i create timeline of number of soldiers deployed abroad @ given time.
operation start end soldiers operation 1 24.03.1999 14.01.2001 447 operation 2 15.05.2004 03.03.2009 880 operation 3 19.12.2006 24.01.2014 4390 for simplicity, assume number of soldiers deployed in operation constant on time.
however, cannot assume soldiers deployed @ same time , stayed entire period. if operation 4 lasted 30 years , consisted of 24 000 soldiers, cannot assume 24 000 deployed 30 years. if need be, can assume each soldier stayed 1 year.
i assume should generate period, , divide number of soldiers number of days.
gen duration = end - start however, not know go here. advice on how (a) identify number of soldiers (on average) deployed in given year , (b) clear way graph appreciated.
ps: 1 problem in past regards graphing operations begin on same day.
this simpler fear, need different data structure. consider number of soldiers abroad increases when operation starts , decreases when ends. whether same people or different immaterial if want count soldiers. alternatively, give no information on base calculations of number of different soldiers.
the technique here written within this paper essence conveyed title alone. given results, graphing immediate using line; connect(j) option may seem congenial.
clear input operation str10(sstart send) soldiers 1 "24.03.1999" "14.01.2001" 447 2 "15.05.2004" "03.03.2009" 880 3 "19.12.2006" "24.01.2014" 4390 end foreach v in start end { gen `v' = daily(s`v', "dmy") drop s`v' } expand 2 bysort operation: gen date = cond(_n == 1, start, end) operation: gen status = (_n == 1) - (_n == 2) format date %td list sort date status gen total = sum(status * soldiers) bysort date : replace total = total[_n] list date status soldiers total, sep(0) +---------------------------------------+ | date status soldiers total | |---------------------------------------| 1. | 24mar1999 1 447 447 | 2. | 14jan2001 -1 447 0 | 3. | 15may2004 1 880 880 | 4. | 19dec2006 1 4390 5270 | 5. | 03mar2009 -1 880 4390 | 6. | 24jan2014 -1 4390 0 | +---------------------------------------+
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