تتطلب عملية الإنتاج استخدام العمال مختلفين في مؤهلاتهم وكفاءتهم وأجورهم , وقد يلجأ قسم الإنتاج في بعض الأوقات إلى إحلال أو استبدال عمال من فيئة ذات مهارات معينة مكان العمال من فيئة ذات مهارات أخرى سبب تغيير بعض العمال أو نتيجةً لتعين عمال في أماكن لا تتناسب مع مؤهلاتهم , مما يؤدي إلى نشوء انحرافات نسبة تركيبة مزيج العمال المنفذين .
ومن جهة اخرى قد يختلف الزمن الفعلي للإنتاج عن زمن المعياري المحدد لهذا الإنتاج , مما يؤدي إلى نشوء انحرف الكفاءة ( العائد ) .
ويتضح مما يتقدم ان انحراف الوقت يمكن تحديده إلى انحراف نسبة التركيبة لليد العاملين وانحراف الكفاءة وذلك اذا نفذه العملية الإنتاجية من مجموعة من العمال ذات فئات مختلفة المؤهلات والأجور . أي اذا كانت العمل منفذ على أساس مجموعات العمل وكانت العمال فيما بينهم يظهرون علاقة إحلال واستبدال.
Example :- 5
A unit of product X is requested three classes of employments .
The following are S. Time , S. Rate & S. cost labor per unit :-


Class S. Time S. Rate S. cost
A 25 H 5 D/H 125 D
B 15 H 4 D/H 60 D
C 10 H 3 D/H 30 D
50 215
Because of absent some employments and replacement another, the records indicated the following actual data :
A.P.V.5 unit from product X through one week . خلال أسبوع واحد
Another actual data :
Class A. Time A. Rate A. cost
A 75 H 5.2 D/H 390 D
B 135 H 4.2 D/H 567 D
C 90 H 3.2 D/H 388 D
300 1245 D
Required : compute the Total variance of labor through a week then analysis it qualitative .


Solution:-
S.T = A.P.V × S.T. per unit
A = 5×25 = 125
B = 5×15 = 75
C = 5×10 = 50
Total standard time = 250


Total labor variance = A.C − S.C
= ( A.H × A.R ) − ( S.H × S.R )
A = ( 75 × 5.2 ) − ( 125 × 5 )
= 390 − 635
B = ( 135 × 4.2 ) − ( 75 × 4 )
567 − 300
C = ( 90 × 3.2 ) − ( 50 × 3 )
288 − 150
= ( 390 + 567 + 288 ) − ( 625 + 300+ 150 )
1245 − 1075 = 170 UF
Time variance
Efficiency. V = ( A.H − S.H ) S.R
A = ( 75 − 125 ) × 5 = 250 F
B = ( 135 − 75 ) × 4 = 240 UF
C = ( 90 − 50 ) × 3 = 120 UF
110 UF
Total labor variance = Rate variance + Time variance
= 60 UF + 110 UF
= 170 UF
W.A = S.C Per unit of production ˍ
S. time per unit of production
= 215 ͇ 4.3
50
Mix . v = ( A.Ts × S.R ) − ( total .A.Ts × W.A )
= ( 75 × 5 + 135 × 4 + 90 × 3 ) − ( 300 × 4.3 )
= 375 + 540 + 270 ) − 1290
= 1185 − 1290 = 105 F
Yield .v = ( Total actual time − Totals .T.S ) W.A
= ( 300 − 250 ) 4.3
= 215 UF
Time .v = Mix. v + yield. v
= 105 F + 215 UF = 110 UF
Example : 6
For alsalam company producing chemical solutions and try to produce a new compound and desired to apply the S.C. system . the manufacturing a new product is completed by adding material A to material B then heating the mixing with another material X after that bottling the produced compound in a plastic Jar of 20 litters each .
The primary compound 20 litter is formed from 27 kg of material A, B13 kg. material B after the cooling add 10 kg from material X .
The final weight of the compound 30 kg the standard price from materials as following :
Material A 250 D/kg
Material B 350 D/kg
Material X 450 D/kg
At the end of one price the records explained :-
1- Actual production volume ( 150 ) Jar
2- Material used :-
Material A 5000 kg at price 260 D/kg
Material B 2500 kg at price 340 D/kg
Material X 1300 kg at price 470 D/kg
Required :-
Analysis the material variance ?
Solution :- بطاقة المعيارية
S.Q S.P S.C ˍ
Material A 27kg 250 D/kg 6750 D
Material B 13 kg 350 D/kg 4550 D
Material X 10 kg 450 D/kg 4500 D
Input 50 kg 15800 D
Output 30 kg 15800 D


S.Q = A.P.V × S.Q. per unit .
Material A = 150 × 27kg = 4050 kg
Material B = 150 × 13kg = 1950 kg
Material X = 150 × 10 kg = 1500 kg
7500




Total variance = A.C − S.C
= ( A.Q × A.P ) − ( S.Q × S.P )
Material A = ( 5000 × 260 ) − ( 4050 × 250 ) = 287500 UF
Material B = ( 2500 × 340 ) − ( 1950 × 350 ) = 167500 UF
Material X = ( 1300 × 470 ) − ( 1500 × 450 ) = 64000 F
8800 7500 391000 UF total variance


Price variance = ( A.P − S.P ) × A.Q
Material A = (260 − 250 ) × 5000 = 50,000 UF
Material B = ( 340 − 350 ) × 2500 = 25000 F
Material X = ( 470 − 450 ) × 1300 = 26000 UF
Total price variance 51000 UF


Efficiency variance = ( A.Q − S.Q ) S.P
Material A = ( 5000 − 4050 ) × 250 = 237,500 UF
Material B = ( 2500 − 1950 ) × 350 = 192,500 UF
Material X = ( 1300 − 1500 ) × 450 = 90,000 F
Total Efficiency variance 340,000 UF


Total variance = price variance + Efficiency variance
= 51000 UF + 340000 UF = 391000UF

W.A = S.C per unit of production ˍ = 15800 D = 316 D/kg
S. Material per unit production 50kg


Mix variance = A.Q × S.P − total A.q.for input of material × W.A

A B X
= (5000 × 250) + (2500 × 350) + (1300 × 450) − ( 8800 × 316)

= 2710000 − 2780800 = 70800 F


Yeild variance = ( total A.q − total S.q ) W.A
( 8800 − 7500 ) 316
= 410800 UF
Efficiency variance = Mix variance + Yield variance
= 70800 F + 410800 UF = 340000 UF
Example : 7
The Colorado company manufactures a singe product whose standard product cost is as follow :-
D.M 24 Ib .@ $3per Ib . 72
D.L 6 Ib. @ $3.25 per hour 19 ˍ
Total unit standard cost 91
Actual date for the month of November :
Planned production 7500 unit
Material put into production 192410 Ibs . @ $3.04 per Ibs.
D.L. 46830 hours @ $3.3 average labor cost
Other data :
the company in applied the FIFO method account for FIFO الشركة تطبق طريقة
opening inventory, work in process 80unit , all material / 50% converted .
closing inventory , work in process 100 unit , all material / 50% converted .
started in process during November 7850 U. الوحدات التي تم البدء بها خلال نفس الشهر
Required :
A variance and analysis of :-
Direct materials تحليل الانحراف مواد المباشر
Direct labor cost تحليل الأجور
Solution: مواد مباشرة تشكيل , أجور مباشرة
Flow of production unit D.M D.L
W.I.P at beginning 50% 80
Started in process during 7850
current period
To account for 7930
Completed and transferred
out during current period:
- From beginning period W.I..P 80 ( 80×0%)0 (80×50%) 40
- Started and completed 7750 ×100% 7750 7750
- Ending W.I.P 50% 100 (100×100%) (100×50%)
- Accounted for 7930 100 50


- Work done in current 7850 7840


































S.Q = A.P.V × S.Q per unit
S.Q = 7850 × 24 = 188400 ID
S.H = 7840 × 6 = 47040 H
Total material variance = A.C − S.C
= ( 192410 × 3.04 ) − ( 188400 × 3 )
= 584926.4 − 565200
= 19726.4 UF
Price variance = ( A.P − S.P ) A.Q
= ( 3.04 − 3 ) 192410
= 7696.4 UF

Q.V = ( A.Q − S.Q ) S.P
= ( 192410 − 188400 ) 3
= 12030 UF
Total variance = 19726.4 UF

Total labor . variance = A.C − S.C
= ( 46830 × 3.3 ) − ( 47040 × 3.25 )
= 154539 − 152880
= 1659 UF


Rate variance = ( A.R − S.R ) A.H
= ( 3.3 − 3.25 ) 46830
=2341.5 UF
Efficiency variance = ( A.H − S.H ) S.R
= ( 46830 − 47040 ) 3.25
= 682.5 F


Total variance = 1659 UF


Example : 8
The Duhok company manufactures a perfume X by mixing 3 types of solution .The following is the percentage of standard mixing for 1 litter :


Material percentage S. price
A 32% 9 D/L
B 17% 12 D/L
C 51% 6 D/L
Xالمزيج المعياري ينتج 82% من
The standard mixing produced 82% about its weighted X perfume .
The following are the actual data for one period since the company is produced 1107 litters from X perfume the used Quantities and price as following :-
Material
A 620 Ls @ 9.25 D/L
B 300 Ls @ 11.8 D/L
C 780 Ls @ 6.1 D/L
Required :-
- A variance analysis of the direct materials . تحليل الانحرافات المواد المباشر
ملاحظة : ان حجم إنتاج الفعلي في السؤال هو ( 1107 ) لتر ولما كانت المواد تفقد من وزنها أو من حجمها أثناء عملية التصنيع ويبقى منها بعد التصنيع فهذا يعني ان حجم الإنتاج الفعلي ( 1107 ) لتر = 82% من حجم المواد الداخلة في الإنتاج وعلى هذا الأساس .
نسبة مقلوبة
Standard Q for A.P.V = A.P.V × revised percentages
= 1107 × 100 = 1350 Ls.
82


S.Q. for material A = 1350 × 32% = 432 litters
B = 1350 × 17% = 229.5 litters
C = 1350 × 51% = 688.5 litters
Standard card
Material S.Q S.P S.C
A 432 9 3888
B 229.5 12 2754
C 688.5 6 4131
Input 1350 10773
Output 1107 10773




Material A.C − S.C V ˍ
A (620 ×9.35)5735 − ( 432 × 9 )3888 1847 UF
B (300×11.8)3540 − (229.5×12)2754 786 UF
C (780×6.1)4758 − ( 688 × 6 ) 4131 627 UF
Total variance 14033 10773 3260 UF


Price. V = ( A.P − S.P ) A.Q
A = ( 9.25 − 9 ) 620 = 155 UF
B = ( 11.8 − 12) 300 = 60 F
C = ( 6.8 − 6 ) 780 = 68 UF
Total price . 173 UF


Q.V = ( A.Q − S.Q ) S.P
A = ( 620 − 432 ) ×9 = 1692 UF
B = ( 300 − 229.5) ×12 = 846 UF
C = ( 780 − 688.5) ×6 = 549 UF
Total Q. variance 3087 UF


Total .V = P.V + Q.V
= 173 UF + 3087 UF
= 3260 UF


W.A = 10773 = 7.98
1350
Mix .V = ( Actual .Qs × S.P ) − ( Total. A.Q × W.A)
= ( 620 + 300 × 12 + 780×6 ) − ( 1700 × 7.98 )
= ( 5580 + 3600 + 4680 ) − 13566
= 13860 − 13566
= 294 UF


Yield .V = ( Total A.Q − Total S.Q ) W.A
= ( 1700 − 1350 ) × 7.98
= 2793 UF


Total .V = Mix .V + Yield .V
= 294 UF + 2793 UF
= 3087 UF
عنوان / تعديل بطاقة المعيارية
تقوم الشركات أحياناً بتعديل بطاقتها المعيارية تعديلاً دائماً أو مؤقتاً ولاسباباً مختلفاً , وعندما يكون التعديل دائمي ( permanent ) في هذه الحالة تمهل البطاقة الأولي الأساسية ويتم إيجاد انحرافات على أساس البطاقة المعدلة أو الجديدة .
اما اذا كانت التعديل مؤقت ( temporary ) في هذه الحالة يتوجب احتساب انحراف الجديد يسمى انحراف التعديل والذي يجب بصيغة التالية :-
Adjusted V. = (Adjusted .S.Q – original .S.Q ) × S.P
وبعد ذلك نجمع انحراف التعديل + إجمالي انحراف على أساس البطاقة المعدلة = إجمالي انحراف على أساس البطاقة الأساسية
Example :9
The following is the standard cost to produce one unit of product M:
Material S.Q S.P S.C .
A 6 kg 3D/kg 18
B 9kg 2D/kg 18
C 5kg 4D/kg 20 .
20 56
Because of unavailable material C the management decided adjusted the standard mixing quantitative with product the quality of product . the new quantities
Material S.Q .
A 6kg
B 10kg
C 4kg

At the end o period the records indicted the following data :-
Completed 75 units
W.I.P. 50 unit 50% material
Material
A 800kg used at 3.1 D/kg
B 900kg used at 1.9 D/kg
C 300kg used at 4.2 D/kg
Required :
Compute the total variance for material on the base the adjusting mixing @ permanent دائمي
@ temporary مؤقتة
ملاحظة : في حالة إن سؤال عن نظام المراحل ولا يتضمن انتاج تحت تشغيل اول المدة , في هذه الحالة لا يمكن إن نحسب الانتاج المكافئ بطرقة FIFO, وعليه يمكن حل السؤال فقط بطريقة المعدل الموزون W.A
وعلى هذا الاساس فان الانتاج المكافئ في هذه السؤال يساوي وحدات التامة × 100% + انتاج تحت تشغيل اخر المدة × النسب التي وصل اليها .
Solution :
A- permanent
A.P.V = 75 + 50 × 50%
= 75 + 25 = 100 units
S.Q = A.P.V × S.Q. per unit
A = 100 × 6 = 600 kg
B = 100 × 10 = 1000 kg
C = 100 × 4 = 400 kg


Details A.C _ S.C Variance
Material A (800*3.1) (600*3)
2480 1800 = 680 UF
Material B (900*1.9) (1000*2)
1710 2000 = 290 F
Material C (300*4.2) (400*4)
1260 1600 = 340 F
Total material .V 5450 5400 50 UF


P.V = ( A.P − S.P ) * A.Q
Material A = ( 3.1 − 3 ) * 800 = 80 UF
Material B = ( 1.9 − 2 ) * 900 = 90 F
Material C =( 4.2 − 4 ) * 300 = 60 F
50 F
Q .V = ( A.Q − S . Q ) * S P
Material A = ( 800 − 600 ) * 3 = 600 UF
Material B = ( 900 − 1000 ) * 2 = 200 F
Material C = ( 300 − 400 ) * 4 = 400 F
Total Q.V = zero
Total .V. = PV + Q.V = 50 F + zero = 50 F

بطاقة معيارية جديدة
New standard card
Material S. Q S.P S.C
A 6kg 3D/kg 18
B 10kg 2D/kg 20
C 4kg 4D/kg 16
20 54
W.A = 54 = 2.7
20
Mix.V = ( A.Qs × S.P ) − ( total A.Qs × W.A )

= ( 800 × 3 )+ ( 900 × 2 )+ ( 300 × 4 ) − 2000 × 2.7
= 5400 − 5400 = zero

Yeild. V = ( total A.Q − total S.Q ) * W.A
Yeild. V = ( 2000 − 2000 ) * 2.7 = zero


Q.V = Mix + Yeild
= zero + zero = zero
ملاحظة :
يلاحظ رغم اختلاف نسب المزيج بالبطاقتين إلا إن انحراف المزيج كان صفرا فهذا يعني إن نسبتين لها نفس التأثير من حيث التكاليف وبالتالي تمثل هذه النقطة نقطة التماثل بالتكاليف وهي نقطة التي تتساوى فيها التكاليف المدائن المختلفة .
B – Temporary
S.Q an base original card
Material A = 100 × 6 = 600
Material B = 100 × 9 = 900
Material C = 100 × 5 = 500
تقرير الأداء
Details A.C S.C variance
600*3
A 2480 1800 = 680 UF
900*2
B 1710 1800 = 90 F
500*4
C 1260 2000 = 740 F
Total material variance 150 F
Adjustment V. = (Adjusted S.Q − original S.Q ) * S.P
Material A = ( 600 − 600 ) * 3 = zero
Material B = ( 1000 − 900 ) *2 = 200 UF
Material C = ( 400 − 500 ) * 4 = 400 F
انحراف التعديل Adjustment variance 200 F


Total variance original card = Adjusted + total variance base adjusted
= 200 F + 50 UF
= 150 F
ـــــــــــــــــــ ـــــــــــــــــــ ــــــــــــــــــــــ ــــــــــــــــــــ Example : 10
AL-ZAB CO. has set up the following standards per finished unit for direct material and direct manufacturing labor :
1- direct materials ;-
material X 10 kg at 0.275 D/kg
material Y 5 kg at 0.64 D/kg


2- direct manufacturing labor :
Excellent class 5H at 1.48 D/H
Ordinary class 3H at 1 D/H


At the end of costing period the actual data were as following :
1- direct material are added 100% at the beginning of product manufacturing process , the conversion cost entered regularly and the moving inventories were as following :-
A – beginning W.I.P 500 units converted 60%
B – New units and started 8500 units
C – completed unit 8000 units
D – ending W.I.P ? converted 30%


2- direct materials used material
Material X 94000 kg at 0.26 D/kg
Material Y 42600 kg at 0.65 D/kg





3- Actual data of direct manufacturing labor following :-
- Excellent class 41875 H at total costs , 62812.5 D
- Ordinary class 25125 H at total costs , 2562.5 D


4-The company has applied weighted average method to account for .


Required : -
A variance analysis of @ direct materials
@ the direct labor cost
Solution:-
Flow of production physical unit D.M D.L
W.I.P beginning 60% 500
Start in process during 8500
current period
To account for 9000

completed and trans period 8000 8000 8000 *100%
W.I.P at ending 30% (9000-8000) ? 1000 1000 300 *30%

To account for 9000
- Work done incurrent only 9000 8300
A.P.V for material A.P.V for labor


S.Q. for material
Material X = 9000 × 10 = 90000 kg
Material Y = 9000 × 5 = 45000 kg


S.H. for labor
Excellent class = 8300 × 5 = 41500 H
Ordinary class = 8300 × 3 = 24900 H


A. labor Rate = A.L costs
A. Hours
Excellent class = 62812.5 = 1.5 D/H
41875


Ordinary class = 25627.5 = 1.02 D/H
25125
تقرير الأداء
Performance Report


Direct materials A.C S.C Variance
A.P. × A.Q S.P × S.Q A.C − S.C
Material X 94000 × 0.26 90000 × 0.275
24440 24750 310 F
Material Y 24600 × 0.26 45000 × 0.64
27690 28800 1110 F
Total material variance 52130 53550 1420 F


Direct labor
Excellent class 41875 × 1.5 41500 × 1.48
62812.5 61420 1392.5 UF
Ordinary class 25125 × 1.02 24900 × 1
25637.5 24900 727.5 UF
Total labor variance 88440 86320 2120 UF

The primary cost variance 140570 139870 700 UF

Price variance = ( A.P – S.P ) × A.Q
Material X = ( 0.26 – 0.275 ) × 94000 = 1410 F
Material y = ( 0.65 – 0.64 ) × 42600 = 426 UF
Total variance 984 F


Q.V = ( A.Q – S.Q ) × S.P
Material X = ( 94000 – 90000 ) × 0. 275 = 1100 UF
Material Y = ( 42600 – 45000 ) × 0. 64 = 1536 F
Total variance 436 F
¬¬¬¬¬
Total materials = P.V + Q.V
= 984 F + 436 F = 1420 F

W.A for material =10 ×0.275 + 5× 0.64 = 5.95 = 0. 346666
15 15
Mix V. = ( A.Qs × S.P ) – ( total A.Qs × w.A )
= ( 94000 × 0.275 + 42600 × 0.64 ) – ( 13660 × 0.396666 )
= ( 25850 + 27264 ) – 54185 = 1071 F
Yield V. = ( total A .Qs – total .S .Qs ) × Q.A
= ( 136600 - 135000 ) × 0.396666 = 635 UF

Q . V = mix V. + yield .V.
= 1071 F + 635 UF = 436 F


2- Total labor .V . Analysis :-
Rate V . = ( A. R – S. R ) × A . H
Excellent class = ( 1.5 – 1.48 ) × 41875 = 837.5 UF
Ordinary class = ( 1.02 – 1 ) × 25125 = 502.5 UF
Total rate V. 1340 UF


Time V. = (A.H – S.H)* S.R
Excellent class = ( 41875 – 41500 )* 1.480 = 555 UF
Ordinary class = ( 25125 – 24600 )* 1 = 225 UF
Total Time V. 780 UF


Total Labor . V = Rate .V + Time .V
= 1340 UF + 780 UF = 2120 UF


W.A .for hours input = 5 × 1.48 + 3 × 1 = 7.4+3 = 10.4 = 1.3
8 8 8
Mix .V = ( A.Hs × S.R ) – ( Total A.Hs × W.A )
= ( 41875 × 1.48 + 25125 ×1 ) – ( 67000 × 1.3 )
= 87000 – 87000 = zero


Yeild .V = ( Total A.Hs – Total S.Hs )× W.A
= ( 67000 – 66400 ) × 1.3 = 780 UF


Time .V = mix .V + yield .V
= zero + 780 UF = 780 UF






Example :- 11
The Lennon corporation operates a machine shop and employs a standard cost system . In September the firm was the Low bidder on contract to deliver 600 kart Z by November , 15 at a contract price of 200$ each . Lennon’s estimate of the cost to manufacture each kart Z was :-


40 Ibs of materials at 1.5 $ per Ibs 60 $
20 Hours of direct labor at 2 per hour 40 $
On September , 1 . when the contract was obtained 30 completed kart Z were still in work in process 70 kart Z were in process with 2800 Ibs of material at a cost of 4200 $ and 60% processed , and 2000 pound of material at a cost 3000 $ were in raw material inventory .


All cost were at standard . In September 500 kart Z were standard in production 480 kart were transferred to finished goods inventory the work in process inventory at September 30 was 10% processed with material added at the short of production .


The material inventory is priced under the FIFO method at actual cost .
The following information is available for the method of September , materials , purchased :-
Pounds amount
8000 12000 $
8000 12000 $
4000 5600 $
Material requisition and put into production 21000 Ibs .
Direct labor pay roll amounted to 18648 $ for 8880 hours .


Required : جدول الإنتاج المكافئ
1- an equivalent production schedule by FIFO method ?
2- variance analysis of A) direct materials ?
B) direct labor ?










Solution: مواد مباشرة تشكيل , أجور مباشرة
Flow of production physical volume D.M D.L
W.I.P at beginning 60% 70
Started in process during current period 500
To account for 570
Completed and transferred
out during current period:
- From beginning period W.I..P 70 ( 70×0%)zero (70×40%) 28
- Started and completed 410 ×100% 410 410
- W.I.P Ending 10% 90 (90×100%) 90 (90×10%) 9
- Accounted for 570
- Work done in current 500 447
A.P.V A.P.V


Material S.Q = A.P.V × S.Q per unit
S.Q = 500 × 40 = 20000 Ibs
S.H = 447 × 20 = 8940 Hs


A.L.R = Total actual D.L = 188648 = 2.1 $/H
Total A.H 8880


Actual material available for use
Q price cost
2000 1.5 $ 3000
8000 1.5 $ 12000
8000 1.5 $ 12000
4000 1.5 $ 5600


Total material .V = A.C _ A.C
= ( A.Q*A.P) – (S.Q*S.P)
= ( 18000 × 1.5 + 3000 × 1.4 ) – (20000 × 1.5 )
= ( 27000 + 4200 ) – 30000
= 31200 − 30000
= 1200 U.F

Price variance = ( A.P – S.P )* A.Q
= ( 1.5 – 1.5 )* 18000 = zero
= ( 1.4 – 1.5 )* 3000 = 300 F
Total price variance 300 F
Quantity variance = ( A.Q – S.Q )*S.P
= ( 21000 – 20000 ) 1.5
= 1500 U.F


Total variance = P .V + Q.V
= 300 F + 1500 UF = 1200 UF


Total labors variance = A.C − S.C
= ( 8880 * 2.1 ) – ( 8940 * 2)
= 18648 – 17880
= 768 U.F

Rate variance = ( A.R – S.R ) A.H
= ( 2.1 – 2 ) 8880 = 888 U.F


Efficiency variance = ( A.H – S.H )*S.R
= (8880 – 8940 ) 2 = 120 F


Total labor .V = R.V + Time .V
= 888 U.F + 120 F
= 768 U.F