping cost will be Rs141. Additional Transportation Cost Rs37.
The carrier of C to II must reduce the cost by Rs4.63 (Rs37/8) so that the total cost of transportation remains the same and clerk can give him business.
(iv) Revised transportation table is shown below-
Warehouse Market supply
I II III IV IV
A
5
12 2
2 4
8
3 22
B
4 8
11
1 6 15 / 11
C 7
4 6 7
1
5 8
Req 7 12 17 / 13 9 45
Since the alterations are restricted to allocated cells only, the present alterations do not disturb the optimal allocation schedule.
(v) In this situation, alterations are not restricted to allocated cells since allocation in cell (A, III) is 2 units only while reduction in
requirement of Market III as well as supply of Warehouse A is 3 units. Therefore, it is essential to make alterations in allocations, check solution for optimality test and iterate if required.
b. Statement of Reconciliation - Budgeted Vs Actual Profit
Particulars D
Budgeted Profit 5,19,000
Less: Sales Volume Contribution Planning Variance (Adverse) 52,125 Less: Sales Volume Contribution Operational Variance (Adverse) 93,825
Less: Sales Price Variance (Adverse) 39,600
Less: Variable Cost Variance (Adverse) 29,700
Less: Fixed Cost Variance (Adverse) 15,000
Actual Profit 2,88,750
Workings Basic Workings
Budgeted Market Share (in %) = 2,00,000 units
4,00,000 units
= 50%
Actual Market Share (in %) = 1,65,000 units
3,75,000 units
= 44%
Budgeted Contribution = D 21,00,000 – D 12,66,000
= D 8,34,000
Average Budgeted Contribution (per unit) = D 8,34,000
D 2,00,000
= D 4.17
Budgeted Sales Price per unit = D 21,00,000
D 2,00,000
= D 10.50
Actual Sales Price per unit = D 16,92,900
D 1,65,000
= D.26
Standard Variable Cost per unit = D 12,66,000
D 2,00,000
= D 6.33
Actual Variable Cost per unit = D 10,74,150
D 1,65,000
= D 6.51
Calculation of Variances Sales Variances :………
Volume Contribution Planning* = Budgeted Market Share % × (Actual Industry Sales
Quantity in units – Budgeted Industry Sales Quantity
in units) × (Average Budgeted Contribution per unit)
= 50% × (3,75,000 units – 4,00,000 units)
× D 4.17
= 52,125(A) (*) Market Size Variance
Volume Contribution Operational** = (Actual Market Share % – Budgeted Market Share%)×
(Actual Industry Sales Quantity in units) × (Average Budgeted Contribution per unit)
= (44%– 50 %) × 3,75,000 units × D 4.17
= 93,825 (A) (**) Market Share Variance
Price = Actual Sales – Standard Sales
= Actual Sales Quantity × (Actual Price – Budgeted Price) = 1,65,000 units × (D 10.26 – D 10.50) = 39,600(A) Variable Cost Variances :………
Cost = Standard Cost for Production – Actual Cost
= Actual Production × (Standard Cost per unit – Actual Cost per unit)
= 1,65,000units × (D 6.33 – D 6.51) = D 29,700(A)
Fixed Cost Variances :……….
Expenditure = Budgeted Fixed Cost – Actual Fixed Cost = D 3,15,000 – D 3,30,000
= D,000 (A)
Question 5
a. (i) Journal Entries for July are as follows
D D
E.1
Material and In-Process Inventory Control Accounts Payable Control
(Direct Materials Purchased)
2,64,000
2,64,000 E.2
Conversion Costs Control Various Accounts (Conversion Cost Incurred)
1,26,600
1,26,600 E.3
Finished Goods Control
Materials and In-Process Inventory Control Conversion Costs Allocated (standard cost of finished goods completed)
3,75,000
2,55,000 1,20,000 E.4
Cost of Goods Sold
Finished Goods Control
(standard cost of finished goods sold)
3,57,000
3,57,000 (ii) Zero inventory is the goal of an ideal JIT production system.
Accordingly, entry (E.3) would be D 3,57,000 finished goods production, not D 3,75,000. If the marketing division could only sell goods costing D 3,57,000, the JIT production system would call for direct materials purchases and conversion costs lower than D 2,64,000 and D 1,26,600, respectively, in entries (E.1) and (E.2).
b. Statement Showing Recommended Selling Price
100 Units (per Unit) 200 units (per Unit) Department KTS:
Direct Labour 36 hrs. × Rs.3
108.00 36Hrs. × 0.80 × Rs.3
86.40
Overtime Premium# 0.00 10.80
Total Labour Cost ...(A)
108.00 97.20
Variable Overheads 36 hrs. × Rs.
10
360.00 36 Hrs. × 0.80 × Rs.
10
288.00
Fixed Overheads 36 hrs. × Rs.
11
396.00 36 Hrs. × 0.80 × Rs.11
316.80
Total Overheads ...(B) 756.00 604.80
Department KTW:
Direct Labour 18 hrs. × Rs.
2.5
45.00 18 Hrs. × 0.70 × Rs.2.5
31.50
Overtime Premium 0.00 0.00
Total Labour Cost ...(C)
45.00 31.50
Variable Overheads 18 hrs. × Rs. 6
108.00 18 Hrs. × 0.70 × Rs.6
75.60 Fixed Overheads 18 hrs. ×
Rs. 7
126.00 18 Hrs. × 0.70 × Rs.7
88.20
Total Overheads ...(D) 234.00 163.80
Special Tool ...(E) Rs.5,500 / 100
55.00 Rs.5,500 / 200
27.50
Direct Material ..(F) 36.00 32.40
Profit on Labour (10%)
Rs. (108 + 45)×
10%
15.30 Rs. (97.20 + 31.50)×10%
12.87
Profit on
Overheads(15%)
(756+234) ×
15% 148.50 (604.80 +163.80) ×
15% 115.29 Total
Profi t ...(G)
163.80 128.16
Recommended Selling Price [(A)+
(B)+(C) +
+(D)+(E)+(F)+(G)]
1,397.80 1,085.36
(#) Statement Showing Overtime Premium
($) 1,440 Hrs × 3 × 50% / 200 Units
Department KTS Department Direct Labour Hours Available 12,000 KTW 8,000
Present workload 7,680 4,200
Balance Direct Labour Hours 4,320 3,800
Hours Required to produce 100 units: 3,600 1,800 Hours Required to produce 200 units: 5,760
(200 x 36 x 0.80)
2,520 (200 x 18 x 0.70)
Overtime Required to produce 100 units
--- ---
Overtime Required to produce 200 units
1,440 Hrs ---
Overtime Premium 10.8$ per Unit ---
Question 6 a.
(i) Simulation Sheet
Patient
Rando m No.
of Arriva
l
Appt.
Time
Arrival Time
of Patient
Cons.
Start Time
Random No. of Cons.
Cons.
time
Cons.
End Time
Patient Waiting Time
Diet’s Idle Time
1 15 10:00 09:58 10:00 17 18 10:18 - - 2 4 10:20 10:17 10:18 15 18 10:36 - - 3 35 10:40 10:39 10:39 12 15 10:54 - 3 4 67 11:00 11:00 11:00 58 25 11:25 - 6 5 75 11:20 11:20 11:25 60 25 11:50 5 - 6 86 11:40 11:40 11:50 72 25 12:15 10 - 7 25 12:00 11:59 12:15 30 20 12:35 15
Total 30 9 Assumption - Patients waiting before the appointment time is not taken for calculation of waiting time.
(ii) The patients are sensitive to waiting time, hence if the appointment interval of 20 minutes will be increased to 25 minutes then there will be no patient waiting time, but the dietician’s idle time will be increased to 24 minutes (refer the working note). Though in the question it is not mentioned whether the dietician is available for any stretch of time without considering her idleness in the clinic.
As a Management Accountant, the ADVICE would be to bring the equilibrium to both the patients waiting time as well as to the dietician’s idle time. The equilibrium falls between appointment interval time of 22 and 23 mins (shown by an arrow in working note).
Working Notes:
Simulation Sheet (Interval Time Increased to 21 Minutes)
Patient
Rando m No.
of Arriva
l
Appt.
Time Arrival Time of Patient
Cons.
Start Time
Random No. of
Cons.
Cons.
Time Cons.
End Time
Patient Waiting Time
Diet’s Idle Time
1 15 10:0 0
09:58 10:00 17 18 10:18 - -
2 4 10:2 1
10:18 10:18 15 18 10:36 - -
3 35 10:4 2
10:41 10:41 12 15 10:56 - 5
4 67 11:0 3
11:03 11:03 58 25 11:28 - 7
5 75 11:2 4
11:24 11:28 60 25 11:53 4 -
6 86 11:4 5
11:45 11:53 72 25 12:18 8 -
7 25 12:0 6
12:05 12:18 30 20 12:38 12 - Total 24 12 Statement Showing “Change in Appointment Interval Time and
Corresponding Waiting/ Idle Time”
Appointmen t Time (mins.)
Patient Waiting Time
(mins.)
Diet’s Idle Time (mins.)
Remark
20 30 9 Refer Simulation Sheet
21 24 12 Refer Simulation Sheet
→ 22 18* 15* *If Change in Appointment Time by 1 min. (↑) than corresponding Patient Waiting Time change by is 6 mins (↓) and Diet’s Idle Time change by 3 mins (↑)
→ 23 12* 18*
24 6* 21*
25 -* 24*
Verification
Simulation Sheet (Interval Time Increased to 25 Minutes) Rando
m No.
Arrival of
Appt.
Time Arrival Time of Patient
Cons.
Start Time
Random No. of Cons.
Cons.
Time Cons.
Time End
Patient Waiting Time
Diet’s Idle Time
1 15 10:00 09:58 10:00 17 18 10:18 - -
2 4 10:25 10:22 10:22 15 18 10:40 - 4
3 35 10:50 10:49 10:49 12 15 11:04 - 9
4 67 11:15 11:15 11:15 58 25 11:40 - 11
5 75 11:40 11:40 11:40 60 25 12:05 - -
6 86 12:05 12:05 12:05 72 25 12:30 - -
7 25 12:30 12:29 12:30 30 20 12:50 - -
Total - 24
b.
(i) Statement Showing “Production Budget”
Particulars ‘RB’
(units)
‘RD’
(units) Inventory (at the end of the year) 31,250 11,250
Add: Projected Sales 75,000 50,000
Total Requirements 1,06,250 61,250
Less: Beginning inventory 25,000 10,000
Production 81,250 51,250
(ii) Statement Showing “Direct Material Purchase Budget”
Particulars Material
‘A’ (Kg.)
Material
‘B’ (Kg.)
Material
‘C’ (Kg.) Requirement for
Production ‘RB’
4,06,250 (81,250 units
× 5 Kg.)
2,03,125 (81,250 units
× 2.50 Kg.)
---
Requirement for Production ‘RD’
2,56,250 (51,250 units
× 5 Kg.)
1,53,750
(51,250 units
× 3 Kg.)
51,250
(51,250 units
× 1 Kg.)
Total Requirement 6,62,500 3,56,875 51,250
Add: Closing inventory 45,000 40,000 8,750
Less: Beginning inventory 40,000 36,250 7,500
Purchase 6,67,500 3,60,625 52,500
(iii) Statement Showing “Direct Labour Hours Required vs Available”
Particulars ‘RB’ Units ‘RD’ Units Total
Maximum Sales 75,000 50,000
Less: From Stock 25,000 10,000
Required Goods for Pdn. 50,000 40,000
Direct Labour Requirement
2,00,000 (50,000 units ×
4 hrs.)
2,40,000 (40,000 units ×
6 hrs.)
4,40,000 hrs.
Direct Labour Available 6,00,000 hrs.
∵ Direct Labour Hrs. (Requirement) is < Direct Labour Hrs. (Availability)
∴ Principal Budget Factor is Sales (units)
Question 7
a. (i) Fitness Solution’s main Critical Success Factors are
(a) Developing and maintaining a high level of customer satisfaction.
(b) Offering facilities that are not much below that offered by competition.
(c) Keeping a tight cap on costs as there is considerable competitive pressure in this industry and entry barriers are not high.
(ii) The following is a possible Balance Scorecard for Fitness Solution Financial
Perspective
Operating expenses relative to budget Cash flow
Total daily operating revenue Customer
Perspective
Turnover rate among members Customer satisfaction rate Internal
Perspective
Number of employee complaints
Number of equipment not available on average day (due to maintenance)
Innovation and Learning
Number of new equipment put into service Number of staff participating in training courses
b. (i) Invalid (ii) Valid (iii) Valid (iv) Valid (v) Valid
c. The production quantity of a given item doubled the cost of that item decrease at a fixed rate. This phenomenon is the basic premise on which the theory of learning curve has been formulated. The distinctive
features of a learning curve are:
(i) Better tooling methods are developed and used.
(ii) More productive equipments are designed and used to make the product.
(iii) Design bugs are detected and corrected.
(iv) Better design engineering reduces material and labour costs.
(v) Early teething problems are overcome. As production progresses management is prompted to achieve better planning and better management.
(vi) Rejections and rework tend to diminish over time.
d. Classification of Items under Cost Reduction (CR)/ Cost Control (CC)
Sl. No. Item Category
CC/ CR (i) Costs exceeding budgets or standards are
investigated
CC
(ii) Preventive function CC
(iii) Corrective function CR
(iv) Measures to standardize for increasing productivity
CR
(v) Provision of proper storage facilities for materials
CC
(vi) Continuous comparison of actual with the standards set
CC
(vii) Challenges the standards set CR
(viii) Value analysis CR
e. The initial solution need not be the same under both methods.
Vogel’s Approximation Method (VAM) uses the differences between the minimum and the next minimum costs for each row and column. This is the penalty or opportunity cost of not utilising the next best alternative.
The highest penalty is given the 1st preference. This need not be the lowest cost.
For example if a row has minimum cost as 2, and the next minimum as 3, penalty is 1; whereas if another row has minimum 4 and next
minimum 6, penalty is 2, and this row is given preference. But Least Cost Method gives preference to the lowest cost cell, irrespective of the next cost. Solution obtained using Vogel’s Approximation Method is more optimal than Least Cost Method.
Initial solution will be same only when the maximum penalty and the minimum cost coincide.
