Transcript future worth amount method

CHAPTER 4
ECONOMIC EVALUATION OF
ALTERNATIVES
TOPICS IN CHAPTER 4
• BASES FOR COMPARISON OF
ALTERNATIVES
• PRESENT WORTH AMOUNT
• CAPITALISED EQUIVALENT AMOUNT
• ANNUAL EQUIVALENT AMOUNT
• FUTURE WORTH AMOUNT
• CAPITAL RECOVERY WITH RETURN
• RATE OF RETURN APPROACH
• INCREMENTAL APPROACH
OBJECTIVES
• TO SELECT THE BEST ALTERNATIVE
ECONOMICALLY
• UNDERSTAND THE VARIOUS BASES
• KNOW HOW REGARDING USAGE OF
VARIOUS METHODS
PRESENT WORTH AMOUNT METHOD
• THE PRESENT WORTH IS A NET EQUIVALENT AMOUNT
AT THE PRESENT THAT REPRESENTS THE DIFFERENCE
BETWEEN
EQUIVALENT
DISBURSEMENTS
AND
EQUIVALENT RECEIPTS OF AN INVESTMENT CASH
FLOW FOR A SELECTED INTEREST RATE
• IT CONSIDERS THE TIME VALUE OF MONEY
• IN A COST DOMINATED CASH FLOW DIAGRAM THE
COST WILL BE ASSIGNED WITH POSITIVE SIGN AND THE
PROFIT ,REVENUE WILL BE ASSIGNED WITH NEGATIVE
SIGN
• IN A REVENUE DOMINATED CASH FLOW DIAGRAM THE
PROFIT ,REVENUE WILL BE ASSIGNED WITH POSITIVE
SIGN AND CASH OUTFLOWS WITH NEGATIVE SIGN
PRESENT WORTH AMOUNT METHOD
• IN REVENUE DOMINATED CASH FLOW DIAGRAM THE
ALTERNATIVE WITH THE MAXIMUM PRESENT WORTH
AMOUNT SHOULD BE SELECTED AS THE BEST
ALTERNATIVE
• IN COST DOMINATED CASH FLOW DIAGRAM THE
ALTERNATIVE WITH MINIMUM PRESENT WORTH
SHOULD BE SELECTED AS THE BEST ALTERNATIVE.
• WE HAVE TWO SITUATIONS
EQUAL LIVED
ALTERNATIVES & UNEQUAL LIVED ALTERNATIVES
PROBLEM 1
•
A COMPANY IS CONSIDERING A PLANT TO MANUFACTURE A
PARTICULAR PRODUCT. THE LAND COSTS RS 3,00,000 AND THE
BUILDING COSTS RS 6,00,000 THE EQUIPMENT COSTS RS
2,50,000 AND RS 1,00,000 WORKING CAPITAL IS REQUIRED. IT IS
EXPECTED THAT THE PRODUCT WILL RESULT IN SALES OF RS
7,50,000 PER YEAR FOR TEN YEARS AT WHICH TIME THE LAND
CAN BE SOLD FOR RS 4,00,000 THE BUILDING FOR RS 3,50,000
THE EQUIPMENT FOR RS 50,000 AND ALL OF THE WORKING
CAPITAL IS RECOVERED. THE ANNUAL OUT OF POCKET
EXPENSES FOR LABOUR,MATERIALS AND ALL OTHER ITEMS
ARE ESTIMATED TO TOTAL OF RS 4,75,000. IF THE INTEREST
RATE IS 25 % PER YEAR DETERMINE WHETHER THE COMPANY
SHOULD INVEST IN THE NEW PRODUCT LINE . USE PRESENT
WORTH METHOD.
PROBLEM 2
•
TWO MACHINES ARE UNDER CONSIDERATION BY A METAL
FABRICATING COMPANY. MACHINE “A” WILL HAVE A FIRST
COST OF RS 15,000 AN ANUUAL MAINTENANCE AND
OPERATION COST OF RS 3000 AND RS 3000 SALVAGE VALUE.
MACHINE “B” WILL HAVE A FIRST COST OF RS 22,000 AN
ANNUAL COST OF RS 1500 AND A RS 5000 SALVAGE VALUE. IF
BOTH MACHINES ARE EXPECTED TO LAST FOR 10 YEARS
DETERMINE WHICH MACHINE SHOULD BE SELECTED ON THE
BASIS OF PRESENT WORTH METHOD USING AN INTEREST
RATE OF 12 % PER YEAR.
PROBLEM 3
•
A PLANT MANAGER IS TRYING TO DECIDE BETWEEN THE
MACHINES DETAILED BELOW : TAKE I = 15%
MACHINE “A”
FIRST COST - RS 49,500
ANNUAL OPERATING COST – RS 15,750
SALVAGE VALUE - RS 4500
LIFE - 6 YEARS
MACHINE “B”
FIRST COST - RS 63,000
ANNUAL OPERATING COST – RS 13,950
SALVAGE VALUE - RS 9000
LIFE - 9 YEARS
PROBLEM 4
•
THE FOLLOWING DATA REFERS TO THE CASH FLOWS OF FIVE
INVESTMENT PROPOSALS. THE TOTAL MONEY AVAILABLE FOR
THE COMPANY TO INVEST IS RS 35,000. THE COMPANY CAN
ACCEPT ANY ONE OF THE PROPOSALS WITH DIFFERENT LETTERS
SUBJECT TO BUDGET AVAILABLE. SELECT THE BEST
ALTERNATIVE BASED ON THE PRESENT WORTH ON TOTAL
INVESTMENT CRITERION . THE RATE OF RETURN IS 8 %
PROPOSAL
FIRST COST
NET INCOME
YEARS 1 TO 10
A1
-10,000
2000
A2
- 12,000
2100
B1
-20,000
3100
B2
- 30,000
5000
C1
- 35,000
4500
ANNUAL EQUIVALENT
METHOD
• THIS IS OTHERWISE CALLED AS EQUIVALENT
UNIFORM ANNUAL WORTH OR EQUIVALENT
UNIFORM ANNUAL COST
• AS ITS NAME SUGGESTS ANNUAL EQUIVALENT
WORTH ANALYSIS IS ALSO A METHOD BY WHICH
WE CAN DETERMINE THE EQUIVALENT ANNUAL
RATHER THAN OVERALL PRESENT OR FUTURE
WORTH OF A PROJECT
• THE ANNUAL WORTH CRITERION PROVIDES A
BASIS
FOR
MEASURING
WORTH
BY
DETERMINING EQUAL PAYMENTS ON AN
ANNUAL BASIS.
PROBLEM 5
• A FIRM IS CONSIDERING THE PURCHASE OF ONE
OF THE TWO NEW MACHINES . THE DATA ON
EACH ARE DESCRIBED BELOW : - IF THE FIRM
MARR IS 12 % WHICH M/C SHOULD BE SELECTED
USING EUAW METHOD ?
MACHINE A
MACHINE B
INITIAL COST
3400
6500
SERVICE LIFE
3 YEARS
6 YEARS
SALVAGE VALUE 100
500
NET OPERATING 2000/YEAR
CASH FLOW
AFTER TAXES
1800/YEAR
PROBLEM 6
COMPARE THE FOLLOWING MACHINES ON THE
BASIS OF THEIR EUAC ,USE I = 18 % PER YEAR
MACHINE X MACHINE Y
FIRST COST
44,000
23,000
ANNUAL
OPERATING
COST
7000
9000
ANNUAL
REPAIR COST
210
350
OVERHAUL
EVERY 2
YEARS
----------
1900
OVERHAUL
EVERY 5
YEARS
2500
--------
SALVAGE
VALUE
4000
3000
LIFE (YEARS)
15
8
PROBLEM 7
THE HEAT LOSS THROUGH THE EXTERIOR
WALLS OF A BUILDING COSTS RS 215 PER YEAR .
INSULATION THAT WILL REDUCE THE HEAT
LOSS COST BY 93% CAN BE INSTALLED FOR RS
127 AND INSULATION THAT WILL REDUCE THE
HEAT LOSS COST BY 89% CAN BE INSTALLED FOR
RS 90. DETERMINE WHICH INSULATION IS MOST
DESIRABLE IF THE BUILDING IS TO BE USED FOR
6 YEARS AND IF THE INTEREST RATE IS 12 % .
FUTURE WORTH AMOUNT
METHOD
• IN
THE
FUTURE
WORTH
METHOD
OF
COMPARISON OF ALTERNATIVES THE FUTURE
WORTH OF VARIOUS ALTERNATIVES WILL BE
COMPUTED.
• THE ALTERNATIVE WITH THE MAXIMUM
FUTURE WORTH OF NET REVENUE OR WITH THE
MINIMUM FUTURE WORTH OF NET COST WILL BE
SELECTED AS THE BEST ALTERNATIVE FOR
IMPLEMENTATION
PROBLEM 8
• CONSIDER THE FOLLOWING 2
ALTERNATIVES
END OF YEAR
0
1
2
3
4
A(RS)
-50 lacs
20 lacs
20 lacs
20 lacs
20 lacs
B(RS)
-45 lacs
18 lacs
18 lacs
18 lacs
18 lacs
ALTERNATIVE
AT I = 18% ,SELECT THE BEST
ALTERNATIVE BASED ON FUTURE WORTH
METHOD OF COMPARISON.
PROBLEM 9
• M/S KRISNA CASTINGS LTD IS PLANNING TO REPLACE ITS
ANNEALING FURNACE. IT HAS RECEIVED TENDERS FROM 3
DIFFERENT ORIGINAL MANUFACTURERS OF ANNEALING FURNACE.
THE DETAILS ARE AS FOLLOWS :
MANUFACT
URER
1
2
3
INITIAL
COST
80 LACS
70 LACS
90 LACS
LIFE
12
12
12
ANNUAL
8 LACS
OPERATION
COST
9 LACS
8.5LACS
SALVAGE
VALUE
4 LACS
7LACS
5 LACS
WHICH IS THE BEST ALTERNATIVE BASED ON FUTURE
WORTH METHOD AT I = 20 %
PROBLEM 10
A COMPANY MUST DECIDE WHETHER TO BUY M/C A OR
M/C B . THE DETAILS ARE AS FOLLOWS
MACHINE A
MACHINE B
INITIAL COST
RS 4 LACS
RS 8 LACS
LIFE
4 YEARS
4 YEARS
SALVAGE VALUE 2 LACS
5.5 LACS
ANNUAL
MAINTENANCE
COST
0
40,000
AT 12 % INTEREST RATE WHICH M/C SHOULD BE
SELECTED ? USE FUTURE WORTH METHOD OF
COMPARISON.
RATE OF RETURN METHOD
• IN THIS METHOD OF COMPARISON THE RATE OF
RETURN FOR EACH ALTERNATIVE IS COMPUTED.
• THE ALTERNATIVE WITH THE HIGHEST RATE OF RETURN
IS SELECTED AS THE BEST ALTERNATIVE
• IN THIS TYPE EXPENDITURES ARE ALWAYS ASSIGNED
WITH A NEGATIVE SIGN AND REVENUES WITH A
POSITIVE SIGN
• START WITH AN INTUITIVE VALUE OF I SATISFYING THE
RELATION
• THE RATE OF RETURN IS DETERMINED
BY
INTERPOLATION IN THE RANGE VALUES OF I
PROBLEM 11
• A PROJECT REQUIRES AN INITIAL INVESTMENT OF RS
2,50,000 AND GENERATES NET CASH FLOWS OF RS
95,000 , 95,000 ,1,00,000 AND 1,12,500 IN THE FIRST
,SECOND,THIRD AND FOURTH YEAR RESPECTIVELY.
CALCULATE THE INTERNAL RATE OF RETURN OF THE
PROJECT.
PROBLEM 12
• A COMPANY IS TRYING TO DIVERSIFY ITS BUSINESS IN A
NEW PRODUCT LINE . THE LIFE OF THE PRODUCT IS 10
YEARS WITH NO SALVAGE VALUE AT THE END OF ITS
LIFE, THE INITIAL OUTLAY OF THE PROJECT IS RS
20,00,000. THE ANNUAL NET PROFIT IS RS 3,50,000. FIND
THE RATE OF RETURN FOR THE NEW BUSINESS.
PROBLEM 13
• A COMPANY IS PLANNING TO EXPAND ITS PRESENT
BUSINESS ACTIVITY. IT HAS TWO ALTERNATIVES FOR
THE
EXPANSION
PROGRAMME
AND
THE
CORRESPONDING CASH FLOWS ARE TABULATED
BELOW . EACH ALTERNATIVE HAS A LIFE OF 5 YEARS
AND A NEGLIBILE SALVAGE VALUE. SUGGEST THE BEST
ALTERNATIVE TO THE COMPANY.
ALTERNATIVES
INITIAL
INVESTMENT
YEARLY
REVENUE
ALTERNATIVE 1
5,00,000
1,70,000
ALTERNATIVE 2
8,00,000
2,70,000
CAPITAL RECOVERY WITH
RETURN METHOD
• CR(i) = (P-F) (A/P,I,N) + FI
WHERE P = FIRST COST OF THE ASSET
F= ESTIMATED SALVAGE VALUE
N= LIFE IN YEARS
PROBLEM 14
• A FIRM REQUIRES POWER SHOVELS FOR ITS OPEN PIT
MINING OPERATION. THE FIRST COST IS RS 2,50,000 AND
THE SALVAGE VALUE IS RS 35,000 AT THE END OF 10
YEARS OF SERVICE. IF THE FIRM USES A RATE OF
INTEREST OF 12 % FOR THE PROJECT EVALUATION HOW
MUCH MUST BE EARNED ON AN EQUIVALENT ANNUAL
BASIS SO THAT THE FIRM RECOVERS ITS INVESTED
CAPITAL PLUS EARNS A RETURN ON THE CAPITAL
COMMITTED TO THE EQUIPMENT DURING ITS LIFE TIME.
PROBLEM 15
• A COMPANY CAN INVEST IN ONE OF THE TWO
ALTERNATIVES . THE LIFE OF BOTH THE ALTERNATIVES
IS ESTIMATED TO BE 5 YEARS WITH THE FOLLOWING
INITIAL INVESTMENTS AND SALVAGE VALUES.
ALTERNATIVE
A
B
INVESTMENT
10,000
12,000
SALVAGE VALUE 1500
3500
A) WHAT IS CR (i) FOR EACH ALTERNATIVE ?
B) DETERMINE THE SALVAGE VALUE AT THE END OF
PROJECT LIFE FOR ALTERNATIVE B WHICH WILL
RESULT IN SAME CR (i) IN BOTH ALTERNATIVES .
ASSUME I = 15 %
CAPITALISED COST
METHOD
• THIS METHOD IS OTHERWISE CALLED AS CAPITALISED
EQUIVALENT AMOUNT METHOD.
• DRAW THE CFD SHOWING ALL NON-RECURRING CASH
FLOW AND ATLEAST TWO CYCLES OF ALL RECURRING
CASH FLOWS.
• FIND THE PRESENT WORTH OF ALL NON- RECURRING
CASH FLOWS USING SINGLE PAYMENT PRESENT WORTH
RELATIONSHIPS.
• FIND THE EQUIVALENT UNIFORM ANNUAL AMOUNT A
OVER ONE LIFE CYCLE FOR ALL RECURRING CASH
FLOWS AND DIVIDE THAT AMOUNT BY THE INTEREST
RATE TO GET THE CAPITALISED COST OF RECURRING
CASH FLOWS.
CAPITALISED COST
METHOD
• DIVIDE ALL THE UNIFORM CASH FLOWS OCCURING
FROM YEAR 1 TO YEAR ∞ BY THE INTEREST RATE TO
GET THE CAPITALISED COST OF THOSE UNIFORM CASH
FLOWS.
• ADD THE VALUES OBTAINED IN THE ABOVE STEPS TO
GET THE TOTAL CAPITALISED COST OF THE GIVEN
INVESTMENT.
PROBLEM 16
• RS 8000 IS TO BE WITHDRAWN FROM A SAVINGS
ACCOUNT AT THE END OF EVERY 10 YEARS. CALCULATE
THE CAPITALISED EQUIVALENT AMOUNT OR THE SINGLE
AMOUNT TO BE DEPOSITED NOW AT AN INTEREST RATE
OF 10 % PER YEAR ?
PROBLEM 17
• CALCULATE THE CAPITALISED COST OF A PROJECT
THAT HAS AN INITIAL COST OF RS 1,50,000 AND AN
ADDITIONAL INVESTMENT OF RS 50,000 AFTER 10 YEARS.
THE ANNUAL OPERATING COST WILL BE RS 5000 FOR
THE FIRST FOUR YEARS AND RS 8000 THEREAFTER. IN
ADDITION THERE IS EXPECTED TO BE A RECURRING
MAJOR REWORK COST OF RS 15,000 EVERY 13 YEARS .
ASSUME I = 15 % PER YEAR.
INCREMENTAL APPROACH
• TWO TYPES OF INCREMENTAL APPROACH – THEY ARE
PRESENT WORTH ON INCREMENTAL APPROACH & RATE
OF RETURN ON INCREMENTAL INVESTMENT.
• PRESENT WORTH ON INCREMENTAL INVESTMENT
• IF THE NEGATIVE CASH FLOWS ARE LESS THAN THE
POSITIVE
CASH
FLOWS
THEN
“DO
NOTHING”
ALTERNATIVE MUST BE CONSIDERED.
• IF THE NUMBER OF NEGATIVE CASH FLOWS ARE MORE
THAN THE POSITIVE CASH FLOWS THEN THERE IS NO
NEED TO CONSIDER THE “ DO NOTHING “ ALTERNATIVE.
• LIST THE ALTERNATIVES IN THE ASCENDING ORDER OF
THEIR INITIAL INVESTMENT
INCREMENTAL APPROACH
• SELECT AS THE INITIAL CURRENT BEST ALTERNATIVE
,THE ONE REQUIRING THE SMALLEST FIRST COST. IN
MOST CASES IT IS THE “DO NOTHING “ ALTERNATIVE.
• COMPARE THE INITIAL BEST ALTERNATIVE AND FIRST
CHALLENGING ALTERNATIVE. THE CHALLENGER IS
ALWAYS THE NEXT HIGHEST ALTERNATIVE IN THE
ORDER OF FIRST COST.
THE COMPARISON IS
ACCOMPLISHED BY EXAMINING THE DIFFERENCES
BETWEEN 2 CASH FLOWS.
• IF THE PRESENT WORTH OF THE INCREMENTAL CASH
FLOW IS GREATER THAN ZERO THE CURRENT BEST
ALTERNATIVE IS REJECTED AND THE CHALLENGER
BECOMES THE NEXT CURRENT BEST ALTERNATIVE.
INCREMENTAL APPROACH
• IF THE PRESENT WORTH OF THE INCREMENTAL CASH
FLOW IS NEGATIVE THEN THE CURRENT BEST
ALTERNATIVE
REMAINS
UNCHANGED
AND
THE
CHALLENGER IS REJECTED.
• THE PROCESS OF COMPARISON AND SELECTION IS
REPEATED.
PROBLEM 18
• SIX MUTUALLY EXCLUSIVE ALTERNATIVES ARE SHOWN
BELOW . SELECT THE BEST ALTERNATIVE USING THE
INCREMENTAL APPEOACH IF THE MARR IS 10% AND THE
LIFE OF THE ALTERNATIVE IS 10 YEARS.
INVESTM
ENT
A
B
900
ANNUAL
150
REVENUE
C
D
E
F
1500 2500
4000
5000
7000
276
925
1125
1425
400
PROBLEM 19
• SELECT THE BEST ALTERNATIVE FROM A SET OF
ALTERNATIVES SHOWN BELOW IF THE MARR IS 10 %.
USE PRESENT WORTH ON INCREMENTAL INVESTMENT.
END OF
YEAR
A
B
C
D
0
-25,000
-30,000
-30,000
-37,500
1
-6250
-3750
-3000
-1000
2
-6250
-3750
-3000
-1000
3
2500
3750
3750
7500
PROBLEM 20
THREE MUTUALLY EXCLUSIVE ALTERNATIVES ARE
SHOWN BELOW . IF THE MARR IS 15 % PER YERAR AND
THE ALTERNATIVE HAVE DIFFERENT LIVES SELECT THE
BEST ALTERNATIVE USING PRESENT WORTH ON
INCREMENTAL INVESTMENT METHOD.
A
B
C
INITIAL
COST
- 21,000
- 24,500
-31,500
SALVAGE
VALUE
0
700
1050
CASH
FLOW
7000
10,500
10,500
LIFE
3
4
6
RATE OF RETURN ON
INCREMENTAL APPROACH
• IF THE RATE OF RETURN RESULTING FROM
INCREMENTAL CASH FLOW IS GREATER THAN MARR THE
INCREMENT
OF
INVESTMENT
IS
CONSIDERED
DESIRABLE AND THE CURRENT BEST ALTERNATIVE IS
DROPPED AND THE CHALLENGER BECOMES THE NEW
CURRENT BEST ALTERNATIVE.
• IF THE RATE OF RETURN ON INCREMENT INVESTMENT IS
LESS THAN MARR THE CURRENT BEST ALTERNATIVE
REMAINS UNCHANGED. IT HAS TO BE COMPARED WITH
THE NEXT CHALLENGER IN THE ORDER OF INITIAL
INVESTMENT REQUIREMENT.
PROBLEM 21
FOUR DESIGNS OF A PRODUCT WITH THEIR ASSOCIATED REVENUE
AND COST ESTIMATES HAVE BEEN PRODUCED TO THE TOP
MANAGEMENT FOR A DECISION. A 10 YEAR STUDY PERIOD WAS
USED. A MARR OF 10 % IS REQUIRED. . BASED ON THE FOLLOWING
PROJECTED CASH FLOWS WHICH OF THE FOUR ALTERNATIVE
DESIGN APPEAR MOST ATTRACTIVE . USE INCREMENTAL RATE OF
RETURN APPROACH.
A
B
C
D
INITIAL
INVESTMENT
1,70,000
2,60,000
3,00,000
3,30,000
ANNUAL
RECEIPTS
1,14,000
1,20,000
1,30,000
1,47,000
ANNUAL
DISBURSEMENTS
70,000
71,000
64,000
79,000
PROBLEM 22
A COMPANY IS GOING TO INSTALL A NEW PLASTIC
EXTRUDING MACHINE . FOUR DIFFERENT TYPES ARE
AVAILABLE. THE COSTS ASSOCIATED WITH EACH
MACHINES ARE SHOWN BELOW :
A
B
C
D
INVESTM
ENT
5,25,000
6,65,000
10,85,000 11,37,500
ANNUAL
DIS
BURSEM
ENTS
POWER
59,500
59,500
1,05,000
1,10,250
LABOUR
5,77,500
5,25,000
3,67,500
3,23,750
MAINTEN
ANCE
35,000
39,375
56,875
43,750
INSURAN
CE
10,500
13,300
21,700
22,750
LIFE
5
5
5
5
SUMMARY
• PRESENT WORTH IS AN EQUIVALENCE METHOD OF ANALYSIS IN
WHICH A PROJECT’S CASH FLOW ARE DISCOUNTED TO SINGLE
PRESENT VALUE.
• THE MARR OR MINIMUM ATTRACTIVE RATE OF RETURN IS THE
INTEREST RATE AT WHICH A FIRM CAN ALWAYS EARN OR BORROW
MONEY.
• THE TERM MUTUALLY EXCLUSIVE MEANS THAT WHEN ONE OF
SEVERAL ALTERNATIVES THAT MEET THE SAME NEED IS SELECTED
,OTHERS WILL BE REJECTED.
• AE ANALYSIS YIELDS THE SAME DECISION AS PRESENT WORTH
ANALYSIS.
• THE CAPITAL RECOVERY FACTOR ALLOWS MANAGERS TO
CALCULATE AN ANNUAL EQUIVALENT COST OF CAPITAL FOR EASE
OF ITEMIZATION WITH ANNUAL OPERATING COSTS.
PROBABLE QUESTIONS
• WRITE DOWN THE FORMULA FOR CAPITAL RECOVERY
WITH RETURN.
• ALL THE WORKED OUT PROBLEMS BY ALL THE METHODS
IN THE CLASS.
FURTHER READING FOR
CHAPTER 4
• ENGINEERING ECONOMY BY TED G ESCHENBACH, OXFORD
PUBLICATIONS.
• PRINCIPLES OF ENGINEERING ECONOMY ANALYSIS BY JOHN .A
WHITE , KENNETH .E.CASE ,DAVID .B.PRATT,MARVIN.H.AGEE ,WILEY
PUBLICATION
• ENGINEERING ECONOMY BY WILLIAM .G.SULLIVAN , JAMES .A.
BONTADELLI, ELIN .M .WICKS, PEARSON EDUCATION
• ENGINEERING ECONOMY BY G.J.THUESEN ,W.J.FABRYCKY ,PHI
PUBLICATION
• CONTEMPORARY ENGINEERING ECONOMICS BY CHAN .S PARK ,PHI
PUBLICATION